UC-NRI
V
-
UNIVERSITY EDITION. -REVISED AND ENLARGED.
THEORETICAL AND PRACTICAL
TREATISE ON
A L G E B R A:
IN WHICH THE EXCELLENCIES OF THE DEMONSTRATIVE METHODS OF THE
FRENCH ARE COMBINED "WITH THE MORE PRACTICAL OPERATIONS
OF THE ENGLISH J AND CONCISE SOLUTIONS POINTED
OUT AND PARTICULARLY INCULCATED.
DESIGNED
SCHOOLS, COLLEGES, AND PRIVATE STUDENTS.
BY H. N. ROBINSON, A. M.
FORMERLY PROFESSOR OF MATHEMATICS IN THE UNITED STATES NAVY J AUTHOR
OF A TREATISE ON ARITHMETIC, ASTRONOMY, GEOMETRY, TRIGONOMETRY,
SURVEYING, CALCULUS, NATURAL PHILOSOPHY, ctC. AC.
TWENTY-EIGHTH STANDARD EDITION.
CINCINNATI:
PUBLISHED BY JACOB ERNST
ANDERSON, GATES & WRIGHT, 112 MAIN STREET.
D. ANDERSON & CO.: TOLEDO, OHIO.
IVISON & PHINNEY, N. YORK.
1858.
I
lift -
Entered, according to Act of Congress, in the year 1847,
BY HORATIO N.ROBINSON,
In the Clerk's Office of the District Court of the State of Ohio.
according to aa of Congress, in the year 1857,
BY H. N. ROBINSON,
in the Clerk's Office of the District Court of the United States far the Northern
District of New York.
EDUCATION OEPT
STEREOTYPED BY JAMH8 & OU.,
CINCINNATI.
PREFACE.
SOME apology may appear requisite for offering a new book to the public on
the science of Algebra especially as there are several works of acknowledged
merit on that subject already before the public, claiming attention.
But the intrinsic merits of a book are not alone sufficient to secure its adop-
tion, and render it generally useful. In addition to merit, it must be adapted
to the general standard of scientific instruction given in our higher schools ; it
must conform in a measure to the taste of the nation, and correspond with the
general spirit of the age in which it is brought forth.
The elaborate and diffusive style of the French, as applied to this science,
can never be more than theoretically popular among the English; and the se-
vere, brief, and practical methods of the English are almost intolerable to the
French. Yet both nations can boast of men highly pre-eminent in this science,
and the high minded of both nations are ready and willing to acknowledge the
merits of the other ; but the style and spirit of their respective productions are
necessarily very different.
In this country, our authors and teachers have generally adopted one or the
other of these schools, and thus have brought among us difference of opin-
ion, drawn from these different standards of measure for true excellence.
Very many of the French methods of treating algebraic science are not to b- ;
disregarded or set aside. First principles, theories and demonstrations, are tb 3
essence of all true science, and the French are very elaborate in these. Yet no
effort of individuals, and no influence of a few institutions of learning, can
change the taste of the American people, and make them assimilate to the
French, any more than they can make the entire people assume French ^ iva-
city, and adopt French manners.
Several works, modified from the French, have had, and now have consid-
erable popularity, but they do not naturally suit American pupils. They are
not sufficiently practical to be unquestionably popular ; and excellent as they
are, they fail to inspire that enthusiastic spirit, which works of a more practi-
cal and English character are known to do.
At the other extreme are several English books, almost wholly practical, with
little more than arbitrary rules laid down. Such books may in time make
good resolvers of problems, but they certainly fail in most instances to make
scientific algebraists.
The author of this work has had much experience as a teacher of algebra
961661
iv
PREFACE
and has used the different varieties of text books, with a view to test their co n
parative excellencies, and decide if possible on the standard most proper to be
adopted, and of course he designed this work to be such as nis experience arid
judgment would approve.
One of the designs of this book is to create in the minds of the pupils a love
for the study, which must in some way be secured before success can be at^
tained. Small works designed for children, or those purposely adapted to per-
sons of low capacity, will not secure this end. Those who give tone to public
opinion in schools, will look down upon, rather than up to, works of this kind,
and then the day of their usefulness is past. On the other hand, works of a
high theoretical character are apt to discourage the pupil before his acquire-
ments enable him to appreciate them, and on this account alone such works
are not the most proper for elementary class books.
This work is designed, in the strictest sense, to be both theoretical and prac-
tical, and therefore, if the author has accomplished his design, it will be fouru\
about midway between the French and English schools.
In this treatise will be found condensed and brief modes of operation, no
nitherto much known or generally practised, and several expedients are system-
atised and taught, by which many otherwise tedious operations are avoided.
Some applications of the celebrated problems of the couriers, and also of the
lights, are introduced into this work, as an index to the pupil of the subsequent
utility of algebraic science, which may allure him on to more thorough investi-
gations, and more extensive study.
Such problems would be more in place in text books on natural philosophy
and astronomy than in an elementary algebra, but the almost entire absence of
them in works of that kinu, is our apology for inserting them here, if
apology be necessary.
Quite young pupils, and such as may not hare an adequate knowledge of
physics and the general outlines of astronomy, may omit these articles of ap-
plication ; but in all cases the teacher alone can decide what to omit and what
to teach.
Within a few years many new text books on algebra have appeared in differ
cut parts of the country, which is a sure index that something is desired
something expected, not yet found. The happy medium between the theo-
retical and practical mathematics, or, rather, the happy blending of the two,
which all seem to desire, is most difficult to attain ; hence, many have failed
in their efforts to meet the wants of the public.
Metaphysical theories, and speculative science, suit the meridians of France
and Germany better than those of the United States. But it is almost impos-
sible to comment on this subject without being misapprehended ; the author
of this book is a great admirer of the pure theories of algebraical science, for
it is impossible to be practically skillful without having high theoretical acquire-
ments. It is the man of theory who brings forth practical respite 1 but it is not
theory alone it is theory long and wel applied.
PREFACE v
Who will contend that Watt, Fitch, or Fulton, \vcre ignorant or inattentive
to every theory concerning the nature and power of steam, yet they are only
known as practical men, and it is almost in vain to look for any benefactors of
mankind, or any promoters of real science from those known only as theo-
rists, or among those who are strenuous contenders for technicalities and forms.
We are led to these remarks to counteract, in some measure, if possible, that
false impression existing in some minds, that a high standard work on aigebra,
must necessarily be very formal in manner and abstrusely theoretical in mat-
ter ; but in our view these are blemishes rather than excellencies.
The author of this work is a great advocate for brevity, when not purchased
at the expense of perspicuity, and this may account for the book appearing
very small, considering what it is claimed to contain. For instance, we have
only two formulas in arithmetical progression, and some authors have 20.
We contend the two are sufficient, and when well understood cover the whole
theory pertaining to the subject, and in practice, whether for absolute use or
lasting improvement of the mind, are far better than 20. The great number
onlv serves to confuse and distract the mind ; the two essential ones, can be
remembered and most clearly and philosophically comprehended. The same
remarks apply to geometrical progression.
In the general theory of equations of the higher degrees this work is not too
diffuse ; at the same time it designs to be simple and clear, and as much is
given as in the judgment of the author would be acceptable, in a work as ele-
mentary and condensed as this ; and if every position is. not rigidly demonstra-
ted, nothing is left in obscurity or doubt.
We have made special effort to present the beautiful theorem of Sturm in
such a manner as to bring it direct to the comprehension of the student, and if
we have failed in this, we stand not alone.
The subject itself, though not essentially difficult, is abstruse for a learner,
and in our effort to render it clear we have been more circuitous and elaborate
than we had hoped to have been, or at first intended.
We may apply the same remarks to our treatment of Horner's method of
solving the higher equations.
Brevity is a great excellence, but perspicuity is greater, and, as a goncral
tiling, the two go hand in hand; and these views have guided us in preparing
the whole work ; we have felt bound to be clear and show the rationale of
every operation, and the foundation of every principle, at whatever cost.
The Indeterminate and Diophantine analysis are not essential in a regular
course of mathematics, and it has rot been customary to teach them in many
institutions ; for these reasons we d > not insert them in our text book. The
teacher or the student, however, will find the-m in a concise form in a key to
thin work.
PREFACE TO THE 27TH (ENLARGED) EDITION.
THIS is a progressive age, and teachers, and schools, and books, rnusl
progress, proportionally, or be left in the rear.
To keep up with the spirit of the times, and make this volume more
complete and valuable, we have added some twenty-four pages of what we
conceive to be very valuable matter.
In the demonstration of the binomial theorem we originally adopted
that method, which, to judge from our own personal experience and obser-
vation as a teacher of others, was the one most easily comprehended by
learners but unfortunately, that method was not complete; it demonstrated
only so far as the actual operation was carried, and the further continua-
tion of the law could only be inferred. This being the case, we have
added another demonstration in this edition, which may require a little
more mental discipline to clearly comprehend, but which is, in itself, per-
fect and complete.
We have enlarged the practical part of logarithms, and inserted two
small tables, which will enable one to obtain the logarithm of any number
whatever ; and by the examples and illustrations here given, a student can
obtain a very good knowledge of logarithms, and their uses, in a smaller
compass than can be found in any other book known to us.
We have also added several new practical artifices in the solution of the
higher equations, which we hope will not be overlooked. And to the
progressive teachers of the United States the whole is respectfully dedi-
cated, by their obedient servant,
THE AUTHOR
EI.CRIDGE, N". Y., September, 1857.
VI
CONTENTS.
INTRODUCTION 9
SECTION I.
Addition .13
Subtraction I (
Multiplication 1 !
Division 2(>
Negative Exponents i27
ALGEBRAIC FRACTIONS 34
Greatest Common Divisor '31
Least Common Multiple 42
Addition of Fractions 44
Subtraction of Fractions 46
Multiplication of Fractions 47
Division of Fractions 49
SECTION II.
Equation of one unknown Quantity 52
Question producing Simple Equations. 62
Equations of two unknown Quantities .70
Equations of three or more unknown Quantities 76
Problems producing Simple Equations of two or more unknown quanti-
ties 87
Interpretation of negative values in the Solution of Problems 94
Demonstration of Theorems 97
Problem of the Couriers 98
Application of the Problem of the Couriers 102
SECTION III.
INVOLUTION , 106
Some application of the Binomial Theorem 108
Evolution 113
Cube Root of Compound Quantities 121
Cube Root of Numerals 123
Brief method of Approximation to the Cube Root of Numbers 124
Exponential Quantities and Surds 128
PURE EQUATIONS 1 33
Binomial Surds 141
Problem producing pure Equations 147
Problem of the Lights ..151
Application of the Problem 152
vii
vin CONTENTS.
SECTION IV.
Quadratic Equations 157
Particular mode of completing a Square, (Art. 99) 160
Special Artifices in resolving Quadratics, (Art. 106) 169
Quadratic Equations containing two or more unknown Quantities 175
Questions producing Quadratic Equations 183
SECTION V.
A rithmetical Progression ] 89
Geometrical Progression 1 95
flarmonical Proportion 199
Problems in Progression and Harmonical Proportion 200
Geometrical Proportion 205
SECTION VI.
Binomial Theorem its Demonstration, &c 218
' " its General Application 223
Infinite Series 227
Reversion of a Series 234
Exponential Equations and Logarithms 239
Use and application of Logarithms 256
Compound Interest and Annuities 258
SECTION VII.
General Theory of Equations 263
Newton's Method of Division, (Art. 167) 276
Equal Roots 279
Transformation of Equations, (Art. 171) 282
Synthetic Division 290
Sturm's Theorem 309
Newton's method of Approximation 317
Homer's Method 31 9
Its application to Numerical Roots 333
Expedients to be used in solving Equations 337
Recurring and Binomial Equations 340
Practical examples for solution 343
General method of Elimination (extra) 345
APPENDIX.
Inequality 349
Differential method of Series , 350
Specific Gravity 357
Maxima and Minima 359
ELEMENTS OF ALGEBRA.
INTRODUCTION.
DEFINITIONS AND AXIOMS.
ALGEBRA is a general kind of arithmetic, an universal analysts,
or science of computation by symbols.
Quantity or magnitude is a general term applied to everything
which admits of increase, diminution, and measurement.
The measurement of quantity is accomplished by means of an
assumed unit or standard of measure ; and the unit must be the
same, in kind, as the quantity measured. In measuring length, we
apply length, as an inch, a yard, or a mile, &c. ; measuring area,
we apply area, as a square inch, foot, or acre ; in measuring
money, a dollar, pound, &c., may be taken for the unit.
Numbers represent the repetition of things, and when no ap-
plication is made, the number is said to be abstract. Thus 5, 13,
200, &c., are numbers, but $5, 13 yards, 200 acres, are quanti-
ties.
In algebraical expressions, some quantities may be known,
others unknown ; the known quantities are represented by the
first or leading letters of the alphabet, a, b, c, d, &c,, and the
unknown quantities by the final letters, z, y, x, u, &c.
THE SIGNS.
(1) The perpendicular cross, thus +, called plus, denotes ad'
dition, or a positive value, state, or condition.
(2) The horizontal dash, thus , called minus, denotes siib-
traction, or a negative value, state, or condition.
(3) The diamond cross, thus X, or a point between two quan-
tities, denotes that they are to be multiplied together.
UO: ; V ! J v INTRODUCTION.
(4) A horizontal line with a point above and below, thus -~,
denotes division. Also, two quantities, one above another, as
numerator and denominator, thus ~, indicates that a is divided
by b.
(5) Double horizontal lines, thus =, represent equality. Points
between terms, thus a : b :: c : d, represent proportion, and ar
read as a is to b so is c to d.
(6) The following sign represents root J ; alone it signifies
square root. With small figures attached, thus *.J 4 ,J 5 N /, &c.,
indicates the third, fourth, fifth, &-., root.
Roots may also be represented by fractions written over a
quantity, as a a 3 ? , &c., which indicate the square root,
the third root, and fourth root of .*
(7) This symbol, of>6, signifies that a is greater than b.
This * , a<^b, signifies that a is less than b.
(8) A vinculum or bar , or parenthesis ( ) is used to con-
nect several quantities together. Thus -j-6-{-cX#,or (a-}-b-{-c)x,
denotes that a plus b plus c is to be multiplied by x. The bar
may be placed vertically, thus, which is the same
as (a d-\-e}y, or the same as ay dy-\-ty without the
d
+4
vinculum.
(9) Simple quantities consist of a single term, as a, b, ab, 3#,
&c. Compound quantities consist of two or more terms coli-
nected by their proper signs, as a-\-x, 3b-\-2y, 7ab 3xy-{-c, <fcc
A binomial consists of two terms ; a trinomial of three ; and a
polynomial of many, or any number of terms above two.
(10) Quantities multiplied together are called factors of the
product, and factors are called coefficient in respect to each other.
Thus, in the expressions 3#, abx ; 3 and x are factors of the
quantity 3a?, and 3 is the coefficient of x. Also, a, b, and x are
factors of abx ; a is the coefficient of bx ; b is the coefficient
of ax ; and ab is the coefficient of x.
(11) The measure of a quantity is some exact fact or of that
quantity : thus, 5 is the measure of 20 or 25.
*The adoption and utility of this last mode of notation, which ought to
be exclusively used, will be explained in a subsequent part of this worU.
EXERCISES ON NOTATION. u
(12) The root of any quantity is some equal factor of that
quantity. The square root is one of two equal factors ; the cube
root one of three equal factors, and so on.
(13) The multiple of any quantity is some exact number of
limes that quantity.
Other terms will be defined as we use them.
AXIOMS.
Axioms are self-evident truths, and of course are above demon
stration ; no explanation can render them more clear. The fol-
lowing are those applicable to algebra, and are the principles on
which the truth of all algebraical operations finally rests:
Axiom 1. If the same quantity or equal quantities be added to
equal quantities, their sums will be equal.
2. If the same quantity or equal quantities be subtracted from
equal quantities, the remainders will be equal.
3. If equal quantities be multiplied into the same, or equal
quantities, the products will be equal.
4. If equal quantities be divided by the same, or by equal
quantities, the quotients will be equal.
5. If the same quantity be both added to and subtracted from
another, the value of the latter will not be altered.
6. If a quantity be both multiplied and divided by another,
the value of the former will not be altered.
7. Quantities which are respectively equal to any other quan-
tity are equal to each other.
8. Like roots of equal quantities are equal.
9. Like powers of the same or equal quantities are equal.
EXERCISES ON NOTATION.
When definite values are given to the letters employed, we
can at once determine the value of their combination in any alge-
braic expression
12 ELEMENTS OF ALGEBRA.
Let a=5 6=20 c=4 d=l
Then a+6 c=5+20 4 or a+6 c=21
a&+aH-d=5X 20+5X4+1 = 121
SECTION I.
ADDITION.
(Art. 1.) Before we can make use of literal or algebraical
quantities to aid us in any mathematical investigation, we must
not only learn the nature of the quantities expressed, but how to
add, subtract, multiply, and divide them, and subsequently learn
how to raise them to powers, and extract roots.
The pupil has undoubtedly learned in arithmetic, that quanti-
ties representing different things cannot be added together ; for
instance, dollars and yards of cloth cannot be put into one sum ;
but dollars can be added to dollars, and yards to yards ; units can
be added to units, tens to tens, &c. So in algebra, a can be
added to a, making 2a ; 3a can be added to 5a, making Sa. As
a may represent a dollar, then 3a would be 3 dollars, and 5
would be 5 dollars, and the sum would be 8 dollars. Again, a
may represent any number of dollars as well as one dollar ; for
example, suppose a to represent 6 dollars, then 3a would be 18
dollars, and 5a would be 30 dollars, and the whole sum would be
48 dollars. Also, 8 is 8 times 6 or 48 dollars ; hence any num-
ber of 's may be added to any other number of 's by uniting
their coefficients ; but a cannot be added to &, or 4 to 36, or to
any other dissimilar quantity, because it would be adding unlike
things, but we can write a-\-b and 3+36, indicating the addi-
tion by the sign, making a compound quantity,
ADDITION. 13
Let the pupil observe that a broad generality, a wide latitude
must be given to the term addition. In algebra, it rather means
uniting-, condensing, or reducing terms, and in some cases, the
sum may appear like difference, owing to the difference of signs.
Thus, 4a added to a is 3a ; that is, the quantities wilted can
make only 3a, because the minus sign indicates that one a must
be taken out. Again, 76-f-36 46, when united, can give only
66, which is in fact the sum of these quantities, as 4b has the
minus sign, which demands that it should be taken out; hence to
add similar quantities we have the following
RULE. Add the. affirmative, coefficients into one sum and the
negative ones into another, and take their difference with the
sign of the greater, to which affix the common literal quan-
tity.
EXAMPLES FOR PRACTICE.
5a llx + 5#6 6a-f56 7cd + 8xy
2a 2x Gab Ga4b 2cd + 3xy
Sum la IQx ab +6 9cd+llxy
5fl-f&
cdy+ax
4x 6
3a-l-c
2cdy '3 ax
2^+10
7o 26+c
4cdy-}-3ax
-3x+ 7
3a--36 4c
Icdy ax
6rc 12
Sum 12 46 2c 9x 1
N. B. Like quantities, of whatever kind, whether of powers or
roots, may be added together the same as more simple or rational
quantities.
Thus 3a 2 and 8a* are 11 2 , and 76 3 -f-36 3 =106 3 . No matter
what the terms may he, if they are only alike in kind. Let the
reader observe that 2(a-\-b~)-\-3(a-{-b) must be together 5(a-\-b),
mat is, 2 times any quantity whatever added to 3 times the
same quantity must be 5 times that quantity. Therefore,
{-y=l t Jx J ry 1 for Jx+y, which represents the
square root of x-\-y, may be considered a single quantity.
(Art. 2.) To find the sum of various quantities we have the
following
14 ELEMENTS OF ALGEBRA.
RULE. Collect together all those that are alike, by willing
their coefficients, and then write the different sums, one after
another, with their proper signs.
EXAMPLES.
1.
2.
3.
3xy
, 9x*y
6xy 12 x?
2ax
7x*y
4^ +3xy
5xy
-\-3axy
~{-4x* 2xy
Gax
~4x*y
- -3^-{-4a? 2
Sum Sax 2xy
3axy 2x?y
4xy 8^
4.
5.
6.
1 4ax 2 x z 9-j-lO Jax 5; y
5ax-{-3xy 2x-}-7 l Jxy-{-5y
Sy z 4ax 5y-\-3,Jax-\-4y
^4-406^ 3a^+26 10 4Jax+4y
7. Add 2xy2a z , 3a z -{-xy, a z -\-xy, 4a? 3xy, 2xy 2a*.
Ans. 4a?-{-3xy.
8. Add 8aV 3xy, Sax 5xy, Qxy 5ax, 2a z x z -\-xy,
5ax 3xy. rfns. I0a z x z -+-5ax xy.
O. Add 2a?* 10y, 3jxy-{-Wx, 2x z y-\-25y, \2xy,Jxy,
Sy-f-nV^.
Ans. 2x z y+12xy-{-Wx-\-
10. Add 26a>-12, 3x*2bx, 5a^ 3 Jx,
a!*+3. dns.
11. Add 106 2 3bx*, 2b z x z b z , 1026^, b z x z 20,
3bx z +b z . Ans. 10^ 2 +36 2 a> 2bx z 10.
12. Add 2a 2 Zax^+x 2 , 2ax* -\3xy-\-9, Wtfxy i.
JJns. 12a 2 ax*+x* Uxy+4.
1. Add 9&c 8 18ac* 156c 3 --ac 9ac 2 246c 3 , 9ac 2 2
. ac 2.
ADDITION. 15
14. Add 3m 2 1, 6am 2w 2 -f-4, 7 8a?/i-}-2w 2 , and
6ro 2 -f2am-H. ^ns 9ro 2 +ll.
15. Add 12a 13a6-i-16a;r, 8 4m+2y, 6a-f-7a6 2 4-
\%y 24, and lab lQax-{-4m.
rfns. 6a 6a6+ 1 4y -f-7o6 2 1 6.
16. Add 72ax 4 8a# 3 , 38z 4 3ay 4 +7ay*, S+l2ay\
-607/4-12 34az 4 +5i/ 3 9oy 4 . ^ns. 2ai/ 8 +20.
Add a-(-6 and 3 56 together.
Add Go; 56-f a+8 to 5a4x-\-4b3.
Add a+26 3c 10 to 36 4a+5c-}-10 and 56 c.
Add 3a-f-6 10 to c da and 4c-J-2 36 7.
Add 3a 2 -l-26 2 c to 206 3 2 -{-6c 6.
(Art. 3.) When similar quantities have literal coefficients, we
may add them by putting their coefficients in a vinculum, and
writing the term on the outside as a factor.
Thus the sum of ax and bx is
Add
EXAMPLES.
1.
ax+bif
2.
ay+cx
3ay-{-2cx
4y -{-Qx
Sum (-}--
Add
lc+4d)x-\-(b-\-3a-{-7)y 2 (
3.
3x-\-2xy
bx-^cxy
(a -\-b}x-{-2cdxy
4.
ax-}-7y
7ax 3y
2x -\-4y
Sum
5. Add 8oH-2(;r+a)-r-36, 9ax+6(x-{-a) 96, and
66 7ax 8(a?+).
6. Add (a-\-b),Jx and (c+2a 6)7a? together.
7. Add 28o 3 (aM-57/)+21, 18 13a 3 (#-f-5i/),
. 18a+13.
16 ELEMENTS OF ALGEBRA.
8. Add 17a(x-h3t/)-J-12aW, 8 18ay
SUBTRACTION.
(Art. 4.) We do not approve of the use of the term subtraction,
as applied to algebra, for in many cases subtraction appears like
addition, and addition like subtraction. We prefer to use the
expression^/u/mg 1 the difference.
What is the difference between 12 and 20 degrees of north
latitude ? This is subtraction. But when we demand the differ-
ence of latitude between 6 degrees north and 3 degrees south, the
result appears like addition, for the difference is really 9 de-
grees, the sum of 6 and 3. This example serves to explain the
true nature of the sign minus. It is merely an opposition to
the sign plus ; it is counting in another direction; and if we
call the degrees north of the equator plus, we must call those
south of it minus, taking the equator as the zero line.
So it is on the thermometer scale ; the divisions above zero
are called p!us t those below minus. Momsy due to us may be
called plus ; money that we owe should then be called minus,
the one circumstance is directly opposite, in effect, to the other.
Indeed, we can conceive of no quantity less than nothing, as
we sometimes express ourselves* It is quantity in opposite cir-
cumstances or counted in an opposite direction; hence the differ'
ence or space between a positive and a negative quantity is
their apparent sum.
As a further illustration of finding differences, let us take the
following examples, which all can understand :
From 16 16 16 16 16 16
Take 12 8 2 2 4
Differ, 4 8 14 16 18 ~ 20
Here the reader should strictly observe that the smaller the
number we take away, the greater the remainder, and when the
subtrahend becomes minus, it must be added.
SUBTRACTION. 17
From I2a 120 12a 120 120 12a
Take 20 16 12a 9 60 a
Diff. _8 4a 30 60 lla
When a greater is taken from a less, we cannot have & posi-
tive or plus difference, it must be minus.
From 200 100 50 50 100
Take lla lla lla lla b b 50
Diff. 90 a 60 lla +b b5a 50
Here it will be perceived, that any quantity subtracted from
zero will be the same quantity with its sign changed.
(Art. 5.) Unlike quantities cannot be written in one sum, (Art
1,) but must be taken one after another with their proper signs :
therefore, the difference of unlike quantities can only be ex-
pressed by signs. Thus the difference between a and b is a ft,
a positive quantity if a is greater than 6, otherwise it is negative
From a take b c, (observe that they are unlike quantities).
OPERATION.
From a-fO-f-0
Take Q+b c
Remainder, or difference, a 6-f-c
This formal manner of operation may be dispensed with ; the
ciphers need not be written, and the signs of the subtrahend need
only be changed.
From the preceding observation, we draw the following
GENERAL RULE FOR SUBTRACTION, OR ALGEBRAIC DIF-
FERENCES.
Change the signs of the subtrahend, or conceive them to be
clanged; then proceed as in addition.
EXAMPLES.
1. 2. 3.
From 4a+2x 3c 3ax-{-2y a-\-b
Take a-\-4x 6c xyly a &
Remainder, 3a 2a?+3c Sax xy-\-^y 26
18 ELEMENTS OF ALGEBRA.
4. 5. 6.
From 2# 2 3x-^-y 2 7a+2 5c <
Take x* 4aH-a a-J-2-1- c %x
Rem. 3r>-l-#-- 2 a 8a * 6c
7. 8.
Fmm 8x 2 3xy-}-2y z -}- c ax-\-bx-\-cx
Take a,' 2 Qxy-\-3y z 2c
Diff.
9.
From ax-\-by-{-cz
Take mx ny pz
Diff.
(Art. 6.) From a take b. The result is 6. The minus
sign here shows that the operation has been performed ; b was
positive before the subtraction ; changing the sign performed
the subtraction; so changing the sign of any other quantity
would subtract it.
11. From 3 take (ab-\-x c ^/), considering the terms in
the vinculum as one term, the difference must be 3a (ab-\-x
c y), but if we subtract this quantity not as a whole, but term
by term, the remainder must be 3a ab aj-j-c-f-y.
That is, when the vinculum is taken away, all the signs
within the vinculum must be changed.
12. From 3Qxy take (40#y 26 2 +3c 4d).
Rem. 2b 2 IQxy 3c-{-4c/.
13. From Jx-\-y+3ax 12 take
Rem. 5ax 3jx-}-v 12 b.
14. Find the difference between 6?/ 2 2i/ 5 and 8^/ 2
4-12.
15. From 3a 6 2o;+7 take
MULTIPLICATION. 19
10. From 3p-\-q-\-r 25 take q8r-}-2s 8.
dns. 3j-f-9r 4s+8
17. From 13a 2 2ax+9x* take 5a z lax x 2 .
An*. Sa*+5ar-\-lOx*
\_
18. From 20xy S^a-f 3i/ take to/+5a 2 y
19. From the sum of Gorfy Ilax*, and 8;r 2 i/+3a;r 3 , take
42^ 4 ax ^ ttf Diff.
20. From the sum of 15a 2 6-r-8ctfo 3 and
take the sum of 12a 2 & 3cdx 8 and 16+cdx 4a z b.
Diff.
21. From the difference between Sab I2cy and 3ab-\-
y take tlie sum of Sab Icy and ab-\-cy
Diff. 5ab Wcy.
From 2+26 take ab.
From ax-}-bx take .T bx.
From a-{-c+6 take a-{-c b.
From 3a?+2y+2 take 5x-\-3y+b.
From 6-f 2j?+c take 5a+6a? 3c.
From 4a 2x 2 take 6a 2a? 2.
From 12^2^7/4-3 take 7+Gy+Wx.
MULTIPLICATION.
(A.rt. 7.) The nature of multiplication is the same in arithmetic
and algebra. It is repeating one quantity as many times as there
are units in another ; the two quantities may be called factors,
and in abstract quantities, either may be called the multiplicand ;
the other will of course be the multiplier.
Thus 4X5. It is indifferent whether we consider 4 repeated
5 times or 5 repeated 4 times ; that is, it is indifferent which we
call the multiplier. Let a represent 4, and b represent 5, then
the product is aXb ; or with letters we may omit the sign and
the product will be simply ab.
20 ELEMENTS OF ALGEBRA.
The product of any number of letters, as ab c d, is abed.
The product of x y z is xyz.
Tn the product it is no matter in what order the letters are
placed ; xy and yx is the same product.
The product of axXby is axby or abxy. Now suppose
=6 and 6=8, then ab=4S, and the product of axXby would
be the same as the product of GxXSy or 4Sxy. From this we
draw the following rule for multiplying simple quantities :
Multiply the coefficients together and annex the letters, one
after another, to the product.
EXAMPLES.
1. Multiply 3x by 7 a. Prod. 21 ax.
2. Multiply 4y by Sab Prod. IZaby.
3. Multiply 36 by 5c, and that product by Wx.
Prod. ISQbcx
4. Multiply Qax by I2by by 7ad. Prod. 5Q4aaxydb.
5. Multiply Sac by 116 by xy.
6. Multiply af by pq by 4.
In the above examples no signs were expressed, and of course
plus was understood ; and it is as clear as an axiom that plus
multiplied by plus must produce plus, or a positive product.
(Art. 8.) As algebraic quantities are liable to be affected by
negative signs, we must investigate the products arising from
them. Let it be required to multiply 4 by 3, that is, repeat
the negative quantity 3 times ; the whole must be negative, as
the sum of any number of negative quantities is negative.
Hence minus multiplied by plus gives minus, aXb gives
ab ; also a multiplied by b must give ab, as we may
Conceive the minus b repeated a times.
(Art. 9.) Now let it be required to multiply 4 by 3, that
is, minus 4 must be subtracted 3 times ; but to subtract minus
4 is the same as to add 4, (Art. 5,) giving a positive or plus
quantity; and to subtract it 3 times, as the 3 indicates, will
give a product of +12.
That is, minus multiplied by minus gives phis.
MULTIPLICATION. 21
This principle is so important that we give another mode of
illustrating it:
Required the product of a b by a c.
Here a b must be repeated a c times.
If we take a b, a times, we shall have too large a product,
as the multiplier a is to be diminished by c.
That is a b
Multiplied by a
Gives ua ab 1 which is too great by a b repeated
c times, or by ac cb, which must be subtracted from the former
product; but to subtract we change signs, (Art. 5,) therefore the
true product must be aa ab ac-}-cb.
That is, the product of minus b by minus c gives plus be,
and, in general, minus multiplied by minus gives plus.
But plus quantities multiplied by plus give plus, and minus by
plus, or plus by minus, give minus ; therefore we may say, in
short,
77m/ quantities affected by like signs, when multiplied toge-
ther, give plus, and when affected by unlike signs give minus.
(Art. 10.) The product of a into b can only be expressed by
ab or ba. The product of abed, &c., is abed; but if b c and d
are each equal to a, the product would be aaaa.
The product of aa into aaa is aaaaa ; but for the sake of
brevity and convenience, in place of writing aaa, we write a 3 .
The figure on the right of the letter shows how many times the
letter is taken as a factor, and is called an exponent. The pro-
duct of a 3 into a 4 is a repeated 3 times as a factor, and 4 times
as a factor, in all 7 times ; that is, write the letter and add the
exponents.
EXAMPLES.
What is the product of a 9 by a 5 ? Jlns. a*.
What is the product of x* by # 8 ? Am. x l \
What is the product of i/ 2 by y* by t/ 5 ? JJns. y w .
What is the product of a n by a m ? Ans. a" + "'.
>Vhat is the product of 6V by bx. Ans. b*x*.
What i s the product of ac by ac 2 by aV ? Ans. ^ 5 c 6 .
22 ELEMENTS OF ALGEBRA.
If adding numeral exponents is a true operation, it must be
equally true when the exponents are literal.
N. B. When the exponent is not expressed, one is understood,
for a is certainly the same as a 1 , or once taken.
(Art. 11.) Every factor must appear or be contained in a pro-
duct. Thus ax z multiplied by bx 3 must be abx 5 . Now if a 6
and b ==10 the product would be 60# 5 .
Multiply 3a 2 by 7 3 . Product 21a 5 .
From this we draw the following rule for the multiplication
of exponential quantities.
Multiply the, coefficients and add the exponents of the same
latter. Jill the letters must appear in the product.
EXAMPLES.
Multiply 4a 4 by 3. Prod. 12 5 .
Multiply 3x z by 2x*. Prod. O.r 5 .
Multiply 3x by 7x* by 3a?y. Prod. 63aVy.
What is the product of 2ax z , 4axy, 7abx ?
Prod. 5Ga?x*by.
What is the product of *2& n , 3a m x, and ax 1
Prod. Ga n + m +\x*.
Multiply 9a?x by 4x. Prod. 36a 2 .r 2 .
Multiply ITtfW by 7ac. Prodf. 119 4 6 2 c 4 .
Multiply Ilfl 5 i 2 c by 10 5 i 8 c 9 , Prod 110a 10 6 10 c 10 .
Multiply 121i 2 c 3 ^ by 5a A bxy 2 . Prod. 605a 4 i 3 c 3 ^y.
Multiply 77a 3 cz 4 by 61 2 ^. Prod. 4Q97a 5 bcx*.
Multiply 117a6 2 c 3 a: by 2 3 i 2 c. Prod. 234aWc*x.
Multiply 9a z x by Gx.
Multiply 9 ax 2 by 7 ax.
Multiply 7 ax by 4.
Multiply Sac by 2cx by 4c. Prod 24c s .z
(Art. 12.) When one compound quantity is to be multiplied
or repeated as many times as there are units in another, it is
evident that the multiplicand must be repeated by every term of
the multiplier.
MULTIPLICATION.
Thus the product of a-\-b-}-c by x-\-y+z.
It is evident that a-\-b-{-c must be repeated x times, then y
times, then z times ; and the operation may stand thus :
Product by x ax-\-bx-\-cx
Product by y ay-{-by-\-cy
Product by z az-}-bz-}-cz
Entire Product ax-{-bx-}-cx-{-ay-}-by-\-cy-}-az-{-bz-{-cz.
From the foregoing articles we draw the following general
rule for the multiplication of compound quantities.
Multiply all the terms of the multiplicand by each terra of
the multiplier, observing that like signs, in both factors, give
plus, and unlike, minus.
Write each term of the product distinctly by itself, with its
proper sign, and afterwards condense or connect the terms as
much as possible, as in addition.
EXAMPLES.
1. 2.
Multiply 2ax 3x 3x-\- 2y
By 2x -\-4y 4x 5y
Partial product 4ax 2 - 6x* l2x*-\- Sxy
2d part. prod. Saxy I2xy 15xy ICty 2
Whole prod. 4ax-{-Saxy6x z I2xy 12^ 7xyWy*
3. Multiply 2x z -{-xy 2?/ 2
By 3x 3^
Partial product Qx*-\-3x~y Gxy 2
2d partial product 6x*y 3xy 2 -}-6y 3
Whole product Gx 3 3x 2 y 9xy*+6y*
4. Multiply 3a 2 2a5 b z by 2 4b.
'Prod 6a 3 16a 2 6-f-6& 2 -f 4#>.
5. Multiply x*xy-\-if by x+y. Prod,
24 ELEMENTS OF ALGEBRA.
6. Multiply a 2 3ac-f-c 2 by a c.
Prod, a 3 4a 2 c+4c 2 c 8 .
7. Multiply a-f-6 by a-\-b. Prod. a 2 +2a&-f-& 2 -
8. Multiply x+y by #+# /Vod. rc 2 +2a < i/+3/ 2 .
9. Multiply a b by a b. Prod. a 2 2ab-\-b':
10. Multiply x y by x y. Prod, x* 2xy-\-y\
(Art. 13.) By inspecting all the problems, from the 7th to the
10th, we shall perceive that they are all binomial quantities, and
the multiplicand and multiplier the same.
But when a number is to be multiplied into itself the product
is called a square. Now by inspecting the products, we find
that the square of any binomial quantity is equal to plus ; the
squares of the two parts and twice the product of the two parts.
N. B. The product of the two parts will be plus or minus,
according to the sign between the terms of the binomial.
Let us now examine the product of o-{-b into a b.
a -}-b 2m +2n
a b 2m 2n
a?-\-ab 4m 2 -\-4mn
ab b 2 4mn 4?i 2
Product a 2 b* 4m* in 2
Multiply 20-J-3& by 2a 3b. Prod. 4a*Qb 2 .
Multiply 3y-\-c by 3yc. Prod. 9j/ 2 c 2 .
Thus, by inspection, we find the product of the sum and
difference of two quantities is tqual to the difference of their
squares.
The propositions included in this article are proved also in
geometry.
(Art. 14.) We can sometimes make use of binomial quantities
greatly to our advantage, as a few of the following examples will
show :
1. Multiply a-r-6-f-c, by a+6 f c.
Suppose a-\-b represented by s, then it will be s-J-c.
MULTIPLICATION. 25
The square of this is s 2 -f-2se-fc- ; restoring the value of s, and
we have (a-}-b)~+2(a+b)c-\-c 2 .
2. Square x-\-y z. Let x-}-y=s.
Then ( z) 2 =s 2 2sz+z*=(x-\-y} 2 2(x+y)z-{-z*.
&. Multiply x-\-y-\-z by x+y z. Prod. (x-\-y} 2 z*.
4. Multiplj 2a; 2 3a:+2 by a? 8.
I 2X 3 19^+260; 10.
5. Multiply ax-\-by by ax+cy.
Prod. d*x 2 +(ab-\-ac)xy+cby*.
6. Multiply %x+y by
7. Multiply a 3 -f 2a 2 6+2a6 2 +6
By a 3
Prod.
8. Multiply ^ ^ a? +f
By \x + 2
Product, la 8 -}- V a^Js-K
9. What is the product of a m +6 m by o n +6 n ?
10. What is the product of z 2 fa? by a; 2 ia? ?
^fns. as* f a?+ 1 a; 2
11. What is the product of 4z 3 -|-8;r 2 -r-16aM-32 by 3x 0?
4 192.
12. What is the product of a*-}-a 2 b-{-ab 2 -{-b 2 by a b ?
w?ns. a 4 **
3
26 ELEMENTS OF ALGEBRA.
DIVISION.
(Art. 15.) Division is the converse of multiplication, the pro*
duct being called a dividend, and one of the factors a divisor. If
a multiplied by b give the product &, then ab divided by a must
give b for a quotient, and if divided by b, give a. In short, if
one simple quantity is to be divided by another simple quantity,
the quotient must be found by inspection, as in division of num-
bers.
EXAMPLES.
1. Divide IGab by 4a. *An&. 4b
2. Divide 2 lacd by 7c. rfns. 3a(l
3. Divide ab*c by ac. Am. 6 2 .
4. Divide Zaxy by 2bc. Ans. ?^
In this last example, and in many others, the absolute division
cannot be effected. In some cases it can be partially effected,
and the quotients must be fractional.
5. Divide 3acx* by ac?/. Ans. - .
* J y
Qb
6. Divide 72te by 8abx. Ans. .
7. Divide 27aby by llabx. Ans.
(Art. 16.) It will be observed that the product of the divisoi
and quotient must make the dividend, and the signs must con-
form to the principles laid down in multiplication. The follow-
ing examples will illustrate :
8. Divide 9y by 3?/. jlns. 3.
9. Divide 9i/ by 3y. Ans. +3.
3O. Divide +9y by 3y. Ans. 3.
* The terra quotient would be more exact and technical here ; but, in re-
sults hereafter, we shall invariably use the term Ans., as more brief and ele-
gant, and it is equally well understood.
DIVISION. 27
(Art. 17.) The product of a 3 into 2 ; s a 5 , (Art. 10,) that is,
in multiplication we add the exponents ; and as division is the
converse of multiplication, to divi le powers of tKe same letter,
we must subtract the exponent of the divisor from that of the
dividend.
Divide 2a by a 4 . Ans. 2a*,
Divide a 7 by a 6 . Jlns. a.
Divide IGr 5 by 4x. tfns. 4x*.
Divide ISaxy* by Say. Jlns. 5xy z .
Divide 63a m by 7a n . Aw. 9"- n .
Divide I2ax n by Sax. JLns.
Divide 7rf6 by
Divide 5a 2 ^ by -7fl 4 a; 2 . ^ns. -^,
Divide 117aW by 78 5 6c 4 . .
o
Divide 96a6c by \Za*bc*d. Ans.
Divide a^c 2 by
n
Divide 27a 3 6 4 cd 2 by 21 abed. Ans. -d>b*d.
Divide 14a6 2 cc? by Ga^c 2 . ns. .
ottC
(Art. 18.) The object of this article is to explain the nature of
negative exponents.
Divide a 4 successively by a, and we shall have the following
quotients :
1 1 1
a 3 , a 2 , a, 1, ~, ^, ^, &c.
Divide fl 4 again, rigidly adhering to the principle that to
divide any power of a by a, the exponent becomes one less, and
we have
28 ELEMENTS OF ALGEUKA.
a , a 2 , a 1 , a, or 1 , or 2 , er 3 , &c.
Now these quotients must be equal, that is, a 3 in one series
equals a 3 in the other, and
rt 2 =a 2 , a=a l , l=a, -=<*- 1 =a~* =a~*
a a 2 a 3
Another illustration. We divide exponential quantities by
subtracting the exponent of the divisor from the exponent of the
dividend. Thus a 5 divided by a 2 gives a quotient of 5 ~ 2 =o 3 .
a 5 divided by a 7 =a 5 ~ 7 =a~ 2 . AVe can also divide by taking the
dividend for a numerator and the divisor for a denominator, thus
i 5 I 1
=, therefore 2 =a~ 2 (Axiom 7.)
a 1 a 2 a 2
From this we learn, that exponential factors maybe changed
from a numerator to a denominator^ and the reverse, by chang-
ing the signs of the exponents.
Thu< -ax~* ~ -=a--
US ' ^~ 3*r'~3a 3 x n
Divide d>bc by a*b 2 c~ l . Am. ar l b~ l c*.
Observe, that to divide is to subtract the exponents.
Divide Uab*cd by 6a 2 bc*. dns. =
(Art. 19.) A compound quantity divided by asimple quantity, is
effected by dividing each term of the compound quantity by the
simple divisor.
EXAMPLES.
1. Divide Sax 15x by 3x. Ana. a 5.
. Divide 8a^+12^ by 4x*. Jlns. 2x+3.
3. Divide 3bcd+l2bcx 96 2 c by 3bc. Jlns. d-\-4x 3b.
4. Divide lax-\-3ay 7bd by lad.
x 3 y , b
tins. - T r^H
d Id a
DIVISION. 29
ft. Divide 15 2 6c \5acx*-\-5ad* by 5ac.
a
fins.
c
6. Divide Wx*I5x 2 25a? by 5x. Jlns. 2x* 3x 5.
V Divide -Wab+GQab* by Qab.
. I Oft 9 .
o
8. Divide 36 2 6 2 -f 60 2 & 6a& by 1206.
O. Divide lOrx cry-{-2crx by cr.
10. Divide 10uy-r-16d by 2d.
11. Divide 6?/ 18aceH-24a by Ca.
12. Divide mx amx-\-m by m.
(Art. 20.) We now come to the last and most important ope-
ration in division, the division of one compound quantity by an-
other compound quantity.
The dividend may be considered a product of the divisor into
the yet unknown factor, the quotient ; and the highest power of
any letter in the product, or the now called dividend, must ba
conceived to have been formed by the highest power of the same
letter in the divisor into the highest power of that letter in the
quotient. Therefore, both the divisor and the dividend must
be arranged according to the regular powers of some letter.
After this, the truth of the following rule will become obvious
by its great similarity to division in numbers.
RULE. Divide thejirst term of the dividend by the first term
of the divisor, and set the result in the quotient.'*
Multiply the whole divisor by the quotient thus found, and
subtract the product from the dividend.
The remainder will form a new dividend, with which pro-
ceed as before, till the first term of the divisor is no longer
contained in the first term of the remainder.
The divisor and remainder, if there be a remainder, are then
* Divide the first term of the dividend and of the remainders by the first
term of the divisor ; be nut troubled about other terms.
30 ELEMENTS OF ALGEBRA.
to be written in the form of a fraction, as in division of num-
bers.
EXAMPLES.
Divide a*+2ab+& by a+b.
Here, a is the leading letter, and as it stands first in both the
dividend and divisor, no change of place is necessary.
OPERATION.
ab
ab+b*
ab+b 2
Agreeably to the rule, we consider that a will be contained in
a 2 , a times ; then the product of a into the divisor is a*-\-ab, and
the first term of the remainder is ab, in which a is contained 6
times. We then multiply the divisor by 6, and there being no
remainder, a-}-b is the whole quotient.
Divide a 3 +3 *#-{- Sao? +# 3 by x+a.
As the highest power of a stands in the first term of the divi-
dend, and the powers of a decrease in regular gradation from
term to term, therefore we must change the terms of the divisor
to make a stand first.
OPERATION.
a 3 -!- a z x
2a z x-\-2ax z
DIVISION. 8]
a. a c) 3 4a 2 c-|-4ac 2 C 3 (fl 2 3ac4-c
a 3 a z c
3 . a 2 40-HK 6a 2 -f- 1 2a 8(02
2a 2 -f- 8a -8
4. Divide 6a: 4 96 by 6z 12. .#n$. a*-f 23*4-43+8.
5. Divide a 2 ft 2 by a 6. ./#ns. a+6.
6. Divide 25z 6 ^ 2x? 8x 2 by 5r 4;r*.
(Art. 21.) We may cast out equal factors from the dividend
and divisor, without changing the value of the quotient; for
amxy divided by am gives xy for a quotient ; cast out either of
the common factors a or m from both dividend and divisor, and
we shall still have xy for a quotient. This, in many instances,
will greatly facilitate the operation. Thus, in the 4th example,
the factor 6 may be cast out, as it is contained in all the terms ;
and in the 6th example the factor a? 2 may be cast out ; the quo-
tients will of course be the same.
T. Divide a 9 -+4ax+4x*+y* by a+2x.
8. Divide 6a 4 -j-9a 2 15a by So 55 3a.
(Observe Art. 21.) rfns. 2 8 +2+5.
9. Divide a,- 6 y 6 by x*+2x*y+2xy z +y*.
Ans. x*2x*y -\-2xy* y*.
32 ELEMENTS OF ALGEBRA.
10. Divide * ( 2 -f l>)tf+b z by ax b.
tfns. x ax b
11. Divide 1 by 1 a. J2ns. 1-f a+a 2 -f a 8 , &c., &c
12. Divide aM-+?+l by +*.
N. B. We may multiply both dividend and divisor by the
same number as well as divide them.
13. Divide 1 5*/-H<ty 2 Wy*+5y 4 y 5 by 1 27/-f-7/ 2 .
Am. 1 3y4-3y 2 2/ 3
14. Divide 4 -f 4/; 4 by a z 2ab+2b 2 .
Am. a z -\-2ab-\-2b z .
15. Divide a? 6 a? 4 -|-a- 3 ^ 2 +2^~ 1 by a^+a? 1.
^/w. a: 4
16. Divide a 5 or 5 by a #.
Am.
JT. Divide & 5 -f?/ 5 by
18. Divide 3 +5a 2 ?/-j-5?/ 2 -{-i/ 3 by a-fy.
Ans.
If more examples are desired for practice, the examples in
multiplication may be taken. The product or answer may be
taken for a dividend, and either one of the factors for a divisor ;
the other will be the quotient.
Also, the examples in division may be changed to examples in
multiplication ; and these changes may serve to impress on the
mind of the pupil the close connection between these two opera-
tions.
(Art. 22.) In the following examples the dividends and divi-
sors are given in the form of fractions, and the quotients are the
terms after the sign of equality. Let the pupil actually divide,
and observe the quotients attentively.
a; 2 -a 2
1. - - =
x a
DIVISION, 33
2. ^Zl^
x a
3.
4.
a? a
Hence we may conclude that in general x m m is divisible by
x a, m being any entire positive number.
That is, ~=x m - l -\-ax m -* J r - - - a rn ^x-\-a m ~ l
The quotient commencing with a power of x, one less than
m, and ending with a power of a, one less than m.
These divisions show, that the difference of two equal powers
of different quantities is always divisible by the difference of
their roots.
(Art. 23.) By trial, that is, actual division, we shall find that
.r 2 a z
-; =x a.
x-\-a
x-}-a
= x 5 ax*+ aV
x-\-a
&c. &c. &c. &c.
From which we learn that the difference of any two equal
powers of different quantities, is also divisible by the sum of their
roots when the exponent of the power is an even number.
(Art. 24.) By actual division we find that
--.
x-\-a
x-\-a
And in general, we may conclude that the sum of any two
equal powers of different quantities, is divisible by the sum of
their roots when the exponent of the power is an odd number.
34 ELEMENTS OF ALGEBRA.
Li Art. 22, if we make a=l the formulas become
x 2 I
=aH-l.
, &c.
8 1
a? 1
If we make a?=l, what will the formulas become ?
Make the same substitutions in articles 23 and 24, and exa-
mine the results.
By inspecting articles 22, 23, and 24, we find that
=a? 3 a s , &c..
for the product of the divisor and quotient must always produce the
dividend. These principles point out an expedient of condensing
a multitude of terms by multiplying them by the roots of the terms
involved. Thus, a^iaa^+aVzta^+a 4 ? can be condensed to
two terms by multiplying them by # a, the root of the first and
last term, with the minus sign where the signs are plus in" the
multiplicand, and with the plus sign where the signs are alter-
nately plus and minus. See examples in Art. 22 and 24.
ALGEBRAIC FRACTIONS.
(Art. 25.) We shall be very brief on the subject of algebraic
fractions, because the names and rules of operations are the same
as numeral fractions in common arithmetic ; and for illustration,
shall, in some cases, place them side by side.
CASE i. To reduce a mixed quantity to an improper frac-
tion, multiply the integer by the denominator of the fraction^
and to the product add the numerator, or connect it with its
proper sign, -{- or ; then the denominator being set under
this sum, will give the improper Jraction required.
ALGEBRAIC FRACTIONS. 35
EXAMPLES.
1. Reduce 2f and G-TT to improper fractions.
Jhu. y and *+*
These two operations, and the principle that governs them,
aie exactly alike.
2. Reduce 5J and +y to improper fractions.
An,. y .ad ?*+*"
3. Reduce 4 | and a -- to improper fractions.
x
, ax b
MS. y and --
x
, T 4 1 3#
. Keduce 5 -- - and 20 -- to improper fractions.
5. Reduce 5-|- y to an improper fraction.
6. Reduce 12-J -- -r to an improper fraction.
7. Reduce 4-f-2a?H to an improper fraction.
c
8. Reduce 5a; - - to an improper fraction.
3
9. Reduce 3 9 -- T-TT to an improper fraction.
-
*
a+3
CASE 2. TVte converse of Case 1. Tb reduce improper
fractions to mixed quantities, divide the numerator by the de-
nominator, as far as possible, and set the remainder, (if any,}
36 ELEMENTS OF ALGEBRA.
over the denominator for the fractional part ; the two joined
together with their proper sign, will be the mixed quantity
sought.
EXAMPLES.
1 . Reduce 4 7 and ^ to mixed quantities.
Jins. 5J and 0-f-y
2. Reduce */ and - to mixed quantities.
dns. 2f and 0-f--
3. Reduce to a mixed quantity.
ab+y
50-
y
4. Reduce - - to a whole or mixed quantity.
Jlns.
5. Reduce to a whole number.
xy
Am. 2(s?+xy+y 2 ) by (Art. 22.)
6. Reduce - to a mixed quantity.
c
1002 4a-j-6
7. Reduce - - -- to a mixed quantity.
5a
8. Reduce - to a mixed quantity.
3
3# 2 I2ax-}-y 9x
9. Reduce - 2 - to a mixed quantity.
(Art. 26.) It is very desirable to obtain algebraic quantities in
their most condensed form. Therefore, it is often necessary to
reduce fractions to their lowest terms ; and this can be done as
in arithmetic, by dividing both numerator and denominator by
ALGEBRAIC FRACTIONS. 37
their obvious common factors, or for their final reduction, by
their greatest common measure. If the terms have no common
measure, the fraction is already to its lowest terms.
The principle on which these reductions rest is that of divi-
sion, explained in (Art. 21).
CASE 3. To find the greatest common measure of the terms
of a fraction, divide the greater term by the less, and the last
divisor by the remainder, and so on till nothing remains ; then
the divisor last used will be the common measure required.
But note, that it is proper to arrange the quantities according
to the powers of some letter, as is shown in division.
N. B. During the operation we may cast out, or throw in a
factor to either one of the terms which is not a factor in the
other, as such a factor would make no part of the common
measure, and the value of quantities is not under consideration.
Thus, the fraction - has a4-b for its greatest common
a 2 b*
measure; and this quantity is not affected by casting out the
factor b from the numerator, and seeking the common measure
a+b
of the fraction .
a 2 b 2
(Art. 27.) To demonstrate the truth of the rule for finding the
greatest common measure, let us suppose D to represent a divi-
dend, and d a divisor, q the first quotient and r the first re-
mainder.
Tn short, let us represent successive divisions as follows :
d)D(q
dq
Now, in division, the dividend is always equal to the product
of the divisor and quotient, plus the remainder, if any.
38 ELEMENTS OF ALGEBRA.
Therefore, r=r'q"
and d=rq'-}-r'
and D=dq -f-r.
As r^=r'q", the last divisor r' is a factor in r (there being no
remainder) ; that is, r' measures r.
Now as r' measures r, it measures any number of times r,
or r<7'-f-r', orrf; therefore r' measures d.
Again, as r' measures d and r, it measures any number of
times d -\-r ; that is, it measures dq-\-r or D.
Hence r', the last divisor, is a common measure to both D
and d, or of the terms of the fraction __
d
We have now to show that r' is not only the common mea-
sure of D and d, but the greatest common measure.
In division, if we subtract the product of the divisor and quo-
tient from the dividend, we shall have the remainder.
That is, D dq=r, and drq'r'.
Now, every common measure of D and d is also a measure
of r, because D dq=r ; for the same reason every common
measure of r and d is also a measure of r'. But the greatest
measure of r' is itself. This final remainder is, therefore, the
greatest common measure of D and d.
EXAMPLES.
1. Find the greatest common measure of the two terms of
the fraction - - and with it, reduce the fraction to its lowest
terms.
CONSIDERATION AND OPERATION.
The denominator has a 3 as a factor to all its terms, which is
not a factor in the numerator ; hence this can form no part of
the common measure, or the common measure will still be there
if this factor is taken away.
We then seek the common measure of a 4 1 and a?-\-l.
ALGEBRAIC FRACTIONS.
a 2 !
Hence a 2 ~\-l is the common measure, which, used as a divisor
to both numerator and denominator, reduces the fraction to
a 2 !
" a 3 '
2. Find the greatest common measure, and reduce the frac-
tion
Divide this rem. by y 2 zy 2 ?/ 3
ay 2/ 2
*y af
3. Find the greatest common measure and reduce the frac-
uon
Common measure # ^y.
Fraction reduced
Ans. Greatest common measure 2 +a^y. Reduced frac-
.ion *&
5axy.
40 ELEMENTS OF ALGEBRA.
Find the greatest common measure of
and a 2 c+2abc+b 2 c.
Reject the common factor c in one of the quantities,
a'-f 2a z b+ ab*
4. Find the greatest common divisor, and reduce the fraction
to its lowest terms.
Here we find that neither term is divisible by the other ; but
if these quantities have a common divisor, such divisor will still
exist if we multiply one of the terms by any number whatever,
to render division possible.
Therefore take 4a 3 2a 2 3a-\- I
Multiply it by 3
2__2a l)12a 3 6a 2 9a+ 3(4a
12a 3 8a 2 4a
2a 2 5a+ 3
Multiply by 3
9(2
4a 2
Divide by 11 lla-j-11
a 1
a 1
Jins Greatest common measure a 1. Reduced fraction
j*<zi:L_
4^+201*
ALGEBRAIC FRACTIONS. 41
5. Find the greatest common measure and reduce the frac-
to its lowest terms '
Jlns. Common divisor a 2 x 2 . Reduced fraction .
a x
<* Find the greatest common measure, and reduce the frac-
tion to its lowest terms.
Jlns. Common divisor a? y 2 . Red. frac.
. Reduce
g. Reduce
,
a 2 x z
to its lowest terms.
tins.
a+x
to its lowest terms. ^ &
3a x
Ans.
9. Reduce
10. Reduce
- to its lowest terms.
So 3
Am.
bx
to its lowest terms.
11. What is the value of --
Ans.
12. Find the greatest common divisor of 12a 4 24a 3 6+ 12a 2 6 2 ,
and 8o 3 6 2 -
(A rt. 28.) We may often reduce a fraction by separating both
numerator and denominator into obvious factors, without the
formality of finding the greatest common divisor. The follow
ing are some examples of the kind :
4
42 ELEMENTS OP ALGEBRA.
1. Reduce -- - to its lowest terms.
a(a 2 b 2 ) _a(ab}(a+b}
2. Reduce - to its lowest terms. Jlns.
- T IV AfcO lUWtOl 1C HHO. VJ.IIO. - -
^2 J ,p |
3. Reduce ; to its lowest terms. Jlns. .
zy+y y
CX-}- CX 2 C-}-CX
4. Reduce to its lowest terms. Jins. .
acx-{-abx ac-\-ab
2# 3 .I6x 6
5. Reduce to its lowest terms. Jlns. I.
3^3 24x 9
(Art. 29.) To find the least common multiple of two or
more quantities.
The least common multiple of several quantities is the least
quantity in which each of them is contained without a remainder.
Thus, the least common multiple of the prime factors, a, b, c,
x, is obviously their product abcx. Now observe that the same
product is the least common multiple also, when either one of
these letters appears in more than one of the terms. Take a,
for example, and let it appear with b, c, or x f or with all of
them, as a, ab, c, ax, or a, 6, c, ax, the product abcx is still
divisible by each quantity. Therefore, when the same factor
appears in any number of the terms, it is only necessary that it
should appear once in the product; that is, once in the least
common multiple. If it should be used more than once, the
product so formed would not be the least common multiple.
From this examination, the following rule for finding the taast
common multiple will be obvious :
RULE, Write the given quantities, one after another, and
draw a line beneath them. Then divide by any prime factor
that will divide two or more of them without a remainder,
bringing down the quotients and the quantities not divisible,
to a line below. Divide this second line as the first, forming
ALGEBRAIC FRACTIONS. 43
a third, fyc., until nothing but prime quantities are left. Then
multiply all the divisors and the remainders that are not divis-
ible, and their product will be the least common multiple.
N. B. This rule is also in common arithmetic.
EXAMPLES. I
1. Required the least common multiple of Sac, 4a 2 ,
8c, and ex.
2a)Sac 4a 2 I2ab Sac ex
2c)4c
2a
Qb
4c
ex
2)2
a
3b
2
X
1
a
36
1
X
Therefore 2a X 2c X 2 X a X 36 X x=24a 2 cbx.
Here the divisor 2c will not divide 2a, but the coefficient of
c will divide the coefficient of a, and we let them divide, for it
is the same as first dividing by 2, and afterwards by c. From
the same consideration we permit 2c to divide ex, or let the let-
ter c in the divisor strike out c before x.
By the rule we should divide by 2 and by c separately ; but
this is a practical abbreviation of the rule.
2. Required the least common multiple of 27 a, 156, 9a6, and
3a 2 . Ans. 135a 2 6.
3. Find the least common multiple of (a 2 x 2 }, 4(a x),
and (a+x). tins. 4(a 2 #*).
Find the least common multiple of ax 2 , bx, acx, and
a 2 x 2 . Am. (a?a?)acbx?.
5. Find the least common multiple of a-f-6, a b, a 2 -{-ab-{-b 2 ,
and a 2 ab+b 2 . Ans. a 6 b 6 .
The least common multiple is useful many times in reducing
fractions to their least common denominator.
CASE 4. To reduce fractions to a common denominator.
(Art. 30.) The rule for this operation, and the principle on
which it is founded, is just the same as in common arithmetic,
merely the multiplication of numerator and denominator by the
44 ELEMENTS OF ALGEBKA.
same quantity. The object of reducing fractions to a common
denominator is to add them, or to take their difference, as diffe-
rent denominations cannot be put into one sum.
RULE. Multiply each numerator by all the denominators,
except its own, for a new numerator, and all the denominators
for a common denominator.
Or, find the, least common multiple of the given denomina-
tors for a common denominator; then multiply each denomi-
nator by such a quantity as will give the common denomina-
tor, and multiply each numerator by the same quantity by
which its denominator was multiplied.
EXAMPLES.
1. Reduce and to a common denominator.
x 2c
4ac 3bx
rfns. - and - .
2cx 2cx
2a 3a-f-26
8. Reduce and - to a common denominator.
b 2c
4ac , Bab+2 I?
Jlns. and ---
2bc 2bc
3. Reduce and , and 4d to a common denominator.
3x 2c
, 9bx , 24cdx
Jins. - and and - .
Gcx Gcx Gcx
4. Reduce 7-, - , ~ , to fractions having a common de-
c x-\-a
Jlns. acx-{-a 2 c (bx-\-b)(x-}-a) bey
bcx -\-abc b ex -\-abc ' bcx-\-abc'
(Art. 31.) CASE 5 Addition or finding the sum of fractions.
RULE. Reduce the fractions to a common denominator ;
and the sum of the numerators, written over the common deno-
minator, will be the sum of the fractions.
ALGEBRAIC FRACTIONS. 45
EXAMPLES.
3x 2x X
1 Add , , and - together
57 *
I28.r
// * j o -. ___ _. _ _ .^ -_-,-._ > _
105 105
, a . a-\-b
2. Add T and ~. Ans.
T ~. . - -
b c be
3. Add -, - and - .
f and .
a 6 a 4- 6
a-f-3 2a 5 ,
5. Add 2a+- - and 4aH --- . .^/?5. 6H 077-
8x* %ax %abx 8cz*
6. Add a r- and b-{ Jins. a-\-o-\ .
b c uc
x 2 2;r 3
7. Add 5.r+ -- and 4x .
o ox
Ans.
3.r b , 6 a?.
8. What is the sum of 26-{ -, 7 - and - ?
5 6 x b
2 5bx
Mt ___ f) Q,j O
9 What is the sum of 5y-l-^ - and 4y -- C 1 ?
d o
4C ELEMENTS OF ALGEBRA.
1.0. What is the sum of 50, , and ^~
Ans.
(Art. 32.) CASE G. Subtraction or finding difference.
RULE. Reduce the fractions to a common denominator, and
subtract the numerator of that fraction which is to be sub-
tracted from the numerator of the other, placing the difference
over the common denominator.
EXAMPLES.
7x 2x 1 2lxAx-\-2 l7x-\-2
1. From -take- . An*. - - - =__.
take . Eq. fractions - 2
Difference or *ftns.
nt O/y rp
3. From 5 take ~. Diff. ^
o 7 1
4. From ^ take . Ans.
7 y 63
2a b Ba 4b
9 From ; subtract -7 .
4c 3o
126c
. llfl 10 , 3a 5.
6. From 3a-\ subtract 2a-\ .
15 7
tins, a-
. From x-\ ~~* subtract ^. fins, x
x 2 xy
ab . 2b4a 5ad 5bd 4bc-\-Sac
8. From - ~ take . Jim. -- -
2c bd IQcd
ALGEBRAIC FRACTIONS. 47
. x x a . cx-\-bx~ ab
9. From 3x-\--r take x . fins. 2x-{ -,
be be
a-\-b a b
10. Find the difference between - r and J-T-.
a b a+b
4ab
tins.
a 2 ft 2
11. prom take . fa. 4.
xy xy
CASE 7. Multiplication of fractions.
(Art. 33.) The multiplication of algebraic fractions is just the
fame in principle and in fact, as in numeral fractions, hence the
rule must be the same.
It is perfectly obvious, that f multiplied by 2 must be -|, and
multiplied by 3 must be f ; and the result would be equally ob-
vious with any other simple fraction ; hence, to multiply a frac-
tion by a whole number, we must multiply its numerator.
It is manifest that doubling a denominator without changing its
numerator halves a fraction, thus ; double the 2, and we have
?, the half of the first fraction.
Also f , double the 5 gives T \, the half of |. In the same
manner, to divide a fraction by 3 we would multiply its denomi-
nator by 3, &c. In general, to divide a fraction by any num-
ber, we must multiply the denominator by that number.
Now let us take the literal fraction -r, and multiply it by c, the
ac
product must be -j- .
Again, let it be required to multiply - by -. Here the mul-
tiplication is the same as before, except the multiplier c is divided
by d ; therefore if we multiply by c we must divide by d. But
the product of - by c is ; this must be divided by d 1 and we
ac
shall have - ; for the true product of ,- by -=.
bd b J d
48 ELEMENTS OF ALMEBliA.
From the preceding investigation we draw the following rule
to multiply fractions :
RULE. Multiply the numerators together for a new nume-
rator, and the denominators together for a new denominator.
N. B. When equal factors, whether numeral or literal,
appear in numerators and denominators, they may be canceled,
or left out, which will save subsequent reductions.
EXAMPLES.
1. Multiply by * and . Al . a ^L.
7 b J x c ex
In this example, b in the denominator of one fraction cancels
in the numerator of another.
iviuiiipiy
<
3. Multiply :
4U Multiply
5. Multiply*
fi TVTiiltiT-wlir
f\f\ U ' J f\ f 1 \
ou o{a~\~x\
**+% bv 2
18
Ans. -.
5x
(a-x)a
2 x 2 , 2a
Vtr
2i/ a-j-a?
ar 2 i/ 2 3? a
2 onrl
y
Ans. a.
, ana
a; 07+y a; y
a?+l , ^ 1
N. B. Reduce mixed quantities to improper fractions.
7. What is the continued product of , - : : and
a-\-b ax-\-x*
ax 2 (rt b)
ax x
4v* 15|/ 30
8. Multiply ^-^ by -* . AM.
* Separate into factors when separation is obvious.
ALGEBRAIC FRACTIONS. 49
9. Multiply - by -775. #*
a-\-b ab b* b
10. Multiply -^P-, by jgg^gj* ^ n5 ' 3 (+ :F )-
11 Required the continued product of -= r-, -^-T j and
w ~~~O flr~~it
.5n*. (+*).
19. Multiply + by -. .*..
_ .
14. Mult,ply - by
4 2 166 2 , 56
15. MulUply --- by
8a . +32afe+32ft .
CASE 8 Division of Fractions.
(Art. 34.) To acquire a clear understanding of division in
fractions, let us return to division in whole numbers.
The first principle to which we wish to call the attention of
the reader, is, that if we multiply or divide both dividend and
divisor of any sum in division, by any number whatever, we do
not affect or change the quotient. (Art. 21.)
Thus, 2)6(3 4)12(3 8)24(3 &c
The second principle to which we would call observation is,
that if we multiply any fraction by its denominator, we have the
numerator for a product.
Thus, 5 multiplied by 3 gives 1, the numerator, and | by 5
gives 2, and -r multiplied by b gives a, <fec.
5
50 ELEMENTS OF ALGEBRA.
Now let it be required to divide j- by -.
The quotient will be the same if we multiply both dividend
and divisor by the same quantity. Let us multiply both terms
by ?, the denominator of the divisor, and we have -=- to be di
vided by the whole number c. But to divide a fraction by a
whole number we must multiply the denominator by that num-
ber. (Art. 33.) Hence j~ is the true quotient required.
We can mechanically arrive at the same result by inverting
the terms of the divisor, and then multiplying the upper terms
together for a numerator, and the lower terms for a denominator;
therefore to divide one fraction by another we have the following
RULE. Invert the terms of the divisor, and proceed as in
multiplication.
EXAMPLES.
a+b . c
1. Divide by p-r. rfns.
c a-\-b
2. Divide 5X by -.
a c
c 5x 5 ex
Operation: divisor inverted -rX = r-
b a ab
10c*
3. Divide - by -- ,.
a x a 2 &
I5ab (a-\-x}(a x) 3b(a+x)
Operation, - X^ - - -- '-. ^ns. ^-- -.
a x lOac 2c
2* 2 7 as
4. Divide - by --s. Ans.
x-\-a J x* -\-1ax-\- a*
Operation,
Divide into factors, in all such cases, and cancel.
ALGEBRAIC FRACTIONS.
51
a 2 -\-ax-}-x 2 '
70# 15
5 7 25 *
/?? ~~
5 5
x
18x 21
Jlns. ^ r .
' x+l J 3 '
1O. Divide hv
3a 2
, l . Dmde ____ + _ by
Am.
12. Divide by
5 5
,_.... na nx . ma mx
13. Divide TT - by i ~ 7 .
-
Ans.
m
14. Divide 12 by
15. Divide
, a
by -.
x J x
,., T,. ., x b 3cx
lo. Divide 5 by r .
J
Jlns.
I2x
a*-fax-\-x*
Jlns. b+-
a
Divide a by the product of : into .
X+y xy
Jlns.
18. Divide ^-i-rr b y the product of ^-^ ; into ~.
ofx._i_A\ J 2 a 4-6
52 ELEMENTS OF ALGEBRA.
SECTION II.
CHAPTER I.
Preparatory to the solution of problems, and to extended in-
vestigations of scientific truth, we commenced by explaining the
reason and the manner of adding, subtracting, multiplying, and
dividing algebraic quantities, both whole and fractional, that the
mind of the pupil need not be called away to the art of perform-
ing these operations, when all his attention may be required on
the nature and philosophy of the problem itself.
For this reason we did not commence with problems.
Analytical investigations are mostly carried on by means
OF EQUATIONS.
(Art. 35.) An equation is an algebraical expression, meaning
that certain quantities are equal to certain other quantities. Thus,
3-f-4=7 ; a-{-6=c ; #4-4=10, are equations, and express that
3 added to 4 is equal to 7, and in the second equation that a
added to b is equal to e, <fcc. The signs are only abbreviations
for words.
The quantities on each side of the sign of equality are called
members. Those on the left of the sign form the first member,
those on the right the second.
In the solution of problems every equation is supposed to con-
tain at least one unknown quantity, and the solution of an equa-
tion is the art of changing and operating on the terms by means
of addition, subtraction, multiplication, or division, or by all these
combined, so that the unknown term may stand alone as one
member of the equation, equal to known terms in the other
member, by which it then becomes known.
Equations are of the first, second, third, or fourth degree,
according as the unknown quantity which they contain is of the
first, second, third, or fourth power.
ax-\-b=3az is an equation of the first degree or simple
equation*
EQUATIONS. 53
ax z -\-bx=3ab is an equation of the second degree or quadratic
equation.
ax s -\-bx 2 -\-cx=2a 4 b is an equation of the third degree.
ax*-{-bx 3J rcx 2 -\-dx=2ab 5 is an equation of the fourth degree.
We shall at present confine ourselves to simple equations.
(Art. 36.) The unknown quantity of an equation may be
united to known quantities, in four different ways : by addition,
by subtraction, by multiplication, and by division, and further
by various combinations of these four ways as shown by the
following equations, both numeral and literal :
NUMERAL. LITERAL.
1st. By addition, aH-6=10 x-\-a=b
2d. By subtraction, x 8 = 12 x c=d
3d. By multiplication, 20#=80 axe
4th. By division, 7~ 16 ~d = ~^~ a '
5th. ar+6 8+4=10+2 3, x+a b+c=d+c, &c.,
are equations in which the unknown is connected with known
quantities by both addition and subtraction.
y* or
2aH-r= 21 > ax-\~Cj are equations in which the unknown
o
is connected with known quantities by both multiplication and
division.
Equations often occur, in solving problems, in which all of
these operations are combined.
(Art. 37.) Let us now examine how the unknown quantity can
be separated from others, and be made to stand by itself.
Take the 1st equation, or other similar ones.
x -f6=10 x-\-a=b
Take equal quantities 6= 6 a=a from both
members, and #=10 6 x=b a the
remainders must be equal. (Ax. 2.) Now we find the term
sdded to x, whatever it may be, appears on the other side with
a contrary sign, and the unknown term x being equal to known
terms is now known.
64 ELEMENTS OF ALGEBRA.
Take the equations
Add equals to both memb.
Sums are equal #=124- 8 x=d-{-c (Ax. 1.)
Here again the quantity united to x appears on the opposite
side with a contrary sign.
From this we may draw the following principle cr rule 3f
operation :
Any term may be transposed from one member of an equa-
tion to the other, by changing its sign.
Now 20#=80. axe. If we divide both members by the
coefficient of the unknown term, the quotients will be equal.
(Ax. 4.) Hence #=f J=4. x=-.
That is, the unknown quantity is disengaged from known
quantities, in this case, by division.
7* "7*
Again, take the equations -=16; ~=g-\-a.
Multiply both members by the divisor of the unknown term,
and we have o?=16X4. x=gd-\-ad. Equations which must
be true by (Ax. 3.), and here it will be observed that x is libe-
rated by multiplication.
From these observations we deduce this general principle :
That to separate the unknown quantity from additional
terms we must use subtraction; from subtracted terms we
must use addition ; from multiplied terms we must use divi-
sion ; from divisors we must use multiplication.
In all cases take the opposite operation.
EXAMPLES.
1. Given 3x 4=7# 16 to find the value of #. Ans. #=3.
2. Given 3#-}-9 1 5x=0 to find the value of x.
Am. x.
3. Given 4y-}-7y-\-2l 3-\-y to find y. Ans. y=5h-
4. Given Sax c=b 3ax to find the value of x.
Jlns. *=*-- C
8a
EQUATIONS. 55
5. Given ax z -{-bx=9x z -{-cx to find the value of x in terms
of a, b, and c. jlns. x= -.
a 9
N. B. In this last example we observe that every term of the
equation contains at least one factor of x ; we therefore divide
every term by x, to suppress this factor.
(Art. 38.) In many problems, the unknown quantity is
often combined with known quantities, not merely in a simple
manner, but under various fractional and compound forms.
Hence, rules can only embody general principles, and skill and
tact must be acquired by close attention and practical application :
but from the foregoing principles we draw the following
GENERAL RULE. Connect and unite as much as possible all
the terms of a similar kind on both sides of the equation. Then,
to clear of fractions, multiply both sides by the denominators,
one after another, in succession. Or, multiply by their con-
tinued product, or by their least common multiple, (when such
a number is obvious,) and the equation will be free of fractions.
Then, transpose the unknown terms to the first member of
the equation, and the known terms to the other. Then unite
the similar terms, and divide by the coefficient of the unknown
term, and the equation is solved.
EXAMPLES.
1. Given x-K#-{-3 7=61, to find the value of x.
Uniting the known terms, after transposition, agreeably to the
rule of addition, we find x-}-zX=9. Multiply every term by
2, and we have 2x-\-x=lS. Therefore x=6.
2. Given 2x-{-$x-{-ix 3a=46-f 30, to find x.
N. B. We may clear of fractions, in the first place, before we
condense and unite terms, if more convenient, and among literal
quantities this is generally preferable.
In the present case let us multiply every term of the equation
by 12, the product of 3X4, and we shall have
24;r-f9a:-|-4tf 360=486+ 36a.
Transpose and unite, and 37#=48&-f-72a
56 ELEMENTS OF ALGEBRA.
Divide by 37, Md *=
3. Given 5r-f-#-f-*#=39, to find the value of x.
Here are no scattering terms to collect, and clearing of frac-
tions is the first operation.
By an examination of the denominators, 12 is obviously their
least common multiple, therefore multiply by 12. Say 12
halves are 6 whole ones, 12 thirds are 4, 12 fourths are 3, &c.
Hence, 6x+4x+ 3#=39 X 12
Collect the terras, 13a?=39X 12
Divide by 13, and a?= 3X12=36, Am.
N. B. In other books we find the numerals actually multiplied
by 12. Here it is only indicated, which is all that is necessary.
For when we come to divide by the coefficient of x, we shall
find factors that will cancel, unless that coefficient is prime to all
the other numbers used, which, in practice, is very rarely the
case.
4. Given a?-hia?-H#=0, to find x.
This example ia essentially the same as the last. It is identi-
cal if we suppose a=39.
Solution,
Or, 13#=12a
12a
Divide and X= -T^
L*i
Now if a be any multiple of 13, the problem is easy and
brief in numerals.
, 3x 11 5x 5 , 97 7#
5. Given 2H -- = - + - to find the value of x.
16 o Z
Here 16 is obviously the least common multiple of the deno-
minators, and the rule would require us to multiply by it, and
such an operation would be correct ; but in this case it is more
easy to multiply by the least denominator 2, and then condense
like terms. Thus,
, EQUATIONS. 57
Multiply by 2, and we have
Recollect that we can multiply a fraction by dividing its deno-
minator. Also observe that we can mentally take away 42 from
both sides of the equation, and the remainders will be equal.
(Ax. 2.)
Then 7=
8 4
Multiply by 8, and
3 X .11 =10a> 10+440 56x ;
Transposing and uniting terms, we have
49#=441 ;
By division, x=9,
6. Given ar+2^ + ll=fa?-f 17, to find x.
If we commence by clearing of fractions, we should make
comparatively a long and tedious operation. Let us first reduce
it by striking out equals from both sides of the equation. We
can take 1 1 from both sides without any formality of transposing
or changing signs ; say drop equals from both sides, (Ax. 2.) and
reduce the fraction f#=|a?.
All this can be done as quick as thought, and we shall have
Multiply by 4, then
+10=3+24, or =a
5 o
Hence, 7#=70, or xlQ Am.
7. Given %x 5-KaM-8-Kx 10=100 6 7 to f.nd the
value of x.
Collecting and uniting the numeral quantities, we have
-^=;
Multiply every term by 60, and we have
Collecting terms, 47a?=94.60
Divide both sides by 47, and x= 2.60=120 Ans.
58 ELEMENTS OF ALGEBRA.
(Art. 39.) When equations contain compound fractions and
simple ones, clear them of the simple fractions first, and unite,
as far as possible, all the simple terms.
EXAMPLES.
6#-f-7 , 7x 13
8. Given ~ ~-+ a , Q = ~- to find the value of a\
y tx-po o
Multiply all the terms by the smallest denominator, 3. That
is, divide all the denominators by 3, and
Multiply by 3 again, and 6a?+7+ "
Drop 6#-f-7, and - - =5.
2>x~\~ 1
Clear of fractions, 21 a: 39=10a?-f 5.
Drop IQx and add 39, and we have 11#=44, or x=4.
9. Given -- - -- =- to find the value of x.
21 4x 11 3
Observe that ~ may be expressed in two parts, thus,
7x 16 7x x
-j-ry Observe also, that -=-. Hence these terms may
2121 2 1 o
be dropped, the remainders must be equal. Transpose the
16 x-\-S
minus term, then = -- .
21 4x 11
Clear of fractions, and 64^11X16=21^4-21X8.
Drop 21a; and observe that 11 X 16 is the same as 22 X 8.
Then 432- 22X8=21X8. Let =8,
Then 43z 22a=21.
Transpose 22a and 43#=43a.
Hence x=a. But a=8. Therefore x=8.
N. B. We operate thus, to call attention to the relation of
quantities, and to form a habit of quick comparison, which will,
in many instances, save much labor and introduce the pupil into
the true spirit of the science
EQUATIONS. 59
1O. Given = r~\"7 to find tne value of or.
36 ox 4 4
9a?
By a slight examination we perceive that is equal to %x.
oD
Hence these terms may be left out, as they balance each other.
Therefore
Clear of fractions, and 25a? 20 =36# 108.
Transpose 25x 36#=20 108.
Unite and change signs, and 11#=88 or a?=8, Jim
. 36 . 5#-}-20 4x . 86 t
11. Gl ven _ + _+ _ is = T +_ to find *.
By taking equals from both sides, we have
5#t-20
- - rz%' By reduction a? =4.
Qx 16
12. Given - --- ^= 6 ^ -- r to find *
42 4
Multiply by 4, to clear of fractions, and
3tf 2#-f-2=24;r 20# 13. Reduced x=5.
(Art. 40.) When a minus sign stands before a compound
quantity, it indicates that the whole is to be subtracted ; but we
subtract by changing signs, (Art. 5). The minus sign before
y. __ I
- in the last example, does not indicate that the x is minus,
but that this term must be subtracted. When the term is multi-
plied by 4, the numerator becomes 2x 2, and subtracting it
we have 2#-|-2.
Having thus far explained, we give the following unwrought
equations, for practice :
O/v n\
13. Given =-+24 to find the value of x. Jns. 19],.
14. Given zx-\-lx=lQ to find the value of x. rfns. 24.
60 ELEMENTS OF ALGEBRA.
15. Given ^-+|.=20^i_ to find z. Ana 9
16. Given ^i+^?=16-? to find *. Am. 13
. Given 2 *-+,5= to find *.
3 5
, a ~. 2#+l a?+3
18. Given x --- _=__ to find x. tins. x=l3
19. Given 5#+5#+*tf+ya?=77 to find * ^s. #=60.
20. Given 5a?+|a>Ha?=130 to find a?. rfns. x=120
21. Given ^af+^+T^a^QO to find x. Jlns. x=l20
22. Given ^+^+^=82 to find y. ^ns. y=84
23. Given 5a?-FIa;+5a?=34 to find x. tins.
24. Given 11^4-+++=315 to find x.
25. Given 3/++++=146 to find y. Ans. y=56
26. Given _29=_-30 to find a?, ^n*. a?=2
ar+2 a? 2
There is a peculiar circumstance attending this 26th example,
and the 4th example of Art. 42, which will cause us to refer to
them in a subsequent part of this work.
N. B. In solving equations 19, 20, 21, 22, and 23, use no
larger numbers than those given, indicating and not performing
numeral multiplications.
(Art. 41.) Every proportion may be converted into an equa-
tion. Proportion is nothing more than an assumption that the
same relation or the same ratio exists between two quantities,
as exists between two other quantities.
That is, Ji is to B as C is to D. There is some relation be-
tween A and B. Let r express that relation, then /?=r*#. But
EQUATIONS. 61
the relation between C and D is the same (by hypothesis) as
between Ji and B. Hence D=rC.
Then in place of A : B : : C:D
we have Ji \rA\\ C:rC
Multiply the extreme terms, and we have
Multiply the mean terms, and we have
Obviously the same product, whatever quantities may be re-
presented by either .#, or r, or C.
Hence, to convert a proportion into an equation, we have the
following
RULE. Place, the product of the extremes equal to the pro-
duct of the means.
(Art. 42.) The relation between two quantities is not changed
by multiplying or dividing both of them by the same quantity.
Thus, a : b :: 2a : 2b, or more generally, a : b :: na : nb, for the
product of the extremes is obviously equal to the product of the
means.
That is, a is to b as any number of times a is to the same
number of times b.
We shall take up proportion again, but Articles 41 and 42 are
sufficient for our present purpose.
EXAMPLES.
1. Given 3.r 1 : 2;r-f-l :: 3x : x to find x.
(By Art. 41.) 3x 2 x=6x*+3x.
Transpose and unite, and we have Q=3x*-{-4x.
Divide by x, and 3x4-4=0 or x f , Jlns.
3 3x
2. Given '- : : : 6 : 5x 4 to find x.
The first two terms have the same relation as 5 : Jar, or oa
2 : x. Hence 2 : x :: 6 : 5x 4.
Product of extremes and means, lOa: S=Gx or x=Z.
X.
Am. #=:3.
02 ELEMENTS OF ALGEBRA.
y K
4. Given - : (#5) : : f : to find x.
fins. -=5,
5. Given a?-{-2 : a : : b : c to find the value of x.
dns. j?= 2.
c
6. Given 2a? 3 : x 1 :: 2x : #-f-l to find the value of x.
Jlns. x=3
7. Given z-J-6 : 38 x : : 9 : 2 to find x. fins. x=30.
8. Given #+4 : x 11 : : 100 : 40 to find x. Jlns. x=2l.
QUESTIONS PRODUCING SIMPLE EQUATIONS.
(Art. 43.) We now suppose the pupil can readily reduce a
simple equation containing but one unknown quantity, and he is,
therefore, prepared to solve the following questions. The only
difficulty he can experience is the want of tact to reason briefly
and powerfully with algebraic symbols ; but this tact can only
be acquired by practice and strict attention to the solution of
questions. We can only give the following general direction :
Represent the unknown quantify by some symbol or letter,
and really consider it as definite and known, and go over the
same operations as to verify the answer when known.
EXAMPLES.
1. What number is that whose third part added to its fourth
part makes 21 1 Jlns. 36.
The number may be represented by x.
Then $x-}-Zx=2l. Therefore x=3G.
2. Two men having found a bag of money, disputed about
the division of it. One said that the half, the third, and the
fourth parts of it made $130, and if the other could tell how
much money the bag contained, he might have it all. How
much money did the bag contain 1 Jlns. $120.
(See equation 20, Art. 40.)
EQUATIONS. 63
3. A man has a lease for 20 years, one-third of the time pas.
is equal to one-half of the time to come. How much of the time
nas passed ?
Let x= the time past.
Then 20 x= the time to come.
x 203?
By the question --= . Therefore x=l2 Am.
o Z
4. What number is that, from which 6 being subtracted, and
the remainder multiplied by 11, the product will be 121?
Let 0:= the number.
Then (a: 6)11=121, or x 6=11 by division.
Hence x=l7.
5. It is required to find two numbers, whose difference is 6,
and if | of the less be added to of the greater, the sum will
be equal to ^ of the greater diminished by of the less.
Let x the less. Then #-f-6= the greater.
By the question %x-\ = x.
5 o
Drop |a? from both sides and add \x to both sides, and we
have ~- =2, or x=2, the less number.
We may clear of fractions in full, and then transpose and unite
terms, but the operation would be much longer.
6. After paying | and ^ of my money, I had $66 left ; how
much had I at first? JLns. $120.
7. After paying away 5 of my money, and | of what remained,
and losing ^ of what was left, I found that I had still $24. How
much had I at first ? rfns. 60.
8. What number is that from which if 5 be subtracted, of
the remainder will be 40 ? Jlns. 65,
9. A man sold a horse and a chaise for $200 ; 5 of the price
of the horse was equal to | of the price of the chaise. What
was the price of each? Jlns. Chaise $120. Horse $80.
(M ELEMENTS OF ALGEBRA.
1O Divide 48 into two such parts, that if the less be divided
by 4, and the greater by 6, the sum of the quotients will be 9.
rfns. 12 and 36.
11. An estate is to be divided among 4 children, in the fol-
lowing manner :
The first is to have $200 more than | of the whole.
The second is to have $340 more than ^ of the whole.
The third is to have $300 more than i- of the whole.
And the fourth is to have $400 more than j of the whole.
What is the value of the estate ? Ans. $4800
12. Find two numbers in the proportion of 3 to 4, whose
sum shall be to the sum of their squares as 7 to 50.
Jlns. 6 and 8.
N. B. When proportional numbers are required, it is generally
most convenient to represent them by one unknown term, with
coefficients of the given relation. Thus, numbers in proportion
of 3 to 4, may be expressed by 3x and 4x, and the proportion
of a to b may be expressed by ax and bx.
13. The sum of $2000 was bequeathed to two persons, so
that the share of A should be to that of B as 7 to 9. What was
the share of each? Jins. #'s share $875, B's share $1125.
14. A certain sum of money was put at simple interest, and
in 8 months it amounted to $1488, and in 15 months it amounted
to $1530. What was the sum 1 Ans. $1440.
Let a?= the sum. The sum or principal subtracted from the
amount will give interest: therefore 1488 x represents the
interest for 8 months, and 1530 x is the interest for 15 months.
Now whatever be the rate per cent, double time will give
double interest, &c. Hence 8 : 15 :: 1488 x : 1530 x.
N. B. To acquire true delicacy in algebraical operations, it is
often expedient not to use large numerals, but let them be repre-
sented by letters. In the present example let =1488 Then
a-f-42=1530, and the proportion becomes 8: 15:: a z:a-\-
42 x.
EQUATIONS. 05
Multiply extremes, &c., 8-f-8-42 Sx=l5aI5x.
Drop Sa and Sx. We then have 8'42=7a 7#.
Dividing by 7 and transposing xa 48=1440, Am.
15. A merchant allows $1000 per annum for the expenses
of his family, and annually increases that part of his capital
which is not so expended by a third of it ; at the end of three
years his original stock will double. What had he at first /
rfns. $14,800.
Let x= the original stock, and a=1000.
To increase any quantity by its part is equivalent to multi-
\ 3C \Ct
plying it by . Hence - is his 2d year's stock.
16. A man has a lease for 99 years, and being asked how
much of it had already expired, answered that f of the time past
was equal to of the time to come. Required the time past and
the time to come.
Assume #=99. Am. Time past, 54 years.
17. In the composition of a quantity of gunpowder
The nitre was lOlbs. more than f of the whole,
~r .
The sulphur 4 Ibs. less than \ of the whole,
The charcoal 2 Ibs. less than \ of the nitre.
What was the amount of gunpowder? Jlns. 69 Ibs.
18. Divide $183 between two men, so that y of what the first
receives shall be equal to T \ of what the second receives. What
will be the share of each ? Ans. 1st, $63 ; 2d, $120.
19. Divide the number 68 into two such parts that the differ-
ence between the greater and 84 shall be equal to 3 times the
difference between the less and 40. Ans. Greater, 42 ; Less, 26.
20. Four places are situated in the order of the letters A, B,
C, D The distance from A to D is 34 miles. The distance
from A to B is to the distance from C to D as 2 to 3. And 4
of the distance from A to J?, added to half the distance from C
to D, is three times the distance from B to C. What are the
respective distances ?
Ans. From A to 7?=12 ; from B to C=4 ; frorp C to//=18,
6
66 ELEMENTS OF ALGEBRA.
21. A man driving a flock of sheep to market, was met by a
party of soldiers, who plundered him of 5 of his flock and
more. Afterwards he was met by another company, who took
5 what he then had and 10 more: after that he had but 2 left.
How many had he at first? fins. 45.
22. A laborer engaged to serve for 60 days on these comli
tions : That for every day he worked he should have 75 cents
and his board, and for every day he was idle he should forfeit 25
cents for damage and board. At the end of the time a settlement
was made and he received $25. How many days did he work,
and how many days was he idle ?
The common way of solving such questions is to let x= the
days he worked ; then 60 x represents the days he was idle.
Then sum up the account and put it equal to $25.
Another method is to consider that if he worked the whole 60
days, at 75 cents per day, he must receive $45. But for every
day he was idle, he not only lost his w^es, 75 cents, but 25
cents in addition. That is, he lost $1 every day he was idle.
Now let x= the days he was idle. Then x= the dollars
he lost. And 45 #=25 or #=20 the days he was idle.
23. A boy engaged to carry 100 glass vessels to a certain
place, and to receive 3 cents for every one he delivered, and to
forfeit 9 cents for every one he broke. On settlement, he re-
ceived 2 dollars and 40 cents. How many did he break ?
rfns. 5.
24. A person engaged to work a days on these conditions :
For each day he worked he was to receive b cents, for each day
he was idle he was to forfeit c cents. At the end of a days he
received d cents. How many days was he idle ?
ab d .
Jlns. -T-, days,
o-f-c
25. It is required to divide the number 204 into two such
parts, that f of the less being taken from the greater, the remain-
der will be equal to f of the greater subtracted from 4 times the
less. Arts. The numbers are 154 and 50.*
EQUATIONS. 67
(Art. 44.) We introduce this, and a few following problems,
to teach one important expedient, not to say principle, which is,
not always to commence a problem by putting the unknown
quantity equal to a single letter. We may take 2;r, 3;r, or nx
to represent the unknown quantity, as well as #, and we may
resort to this expedient when fractional parts of the quantity are
called in question, and take such a number of ar's as may be
divided without fractions.
In the present example we do not put x= to the less part, as
we must have f of the less part. It will be more convenient to
put 5x= the less part. Then f of it will be 2x. Put a=204.
26. A man bought a horse and chaise for 341 () dollars.
Now if J of the price of the horse be subtracted from twice the
price of the chaise, the remainder will be the same as if | of the
price of the chaise be subtracted from 3 times the price of the
horse. Required the price of each.
Ans. Horse $152. Chaise $189.
N. B. Let Sx= the price of the horse.
Or let 7x= the price of the chaise.
Solve this question by both of these notations.
27. From two casks of equal size are drawn quantities, which
are in the proportion of 6 to 7; and it appears that if 16 gallons less
had been drawn from that which now contains the less, only one
half as much would have been drawn from it as from the other.
How many gallons were drawn from each 1 Jlns. 24 and 28.
N. B. Let Qx and 7x equal the quantities drawn out.
28. Divide $315 among four persons, ./?, B, C, and D, giving
B as much and more than Jl ; C 5 more than Jl and B toge-
ther ; and D | more than A, B and C. What is the share of
eadi ? Ans. ./? $24. B $36. C $80, and D $175.
If we take x to represent #'s share, we shall have a very
complex and troublesome problem.* But it will be more simple
by making 6x=J2's share.
* Taking x for yTs share, and reducing their sum, gives Equation 24,
Art. 40.
68 ELEMENTS OF ALGEBRA.
Thus, let 6x=tfs share.
Then 9xJB % s share.
And l5z-\-5x=C's share.
Also 35#-j =Z)'s share.
Sum 70*4^=315
280oH-35#=315X4
315^=315X4
x=4 Hence 6a?=24, *#'s sh.
549. A gamester at play staked ^ of his money, which he lost,
but afterwards won 4 shillings ; he then lost ? of what he had,
and afterwards won 3 shillings ; after this he lost ^ of what he
had, and finding that he had but 20 shillings remaining, he left
off playing. How much had he at first ? rfns. 30 shillings.
30. A gentleman spends of his yearly income for the sup-
port of his family, and J of the remainder for improving his
house and grounds, and lays by $70 a year. What is his in-
come? Jlns. 9X70 dollars, or more generally, 9 times the
sum he saves.
31. Divide the number 60 (a) into two such parts that their
product may be equal to three times the square of the less num-
ber ? Ans. 15 and 45, or |a= the less part.
32. After paying away ? and \ of my money, I had 34 (a)
dollars left. What had I at first ?
rfns. 56 dollars. General answer y^X28.
33. My horse and saddle are together worth 90 (a) dollars,
and my horse is worth 8 times my saddle. What is the value
of each ? rfns. Saddle $10. Horse $80.
34. My horse and saddle are together worth a dollars, and
my horse is worth n times my saddle. What is the value of
each ? Jlns. Saddle r Horse
EQUATIONS. 69
35. The rent of an estate is 8 per cent greater this year than
last. This year it is 1890 dollars. What was it last year?
Ans. $1750.
36 The rent of an estate is n per cent, greater this year than
last. This year it is a dollars. What was it last year 1
100
dollare '
37. A and B have the same income. A contracts an annual
debt amounting to 1 of it ; B lives upon | of it ; at the end of
two years B lends to A enough to pay off his debts, and has 32
(a) dollars to spare. What is the income of each ?
Ans. $280 or (35a).
38. What number is that of which |, | and f added together
make 73 (a) ? . a 84a
v ' Am. 84. General Ans. .
73
39. A person after spending 100 dollars more than I of his
income, had remaining 35 dollars more than 5 of it. Required
his income. Ans. $450.
40. A person after spending (a) dollars more than \ of his
income, had remaining (6) dollars more than of it. Required
his income.
41. There are two numbers in proportion of 2 to 3, and if 4
be added to each of them, the sums will be in proportion of 5
to 7? Ans. 16 and 24.
42. It is required to find a number such, that if it be increased
by 7, the square root of the sum shall be equal to the square
root of the number itself, and 1 more. Ans. 9.
43. A sets out from a certain place, and travels at the rate of
7 miles in 5 hours ; and 8 hours afterward B sets out from the
same place in pursuit, at the rate of 5 miles in 3 hours. How
long and how far must B travel before he overtakes A ?
Ans. 42 hours, and 70 raijes
70 ELEMENTS OF AUJEURA.
SIMPLE EQUATIONS.
CHAPTER II.
(Art. 45.) We have given a sufficient number of examplos t
and introduced the reader sufficiently far into the science pre-
vious to giving instructions for the solution of questions contain-
ing two or more unknown quantities.
There are many simple problems which one may meet with
in algebra which cannot be solved by the use of a single un-
known quantity, and there are also some which may be sohid
by a single unknown letter, that may become much more simple
by using two or more unknown quantities.
When two unknown quantities are used, two independent
equations must exist, in which the value of the unknown letters
must be the same in each. When three unknown quantities are
used, there must exist three independent equations, in which the
value of any one of the unknown letters is the same in each.
In short, there must be as many independent equations as
unknown quantities used in the question.
For more definite illustration let us suppose the following
question :
ji merchant sends me a bill of 16 dollars for 3 pair of shoes
and 2 pair of boots ; afterwards he sends another bill of 23
dollars for 4 pair of shoes and 3 pair of boots, charging at the
same rate. Wliat was his price for a pair of shoes, and what
for a pair of boots ?
This can be resolved by one unknown quantity, but it is far
more simple to use two. ,
Let x= the price of a pair of shoes,
And y= the price of a pair of boots.
Then by the question 3a?-|-2y=16
And 4*-}-3r/=23.
These two equations are independent ; that is, one cannot be
converted into the other by multiplication or division, notwith-
standing the value of a: and of y is the same in both equations.
Having intimated that this problem can be resolved with one
EQUATIONS. 7!
unknown quantity, we will explain in what manner, before we
proceed to a general solution of equations containing two un-
known quantities.
Let #= the price of a pair of shoes.
Then 3x= the price of three pair of shoes.
And 16 3x the price of two pair of boots.
I / O/y
Consequently - = the price of one pair of boots.
Now 4 pair of shoes which cost 4#, and 3 pair of boots which
cost - being added together, must equal 23 dollars.
That is, 4x4-24^=23.
Or, 1 #=0. Therefore a? 2 dollars, the price
of a pair of shoes. Substitute the value of x in the expression
1 A Q'v*
- and we find 5 dollars for the price of a pair of boots.
2
Now let us resume the equations,
=23 (B)
FIRST METHOD OF ELIMINATION.
(Art. 46.) Transpose the terms containing y to the right hand
sides of the equations, and divide by the coefficients of x, and
From equation (rf) we have x= - - (C)
o
And from (B) we have x=- ~^-^- (B)
4
Put the two expressions for x equal to each other. (Ax. 7.)
And == - .
An equation which readily gives y=5, which, taken as the
value of y, in either equation (C) or (Z>) will give #=2.
This method of elimination, just explained, is called th >
method by comparison.
72 ELEMENTS OF ALGEBRA.
SECOND METHOD OF ELIMINATION.
(Art. 47.) To explain another method of solution, let us again
resume the equations :
3x-\-2y=lG (A)
4#-f3i/=23 (if)
The value of x from equation (#) is x=i(lQ 2y).
Substitute this value for x in equation (2?), and we have
4X (16 2y)+3y^=23, an equation containing only y.
Reducing it, we find y=5 the same as before.
This method of elimination is called the method by substitu-
tion, and consists in finding the value of one unknown quantity
from one equation to put that value in the other which will cause
one unknown quantity to disappear.
THIRD METHOD OF ELIMINATION.
(Art. 48.) Resume again Sx-\-2y=l6 (.#)
4x+3y=23 ()
When the coefficients of either x or y are the same in both
equations, and the signs alike, that term will disappear by sub-
traction.
When the signs are unlike, and the coefficients equal, the term
will disappear by addition.
To make the coefficients of x equal, multiply each equation
by the coefficient of x in the other.
To make the coefficients of y equal, multiply each equation
ly the coefficient of y in the other.
Multiply equation (A) by 4 and l2x-\-8y=Q4
Multiply equation (E) by 3 and 12#-|-9i/=69
Difference y = 5 as before.
To continue this investigation, let us take the equations
2x+3y=23 (A)
5x 2y=iO (JB)
Multiply equation (A) by 2, and equation (B) by 3, and we
have 4x-{-6y=46
EQUATIONS, 73
Equations in which the coefficients of y are equal, and the
signs unlike. In this case add, and the y's will destroy each
other, giving 19#-=76
Or a?=4.
This method of elimination is called the method by addition
and subtraction.
FOURTH METHOD OF ELIMINATION.
(Art. 48.) Take the equations 2x-\-3y=23. (.#)
And 5x 2t/=10. ()
Multiply one of the equations, for example (,/#), by some inde-
terminate quantity, say m.
Then 2mx-\-3my=23m
Subtract () 5x 2y=lO
Remainder, (C) (2m 5)a?+ (3m+2)y=23m 10
As m is an indeterminate quantity, we can assume it of any
value to suit our pleasure, and whatever the assumption may be,
the equation is still true.
Let us assume it of such a value as shall make the coefficient
of y, (3m-l-2)=0.
The whole term will then be times ?/, which is 0, and equa-
tion (C) becomes
(2m -5)#=23m 10
23m 10 , nx
f *=-2^f W
But 3w-f2=0. Therefore i= |.
Which substitute for m in equation (/)), and we have
__23 X f 10 _ 23 X 2^30_ 76^ i
x ~ 2Xf -5 = ~ 2X2 15~ -19~~
This is a French method, introduced by Bezout, but it is too
indirect and metaphysical to be much practised, or in fact much
known.
Of the other three methods, sometimes one is preferable and
sometimes another, according to the relation of the coefficients
and the positions in which they stand.
7
74 ELEMENTS OF ALGEBRA.
No one should be prejudiced against either method, and in
practice we use either one, or modifications of them, as the case
may require. The forms may be disregarded when the princi-
ples are kept in view.
(Art. 49.) To present these different forms in the most general
manner, let us take the following general equations, as all par-
ticular equations can be reduced to these forms.
ax-\-by =c (J3)
Observe that a and a', may represent very different quantities,
BO b and b' may be different, also c and c' may be different. In
special problems, however, a may be equal to a', or be some
multiple of it ; and the same remark may apply to the other
letters. In such cases the solution of the equations is much
easier than by the definite forms. Hence, in solving definite
problems great attention should be paid to the relative values of
the coefficients.
First method.
Transpose the terms containing y and divide by the coeffi-
cients of a?, and
also x= c ^y (C)
a'
Therefore -=- (Axiom 7.)
a a
Clearing of fractions, give a'c a'by=ac f ab'y.
Transpose, and (ab' a'b)y=ac' a'c.
T j- ac ' a ' c
By division yr, 77-
y ab' ab
When y is determined, its value put in either equation marked
(C) will give x.
Second method.
From equation (.#) ' '
EQUATIONS. 75
Which value of x substitute in equation (B) and
a'c a'by
Clearing of fractions and transposing a'c, we have
ab 'y a'by=ac' a'c
~ ac' a'c
Ot y=^=b
The same value of y as before found.
Third method.
Multiply equation (rf) by ', and equation (B) by a.
And a'ax-\-a'by=a'c.
Also a'ax-}-ab'y=ac'
Difference (ab' ~a'b}y=ac' a'c
_ ac' a'c .
Or y=r, -- TL same va ^ ue as by the
ab' ab
Dther methods.
Fourth method.
Multiply equation (JT) by an indefinite number m,
And amx-}-bmy=mc
Subtract (B] a'x-\- b'y=c'
And (am a']x-}-(bm b']y=mc c'.
Now the value of m may be so assumed as to render the
coefficient of #=0, or am a'=0.
Then (bm b')y=mcc'
But am a'=0, or m= .
a
76 ELEMENTS OF ALGEBRA.
Put this resultant value of m, in equation ((7), and
cX
ca' ac'
bX--b>
a
by multiplying both numerator and denominator by a.
(Art. 50.) The principles just explained for elimination be-
tween two quantities may be extended to any number, where the
number of independent equations given are equal to the number
of unknown quantities. For instance, suppose we have the
three independent equations :
ax+by+cz=d (A)
a'x+b'y+c'z=d' (B)
a"x+b"y-}-c"z=d" (C)
We can eliminate either a?, or y, or z, (whichever may be
most convenient in any definite problem) between equations (A]
and (J5,) and we shall have a new equation containing only two
unknown quantities. We can then eliminate the same letter be-
tween equations (B} and ((7,) or (A] and (C*,) and have another
equation containing the same two unknown quantities.
Then we shall have two independent equations, containing two
unknown quantities, which can be resolved by either of the four
methods already explained.
(Art. 51.) Another theoretical method, from the French, we
present to the reader, more for curiosity than for any thing else.
Multiply the first equation (./?,) by an indefinite assumed num-
ber m. Multiply the second equation (B] by another indefinite
number n, and add their products together. Their sum will be
(am-}-a'n)x-\-(bm-\-b'n)y-i-(cm-{-c'n}z=dm-i-nd'
Subtract eq. (C) a"x +b"y c"z=d".
And (am-\-a'n a"}x-\-(bm-\-b'n &")7/-j-(cra4-c'n c"}z
=dm-\-nd' d"
EQUATIONS. 77
As m and n are independent and arbitrary numbers, they can
be so assumed that
am-\-a'n "=0 and bm+b'n &"=0.
Then am-\-a'n=a" (1) and bm-{-b'n=b" (2)
And z= - : -- - (3)
cm+c'n c"
From equations (1) and (2) we can find the values of m and
, which values may be substituted in equation (3,) and then z
will be fully determined.
EXAMPLES FOR PRACTICE.
1. Given \ , > to find the values of x and y.
? 12#+7=1005
We can resolve this problem by either one of the four methods
just explained. But we would not restrict the pupil to the very
letter of the rule, for that in many cases might lead to operations
unnecessarily lengthy.
If we take the third method of elimination, we should multi-
ply the first equation by 12, the second by 8 ; but as the coeffi-
cients of x contain the common factor 4, we can multiply by 3
and 2, in place of 12 and 8. That is, multiply by the fourth
part of 12 and 8.
In practice even this form need not be observed, we may de-
cide on our multipliers by inspection only.
Three times the 1st gives 24#-f-15?/=204
Twice the 2d gives 24a?-J-14i/=200
Difference gives y4
Substituting this value of y in first equation, and
8#-f-20=68 or x=Q.
In solving this, we have used modifications of the 3d and 2d
formal methods.
78 ELEMENTS OF ALGEBRA.
For exercise, let us use the 4th method.
8 mx -|- 5my = 68m
Take l2x-\- 7y = 100
7)y=68m -100
Assume 8m 12=0.
68m 100
Then =-- But m== -
,, P 68X1100 204200
Therefore =-__ ==
2. Given \ ^J= J 9 J to find * and y.
If we multiply the first of these equations by 3, the coefficients
of y will be equal, and the equations become
15a?-r-6y=57,
And 7a? 6?/=9.
To eliminate y, we add these equations (the signs of the terms
containing y being unlike), and there results
This value of x put in the 1st equation gives
And 2/= 2 -
3. Given ?-t^-J-6?/=21 and 01^4-5^=23 to find x and y
Clear of fractions and reduce. We then have
#-{-24^=76
And 15ff-f y =63.
In this case there are no abbreviations of the rules, as the
coefficients of the unknown terms are prime to each other.
Continuing the operation, we find a?=4, y=3.
EQUATIONS. 79
2^? 3t/
4. Given #-{-?/= 17 and ~~T to & n & x an( ^ y
Owing to the peculiarity of form in the 2d equation, it is most
expedient to resolve this by the 2d method.
From the 2d, x=^. Then ^-f y=17.
O o
Clearing of fractions, 9*/+8y = 17 X 8.
Or, 17y=17X8, or y=8.
Hence, a?=9.
C ^a:+8i/=1947
5. Given < J isi S to values of a? and y.
Here we observe that both x and y are divided by 8, x in one
equation, and y in the other ; also, x and y are both multiplied
by 8.
(Art. 51.) All such circumstances enable us to resort to many
pleasant expedients which go far to teach the true spirit of al-
gebra.
Add these two equations, and ^-f-8(#-}-y)=325.
8
Assume x-\-ys.
Or let s represent the sum of x-\-y, then js-f-8s=325.
Clear of fractions, and s-{- 64s =325X8.
Unite and divide by 65 and 5=5X8.
Or x-{-y=5a. (A.) By returning to the value of s, and put-
ting a=8.
Multiply the 1st equation by 8, and
Subtract (rf) x + i/=5a
Rem.
Divide by 63 and y3a=24. Whence #=2a=16.
Let the pupil take any one of the formal rules for the solution
of the preceding equations, and mark the difference.
6. Given $x-{-3y=2} and ^y-J-8.r=29 to find x and y.
Jins. o:=9. =6
80 ELEMENTS OF ALGEBRA.
7. Given 4#-f-t/=34 and 4y-\-x=l6, to find ;raiid^.
Ans. x=S. y=2
8. Given 5#-{-5?/=14 and 5a;+5y=ll, to find x and y.
?. #=24. v=6
9. Giver. #-Hi/=8 and zX-}-y=7 to find a 1 and y.
Ans. z=Q. 2/=4.
10. Given \x+7y=99 and iy-}-7,r=51 to find x and y.
flns. x=7. =14
11. Given
4
fins. x=6Q. i/=40
12. Given -+-=6 and | =10, to find x and y.
x y x y
Multiply the first equation by c, the second by a, and we shall
have
etc , be
] --- =6c
x y
ac . ad
-- --- =10tf.
By subtraction (be ad)-Qc 10a
f be ad
Therefore =v.
6c 10a y
13. Given =28 (Jft and + = (B) to find
x y x y 3 ^ '
tlie values of x and y.
21 21
Divide equation (#) by 7, and =4.
x y
Divide this result by 21, and =-- (C\
x y 21 ^ J
EQUATIONS. 81
Multiply (C) by 17, gives ^ H =5? (/>)
M>Q 210
Subtract () from () and we have = -.
1 3 1
Divide by 73, and -= =- or y=7.
Putting this value in equation (C) and reducing we find x=3.
14. Given -+-=- 1 and -+-=-+- to find the values
x^y y x^y 0^2
of * and y Jns. *=4 and y=2.
?. #=300. 7/=350.
$3? 24v 2 -H
. Given 3x+Gy+l= ^ ^ ,
to find a: and y.
151 1 6.r , 9^/110
And 3o?= -\ H
4y 1 3y 4
.tfns. a?=9. y=2.
In the first equation actually divide the numerator by the de-
nominator, then drop equals from both sides.
17. Given ^ ' ~j!^"~L-oo C t0 find the values of x and ^*
^ns. a:=2. i/=5
18. Given < ^ . I to find x and y.
19. Given x-}-y =8 and a? 2 y 2 =16 to find a: and t/.
20. Given 4(#-f-2/)=9(a; y) and a? 2 y 2 =36 to find * and y.
82 ELEMENTS OF ALGEBRA.
21. Given x ly :: 4 : 3 and a? ^2/ 3 =37, to find # and y.
Ans. #=4. y=3.
22. Given #-fr/=a and x* y z =ab to find x and y.
a-\-b a />
Ans. a?= . 2/ = -2~
23. Given -- = and - -j =| to find a; and y.
Jlns. #=4. lo
24. Given $(a:+2)+ 8y=31 and 4(y+5)-}-10^=192 to
find the values of a; and y. .tfns. x=19. y=3
25. Given 3a?+7y=79 and 2t/-j-5iC=19 to find the values
of x and y. Jlns. #=10. y=7.
26. Given |(#-f t/)+25=a? and %(x-}-y) 5=y to find
the values of x and y. Jlns. #=85. y=35.
27. Given x 4=y+l and 5#-^=^ ^+37 to find the
o 4 6
values of x and y. Ans. x=S. y3
28. Given 4 ^?=T/ 17| and ^=f+2 to find the
D O O
values of a? and y. jJns. x=W. y=2Q.
7x 21 . x . 3x 19 2a?+v
29. Given - }-y_=4-{ ^ and -f-
9# 7 3y4-9 4#+5y
-- =-*-T --- rtr 3 to find a? and y. Ans. #=9. y=4.
3O. Given
23 # 2
73 3y
--- 2
to find x and
3
Jlns. #=21.
EQUATIONS. 83
CHAPTER HI.
Solution of Equations involving three or more unknown
quantities.
(Art. 52.) No additional principles are requisite to those given
in articles 49 and 50.
EXAMPLES.
f X + y+ z = 9 ]
1. Given \ #+2?/+3z 16 [ to find a?, y, and z.
[ a?+3y+4*=21 J
By the 1st method, transpose the terms containing y and z in
each equation, and
x= 9 T z,
Then putting the 1st and 2d values equal, and the 2d and 3d
values equal, gives
9 y z=16 2y 3z,
1 6 2y 3z =2 1 3y 40.
Transposing and condensing terms, and
y=72z,
Also, 2/ = 5 z,
Hence, 5 z=7 2z, or z=2
Having z=2, we have 2/=5 ^z=3, and having the values of
both z and y, by the first equation we find x=4.
f 2x-\-4y 3z=22 1
a. Given < 4x2y-}-5z=l8 [to find values of a?, y and z
[ Gx-}-7y z=63 J
Multiplying the first equation by 2, 4x-}-Sy 6z~44
And subtracting the second, 4x 2y-\- 5z=18
The result is, (rf l( llz=26
54 ELEMENTS OF ALGEBRA.
Then multiply the first equation by 3, Gx-{-l2y 9z=GG
And subtract the third, 6#-|- 7y z =63
The result is, (B) 5y Sz= 3
Multiply the new equation (B) by 2, Wy 16z= 6
And subtract this from equation (.#) IQy llz=26
The result is, 5z=20
Therefore z= 4
Substituting the value of z in equation (E] and we find y=7.
Substituting these values in the first equation, and we find x=3.
( 3x+9y-\-8z=4l 1
3. Given \ 5x-\-4y 2z=2Q \ to find a?, y and z.
[ llx+ty 6^=37 J
To illustrate by a practical example we shall resolve this by
the principles explained in (Art. 51.)
3mx-\-9my -\-Smz = 41m
5nx -\-4ny 2nz =2Qn
Sum (3ra+5n)#+(9m-f4n)?/-f-(8w 2n)z=41m+20w
Take llx +7 6z=37
Rem. (3m4-5n 11) a; (7 9m 4n)y+(8m 27i+6)z=
Assume 3m-}-5n=ll (1)
And 9m-f4n=7 (2)
From equations (1) and (2) we find wi= T 3 T and w=f i
These values substituted in equation (3) we have
_ 41 x T 3 T 4-2QX f f -37
~ ~8X T 3 T 2Xff+ 6
Multiply both numerator and denominator by 11, and we shall
123+520407 10
have z=-
24 52-r- 66 10
EQUATIONS. 85
Putting this value of z in the 1st and 2d equations, we shall
have only two equations involving x and y, from which the
values of these letters may be determined.
These equations can be resolved with much more facility by
multiplying the 2d equation by 4, then adding it to the 1st to
destroy the terms containing z.
Afterwards multiplying the 2d equation by 3, and subtracting
the 3d equation, and there will arise two equations containing x
and ?/, which may be resolved by one of the methods already
explained.
(Art. 53.) When three, four, or more unknown quantities with
as many equations are given, and their coefficients are all prime
to each other, the operation is necessarily long. But when sev-
eral of the coefficients are multiples, or measures of each other,
or unity, several expedients may be resorted to for the purpose
of facilitating calculation.
No specific rules can be given for mere expedients. Exam-
ples alone can illustrate, but even examples will be fruitless to
one who neglects general principles and definite theories. Some
few expedients will be illustrated by the following
EXAMPLES.
1. Given x+y z=25 to find x, y, and z.
[ x yz= 9 J
Subtract the 2d from the 1st, and 2z=6.
Subtract the 3d from the 2d, and 2?/=16.
Add the 1st and 3d, and 2#=40.
(
2. Given I x y 4 to find x, y and z.
I*** = 6 j
Add all three, and 3#=3 6 or #=12.
{ xy-z= 6 I
3. Given < 3y x z=12 to find a?, y and z
86 ELEMENTS OF ALGEBRA.
Assume x-\-y-\-z=s. Add this equation to each of the given
equations, and we then have
2#= 6+5, (A)
4</=12+5, (B)
8z=24+5. (C)
Multiply (.#) by 4, and (B) by 2, and we have
8#=24+4s,
82=24+ s.
By addition, 85=3 X 24-}- -7s. Or 5=72.
Put this value of s in equation (.#) and we shall have
22=6+72. Or #=3+36=39, &c.
4. Given #+T/=100, 3/+|z=100, z+4#=100 to find
ar, t/, and z.
Put =100. Ans. #=64, y=72, and z=84.
r = 10
5 Given
u-\-v-\-y+z=l2
to find the value of each.
v+#+t/+2'=14
Here are Jive letters and five equations. Each letter has the
same coefficient, one understood. Each equation has 4 letters,
z is wanting in the 1st equation, y in the 2d, <fec.
Now assume w+v+#+?/+;z=5.
Then 5 z=10 (\
s #=12
s v=13
s w=14
Add, and 5s -5=60 Or 5=15.
Put this value of s in equation (.#), and z=5, &c.
6. Given #+y=a, #+z=&, y+z=c.
Add the 1st and 2d, and from the sum subtract the 3d.
Jlns. #=
EQUATIONS
8?
Given
8. Given
O. Given
1O. Given -
2ati+yf* I
3y=u-}-x-\-z I to find the value of w, x, y,
4z=u+x+y [ and z.
u=x 14 J
j. w=26, #=40, y=30, z=24.
2 f #=24.
7 rfns. J y=60.
3 z=120-
f ar=i l T-
\ 2/ = T 5 T
I ^-T 7 i
x=20
x-\-a=
y 2z=40
4y x-\-3z=35
t=l3
u 49
(Art. 54.) Problems producing simple equations involving two
or more unknown quantities.
1. Find three numbers such, that the product of the 1st and
2d, shall be 600 ; the product of the 1st and 3d, 300 ; and the
product of the 2d and 3d, 200.
rfns. The numbers are, 30, 20, and 10.
2. Find three numbers, such that the^rstf with 5 the sum of
the second and third shall be 120 ; the second with | the differ-
ence of the third and first shall be 70 ; and the sum of the three
numbers shall be 190. Ans. 50 ; 65 ; 75.
3. A certain sum of money was to be divided among three
persons, .#, B, and C, so that #'s share exceeded j of the shares
of B and C by $120 ; also the share of B exceeded f of the
shares of A and C by $120; and the share of C, likewise, ex-
ceeded f of the shares of *# and B by $120. What was each
person's share?
tins. JT* share, $600 ; 's, 480 ; and C"s 360.
88 ELEMENTS OF ALGEBRA.
4. Jl and B, working at a job, can earn $40 in 6 days ; A and
C together can earn $54 in 9 days; and B and C$80 in 15
days. What can each person alone earn in a day?
Let A earn x, B earn y, and C earn z dollars per day, then,
By the question, 6x-J- 6y=40
9.2=54
Dividing the equations by the coefficients of the unknown
quantities, we have, , _ AO
-
See Problem 6. (Art. 53.)
A man has 4 sons. The sum of the ages of the first, second
and third is 18 years ; the sum of the ages of the first, second
and fourth is 16 years ; the sum of the ages of the first, third
and fourth is 14 years ; the sum of the ages of the second, third
and fourth is 12 years. What are their ages? See Problem 5.
(Art. 53.) rfns. Their ages are, 8, 6, 4, 2.
5. .#, B and C sat down to play, each one with a certain num-
ber of shillings ; Jl loses to B and C as many shillings as each
of them has. Next B loses to Jl and C as many as each of
them now has. Lastly, C loses to Jl and B as many as each of
them now has. After all, each one of them has 16 shillings.
How much did each one gain or lose ?
Let x= the number of shillings Jl had at first
y B's shillings, and
z C's shillings.
Then, by resolving the problem, we shall find #=26, y=14
and z=S. Therefore, Jl lost 10 shillings, B gained 2, and (78.
N. B. When the equations are found, divide the 1st by 4, the
2d by 2, and then compare them with Ex. 3. (Art. 53.)
6. A gentleman left a sum of money to be divided among four
servants, so that the share of the first was the sum of the
shares of the other three ; the share of the second, | of the sum
of the other three ; and the share of the third, k the sum of the
EQUATIONS. 89
other three ; and it was found that the share of the last was 14
dollars less than that of the first. What was the amount of
money divided, and the shares of each respectively 1
Ans. The sum was $120 ; the shares 40, 30, 24 and 26.
Observe Prob. 7. (Art. 53,) in connection with this problem.
7. A jockey has two horses, and two saddles which are worth
15 and 10 dollars, respectively. Now if the better saddle be put
on the better horse, the value of the better horse and saddle
would be worth of the other horse and saddle. But if the
better saddle be put on the poorer horse, and the poorer saddle
on the better horse, the value of the better horse and saddle is
worth once and f^ the value of the other. Required the worth
of each horse ? Ans. 65 and 50 dollars.
8. A merchant finds that if he mixes sherry and brandy in
quantities which are in proportion of 2 to 1, he can sell the mix-
ture at 78 shillings per dozen; but if the proportion be 7 to 2 he
can sell it at 79 shillings per dozen. Required the price per
dozen of the sherry and of the brandy ?
Ans. Sherry, 81s. Brandy, 725.
In the solution of this question, put a=78. Then a-f-l=79.
9. Two persons, A and J9, can perform a piece of work in
16 days. They work together for four days, when A being
called off, B is left to finish it, which he does in 36 days. In
what time would each do it separately 7
Ans. A in 24 days ; B in 48 days.
1O What fraction is that, whose numerator being doubled,
and denominator increased by 7, the value becomes f ; but the
denominator being doubled, and the numerator increased by 2,
the value becomes |? Ans. .
11. Two men wishing to purchase a house together, valued
at 240 (a) dollars ; says A to J3, if you will lend me of your
money I can purchase the house alone ; but says B to A, if you
lend me J of yours, I can purchase the house. How much
money had each of them ? Ans. A had $160. B $120.
8
90 ELEMENTS OF ALGEBRA.
12. It is required to divide the number 24 into two such parts,
that the quotient of the greater part divided by the less, may be
to the quotient of the less part divided by the greater, as 4 to 1
rfns. 16 and 8.
13. A certain company at a tavern, when they came to settle
their reckoning, found that had there been 4 more in company,
they might have paid a shilling a-piece less than they did; but
that if there had been 3 fewer in company, they must -have paid
a shilling a-piece more than they did. What then was the num-
ber of persons in company, and what did each pay ?
rfns. 24 persons, each paid 7s.
14. There is a certain number consisting of two places, a unit
and a ten, which is four times the sum of its digits, and if 27 be
added to it, the digits will be inverted. What is the number?
Ans. 36.
NOTE. Undoubtedly the reader has learned in arithmetic that
numerals have a specific and a local value, and every remove
from the unit multiplies by 10. Hence, if x represents a digit
in the place of tens, and y in the place of units, the number must
be expressed by lQx-\-y. A number consisting of three places,
with x, y and z to represent the digits, must be expresset^ by
15. A number is expressed by three figures ; the sum of these
figures is 1 1 ; the figure in the place of units is double that in
the place of hundreds, and when 297 is added to this number,
the sum obtained is expressed by the figures of this number re-
versed. What is the number ? Jlns. 326.
10. Divide the number 90 into three parts, so that twice
the first part increased by 40, three times the second part in-
creased by 20, and four times the third part increased by 10, may
be all equal to one another.
. *ftns. First part 35, second 30, and third 25.
17. A person who possessed $100,000 (,) placed the greater
part of it out at 5 per cent, interest, and the other part at 4 per
EQUATIONS. 91
cent. The interest which lie received for the whole amounted
to 4640 (b~) dollars. Required the two parts.
Am. 64,000 and 36,000 dollars.
General Ansiver. (1006 4) for the greater part, and (5
1006) for the less.
18. A person put out a certain sum of money at interest at a
certain rate. Another person put out $10,000 more, at a rate 1
per cent, higher, and received an income of $800 more. A third
person put out $15,000 more than the first, at a rate 2 per cent,
higher, and received an income greater hy $1,500. Required
the several sums, and their respective rates of interest.
Jlns. Rates 4, 5 and 6 per cent. Capitals $30,000, $40,000
and $45,000.
19. A widow possessed 13,000 dollars, which she divided
into two parts and placed them at interest, in such a mannei, that
the incomes from them were equal. If she had put out the first
portion at the same rate as the second, she would have drawn
for this part 360 dollars interest, and if she had placed the se-
cond out at the same rate as the first, she would have drawn for it
490 dollars interest. What were the two rates of interest ?
Ans. 7 and 6 per cent.*
20. There are three persons, .#, B and C, whose ages are as
follows : if J5's age be subtracted from ./2's, the difference will
be C"s age ; if five times J5's age and twice C"s age, be added
together, and from their sum JFs age be subtracted, the remain-
der will be 147. The sum of all their ages is 96. What are
their ages ? Jlns. 's 48, J5's 33, <7's 15.
21. Find what each of three persons, ./?, B and (7, is worth,
from knowing, 1st, that what Ji is worth added to 3 times what
B and C are worth, make 4700 dollars ; 2d, that what B is worth
added to four times what A and C are worth make 5800 dollars;
3d, that what C is worth added to five times what Ji and B are
worth make 6300 dollars.
tins. A is worth 500, B 600, C 800 dollars.
tt*;e brief solution to these two problems, 18 and 19, in Key to Algebra.
92
ELEMENTS OF ALGEBRA.
Put 5= the sum that v?, B and C are worth, to make an aux-
iliary equation.
22. Find what each of three persons, /?, B, C, is worth,
knowing, 1st, that what Jl is worth added to / times whauE? and
C are worth is equal to p; 2d, that what B is worth added to m
times what Ji and C are worth is equal to q; 3d, that what C is
worth added to n times what Ji and B are worth is equal to r.
Let #=r#'s capital, y=B <> 8 1 and 2 = C"s.
Then x-^ly-^-lzp^ is the first equation.
Assume x-{- y+ *=* Multiply this equation by / and
subtract the former, and (/ \}x=ls p.
Is p
r> -i * ms *7
By a similar operation, y= J
And,
m 1
ns r
(A)
By addiuon,
ns r
This equation may take the following form :
=(- 1
+-^+-
Now as the terms in parenthesis are fully determined, of
known value, we may represent the first by a, the second by b t
and this last form becomes
s=5 b
By transposition, &c. (a l)s=6
b
Therefore
*=
a 1
This known value of * put in each of the equations marked
(A), and the values of x, y and z will be theoretically deter-
mined.
EQUATIONS. 93
23. Three brothers made a purchase of $2000 (a;) the first
wanted in addition to his own money the money of the second,
the second wanted in addition to his own 3 of the money of the
third, and the third required in addition to his own of tho
money of the first. How much money had each ?
Jlns. 1st had, $1280; 2d, $1440; and the 3d, $1680.
Gen. Jlns. 1st had if a; 2d, if a; and the 3d \ \a.
See Prob. 6. (Art. 53.)
24. Some hours after a courier had been sent from A to B,
which are 147 miles distant, a second was sent, who wished to
overtake him just as he entered B, and to accomplish this he
must perform the journey in 28 hours less time than the first did
Now the time that the first travels 17 miles added to the time the
second travels 56 miles is 13f hours. How many miles does
each go per hour ? Jlns. 1st 3, the 2d, 7 miles per hour.
25. There are two numbers, such that the greater added to
I the lesser, is 13 ; and if the lesser is taken from 5 the greater,
the remainder is nothing. Required the numbers.
rfns. 18 and 12.
26. Find three numbers of such magnitude, that the 1st with
the 5 sum of the other two, the second with of the other two,
and the third with * of the other two, may be the same, and
amount to 51 in each case. J%ns. 15, 33, and 39.
27. Jl said to B and C t " Give me, each of you, 4 of your
sheep, and I shall have 4 more than you will have left." B said
to A and C, " If each of you will give me 4 of your sheep, I
shall have twice as many as you will have left." C then said
to A and B, " Each of you give me 4 of your sheep, and I shall
have three times as many as you will have left." How many
had each ? rfns. Ji 6, B 8, and C 10.
28. What fraction is that, to the numerator of wnich if 1 be
added, the fraction will be j : but if 1 be added to the denomina-
tor, the fraction will be ? 1 dns. T 4 T .
29. What fraction is that, to the numerator of which if 2 be
U4 ELEMENTS OF ALGEBRA.
added, the fraction will be f ; but if 2 be added to the denomina-
tor, the fraction will be } ? Ans. f .
SO. \Vhat fraction is that whose numerator being doubled, and
its denominator increased by 7, the value becomes f ; but the do-
nominator being doubled, and the numerator increased by 2, the
value becomes | ? Ans. J.
31. If A give B $5 of his money, B will have twice as much
money as A has left; and if B give A $5, A will have thrice as
much as B has left. How much had each ?
Ans. .tf $13, and .#$11.
32. A corn factor mixes wheat flour, which cost him 10 shil-
lings per bushel, with barley flour, which cost 4 shillings per
bushel, in such proportion as to gain 43f per cent, by selling the
mixture at 11 shillings per bushel. Required the proportion.
Am. The proportion is 14 bushels of wheat flour to 9 of
barley.
33. There is a number consisting of two digits, which num-
ber divided by 5 gives a certain quotient and a remainder of one,
and the same number divided by 8 gives another quotient and a
remainder of one.. Now the quotient obtained by dividing by 5
is double of the value of the digit in the ten's place, and the quo-
tient obtained by dividing by 8 is equal to 5 times the unit digit.
What is the number ? Ans. 41.
Interpretation of negative values resulting from the solution
of equations.
(Art. 55.) The resolution of proper equations drawn from
problems not only reveal, the numeral result, but improper enun-
ciation by the change of signs. Or the signs being true algebraic
language, they will point out errors in relation to terms in com-
mon language, as the following examples will illustrate :
1. The sum of two numbers is 120, and their difference is
100 ; what are the numbers ?
Let x be the greater and y the less. Then
a?+y=120 (1)
a^=160 (2)
The solution gives #=140, and ?/ 20.
EQUATIONS. 95
Here it appears that one of the numbers is greater than the
sum given in the enunciation, yet the sum of x and y, in the al-
gebraic sense, is 120.
There is no such abstract number as 20, and when minus
appears it is only relative or opposite in direction or condition
to plus, and the problem is susceptible of interpretation in an al-
gebraic sense, but not in a definite arithmetical sense.
Indeed we might have determined this at once by a considera-
tion of the problem, for the difference of the two numbers is
given, greater than their sum. But we can form a problem, an
algebraic (not an abstract) problem that will exactly correspond
with these conditions, thus :
The joint property of two men amounts to 120 dollars, and
one of them is worth 160 dollars more than the other. What
amount of property does each possess ?
The answer must be -f-140 and 20 dollars ; but there is no
such thing as minus $20 in the abstract ; it must be interpreted
debt, an opposite term to positive money in hand.
2. Two men, A and B, commenced trade at the same time ;
Ji had 3 times as much money as B, and continuing in trade, Jl
gains 400 dollars, and J9 150 dollars; now A has twice as much
money as B. How much did each have at first?
Without any special consideration of the question, it implies
that both had money, and asks how much. But on resolving the
question with x to represent ^?'s money, and y B's, we find
#=300
And y= 100 dollars.
That is, they had no money, and the minus sign in this case
indicates debt; and the solution not only reveals the numerical
values, but the true conditions of the problem, and points out the
necessary corrections of language to correspond to an arithmeti-
cal sense, thus :
A is three times as much in debt as B ; but A gains 400 dol-
lars, and B 150 ; now A has twice as much money as B. How
much were each in debt ?
As the enunciation of this problem corresponds with the real
96 ELEMENTS OF ALGEBRA.
circumstance of the case, we can resolve the problem without a
minus sign in the result. Thus :
Let z= B'B debt, then 3x= J?s debt
150 x J?'s money, 400 3x= .#'s money
Per question, 400 3#-=300 2x. Or a?=100.
3. What number is that whose fourth part exceeds its third
part by 12 ? J2ns. 144.
But there is no such abstract number as 144, and we cannot
interpret this as debt. It points out error or impossibility, and
by returning to the question we perceive that a fourth part of any
number whatever cannot exceed its third part; it must be, its third
part exceeds its fourth part by 12, and this enunciation gives the
positive number, 144. Thus do equations rectify subordinate
errors, and point out special conditions.
4L A man when he was married was 30 years old, and his
wife 15. How mapy years must elapse before his age will be
three times the age of his wife?
Jlns. The question is incorrectly enunciated ; 7 years before
the marriage, not after, their ages bore the specified relation.
5. A man worked 7 days, and had his son with him 3 days ,
and received for wages 22 shillings. He afterwards worked 5
days, and had his son with him one day, and received for wages
18 shillings. What were his daily wages, and the daily wages-
of his son ?
Jfns. The father received 4 shillings per day, and paid 2 shil-
lings for his son's board.
6. A man worked for a person ten days, having his wife with
him 8 days, and his son 6 days, and he received $10.30 as com-
pensation for all three; at another time he wrought 12 days, his
wife 10 days, and son 4 days, and he received $-13.20; at an-
other time he wrought 15 days, his wife 10 days, and his son 12
days, at the same rates as before, and he received $13.85. What
were the daily wages of each ?
rfns. The husband 75 cts., wife 50 cts. The son 20 cts. ex-
pense per day.
EQUATIONS. 97
7. A man wrought 10 days for his neighbor, his wife 4 days,
and son 3 days, and received $11.50 ; at another time he served
9 days, his wife 8 days, and his son 6 days, at the same rates as
before, and received $12.00 ; a third time he served 7 days, his
wife 6 days, and his son 4 days, at the same rates as before, and
he received $9.00. What were the daily wages of each ?
*fins. Husband's wages,$ 1.00; wife ; son 50 cts.
8. What fraction is that which becomes f when one is added
to its numerator, and becomes | when 1 is added to its denomi-
nator ?
Jlns. In an arithmetical sense, there is no such fraction. The
algebraic expression, ~y|, will give the required results.
(Art. 58.) By the aid of algebraical equations, we are enabled
not only to resolve problems and point out defects or errors in
their enunciation, as in the last article, but we are also enabled
to demonstrate theorems, and elucidate many philosophical truths.
The following are examples :
Theorem 1. It is required to demonstrate, that the half sum
plus half the difference of two quantities give the greater of the
two, and the half sum minus the half difference give the less.
Let x= the greater number, y= the less,
s= their sum, d their difference.
Then x+y= s (A]
And xy=d (B)
By addition, 2a?= s-f- d
Or x=*s-}-zd
Subtract (S) from (.#) and divide by 2, and we have
y=%s |c?
These last two equations, which are manifestly true, demon-
strate the theorem.
Theorem 2. Four times the product of any two numbers, is
equal to the square of their sum, diminished by the square of
their difference.
9
98 ELEMENTS OF ALGEBRA
Let x= the greater number, and y= the less, as in the lasi
theorem. 2x=s -\-d
And 2=s d
By multiplication 4xy=s 2 d 2 a demonstration of the
theorem.
Many other theorems are demonstrable by algebra, but we de-
fer them for the present, as some of them involve quadratic equa-
tions, which have not yet been investigated ; and we close the
subject of simple equations by the following quite general prob-
lem in relation to space, time and motion.
To present it at first, in the most simple and practical manner,
let us suppose
Two couriers, A and B, 100 miles asunder on the same road
set out to meet each other, A going 6 miles per hour and B 4.
How many hours must elapse before they meet, and how far
will each travel?
Let x= Jl's distance, y j#'s, and t the time.
Then ^-}-?/=100 ' (1)
As the miles per hour multiplied by the hours must give the
distance each traveled, therefore,
x=6t and y=U (2)
Substitute these values in equation (1) and
(6+4)f= 100
1 rjn
Therefore, *=r ( 3 )
100X6 100X4
And *=8l=- 2/ =4/=
From equation (3,) we learn that the time elapsed before the
couriers met was the whole distance divided by their joint mo-
tion per hour, a result in perfect accordance with reason. From
equations (4,) we perceive that the distance each must travel is
.he whole distance asunder multiplied by their respective mo-
tions and divided by the sum of their hourly motions.
Now let us suppose the couriers start as before, but travel in
the same direction, the, one in pursuit of the other. B having
EQUATIONS. 99
IQO miles the start, traveling four miles per hour, pursued by
A, traveling 6 miles per hour. How many hours must elapse
before they come together, and what distance must each travel?
Take the same notation as before.
Then x y=100 (1.) As A must travel 100 miles more than
B. But equations (2,) that is, x=Qt and y=t, are true under
all circumstances.
Then (64)*= 100
And i
The result in this case is as obvious as an axiom. A has 100
miles to gain, and he gains 2 miles per hour, it will therefore re-
quire 50 hours.
But it is the precise form that we wish to observe. It is the
fact that the given distance divided by the difference of their ma
tions gives the time, and their respective distances must be this
time multiplied by their respective rates of motion.
Now the smaller the difference between their motions, the
longer the time before one overtakes the other ; when the differ
ence is very small, the time will be very great ; when the differ
ence is nothing, the time will be infinitely great ; and this is in
perfect accordance with reason ; for when they travel equally
fast one cannot gain on the other, and they can never come to-
gether.
If the foremost courier travels faster than the other, they must
all the while become more and more asunder ; and if they have
ever been together it was preceding their departure from the
points designated, and in an opposite direction from the one they
are traveling, and would be pointed out by a negative result.
(Art. 59.) Let us now make the problem general.
Two couriers, A and B, d miles asunder on the same road,
set out to meet each other; A going a miles per hour, B going
b miles per hour. How many hours must elapse before they
meet, and how far will each travel?
100 ELEMENTS OF ALGEBRA.
Taking the same notation as in the particular case,
Let x *#'s distance, y= .5's, and /= the time.
Then x-\-y=d (I) x=at ybt (2)
Therefore (a-\-b)t=d Or t==i ~TJ ) ( 3 )
ad bd
If a=b, then x=kd and y=%.d. A result perfectly obvious,
the rates being equal. Each courier must pass over one half
the distance before meeting.
If =0 #=== , . =0 and y=-=d. That is, one
0+6
will be at rest, and the other will pass over the whole distance.
(Art. 60.) Now let us consider the other case, in which one
courier pursues the other, starting at the same time from dif-
ferent points.
Let the line CD represent the space the couriers are asunder
when the pursuit commences, and the point E where they come
together. C D E
j j j
The direction from C towards D we call plus, the other direc-
lion will therefore be minus.
Now as in the 2d example, (Art. 58.)
Put x = CE =^'s distance
y=DE=JS 1 s distance
Then xy=CD=d (1)
As before, let t= the time. Then
Therefore at bt=d
And t=^ (3)
For the distances we have
ad , . t/tt . N
x= (4) and y= , (o)
06 v ' 9 " h
EQUATIONS. 101
By an examination of these equations, it will be perceived fazA
x and y will be equal when a is equal to b, yet d still exists as
a difference between them. This is in consequence of x and y
in that case being so very great that d is lost in comparison. So
all values are great or small only in comparison with others or
with our scale of measure.
To make this clear, let us suppose two numbers differ by one
and if the numbers are small, the difference may be regarded as
considerable; if large, more inconsiderable; if still larger, still
more inconsiderable, &c. If the numbers or quantities be infin-
itely great, the comparative small quantity may be rejected.
Thus :
5 and 6 differ by 1, and their relation is as 1 to 1.2.
Also, 50 and 51 differ by 1, and their relation is as 1 to 1.02.
500 to 501 are as 1 to 1.002, <fcc. The relation becoming nearer
and nearer equality as the numbers become larger, and when the
numbers become infinitely great the difference is comparatively
nothing.
When a=b a ft and x ~~ft a symbol of infinity.
If we suppose b greater than a, a b will become negative, and
as x and y refer to the same point, that point must then be in the
backward direction from that we suppose the couriers are mo-
ving, and will show how far they have traveled since that event.
If in the equations (3), (4) and (5), c?=0, and at the same
time a=bj then we shall have /=-, #=- and 2/=~ ' which
shows that - is a symbol of indetermination, it being equal to
several quantities at the same time. If e?=0 the two couriers
were together at commencement ; and if they travel in the same
direction, and equally fast, they will be together all the while,
and the distances represented by x and y will be equal, and of
all possible values. Hence - may be taken of any value what-
102 CLEMENTS OF ALGEBRA
evsr^ amV-may Ve'-vntidMo take a particular value, to correspond
to any other circumstance or condition.*
APPLICATION.
(Art. 61.) We have hitherto considered CD a right line; but
the equations would be equally true, if we consider CD to be
curved, and indeed we can conceive the line CD to wind about
a perfect circle just forming its circumference, and the point E
upon the circle, CE being a little more than one circumference.
This being understood, Equation (3,) (Art. 60,) gives us a so-
lution to the following problems.
1. The hour and minute hands of a clock are together at 12
o'clock. When are they next together ?
* The 26th equation (Art. 40), if resolved in the briefest manner, will show
the influence of the factor rr. In the equation referred to, add 30 to both
members and divide the numerator of the second member by its denominator,
5^4-5
and we have - -- [-1=6. Drop 1, and divide both members by 5, we then
x4-\
have =1, or x-\-l=x-}-2. Hence 1=2, a manifest absurdity
But all our operations, yea, and all our reasonings have been correct, but
we did not pay sufficient attention to dividing the numerator by the denomi-
nator, which was -7 - . Taking 6 for the quotient, which it would be in
(x t)
every case except when x 2=0, leads to the absurdity ; which absurdity,
in turn, shows that x 2=0, or #=2.
As another illustration of the influence of this symbol, take the identical
equation 100=100, or any other similar one.
This is the same as 96-f-4=96-f-4
Transposing, . 4 4=96 96
Resolving into factors, 1(4 4) =24 (4 4)
Dividing by the common factor, and 1=24 ;
1
But, to restore equality, - in this case must equal , or 24.
Hence we perceive that - is indeterminate, in the abstract, but may be ren-
dered definite in particular cases.
EQUATIONS. 103
By the equation, t j- this problem and all others like it
are already resolved. All we have to do is to determine the
values of </, , and b.
There are 12 spaces (hours) round the dial plate of a clock ;
hence d may represent 12. a and b are the relative motions ol
the hands, a moves 12 spaces or entirely round the dial plate
while b moves one space. Hence a=12, 6=1, and a 6=11.
Consequently, t=\\lh. 5m. 27 T \s.
Again. We may demand what time the hour and minute
hands of a clock are together between 3 and 4.
From 12 o'clock to past 3 o'clock there are 3 revolutions to
q v/ i o
pass over in place of one, and the solution is therefore t
and so on for any other hour.
2. WJiat time between two and three o'clock will the hour
and minute hands of a clock make, right angles with each other?
Here the space that the one courier must gain on the other is
two revolutions and a quarter, or 21 d.
Hence t=\\Wk=\\ X J=H=2h. 27 T 3 T m.
3. What time between 5 and 6 will the two hands of a clock
make a right line ?
Here one courier must gain 5 revolutions, or d in the equa-
tion must be multiplied by 5 = y .
Hence, t=\\ X V =6h.
That is, the hands make a right line at 6 o'clock, a result man-
ifestly tri'3.
This simple equation enables us to determine the exact time
when the two hands of a clock shall be in any given position.
We may apply this equation to a large circle, as well as to a
small one; it may apply to the apparent circular course of the
heavens, as well as to a dial plate of a clock ; and the application
is equally simple.
The circle of the heavens, like all other circles, is divided into
104 ELEMENTS OF ALGEBRA.
360 degrees ; and the sun and moon apparently follow each other
like two couriers round the circle.
In one day the moon moves on an average 13. 1764, (divisions
of the circle,) and the sun apparently 0. 98565, or not quite one
division of the circle. The moon's motion being most rapid.
corresponds to a in the equation, and the sun's apparent motion
to b. Then &=13.1764 0.98565=12.19075 ; and the
time required for one courier to gain on the other the required
space, in this case a revolution of 360 degrees, or / .-=
360 a ~" b
which gives 29.5305887 days, or 29 days, 12 hours,
\Z. 19075
44 minutes, 3 seconds, which is the mean time from one change
of the moon to another, called a synodic revolution.
These relative apparent motions of the sun and moon round
the circular arc of the heavens, are very frequently compared to
the motions of the hour and minute hands of a clock round the
dial plate ; and from the preceding application of the same equa-
tion we see how truly.
We may not only apply this equation to the mean motions of
the sun and moon, but it is equally applicable to the mean mo-
tions of any two planets as seen from the sun. To appearance,
the two planets would be nothing more than two couriers mo-
ving in a circle, the one in pursuit of the other, and the time be-
tween two intervals of coming together, (or coming in conjunc-
tion, as it is commonly expressed,) will be invariably represen-
ted by the equation
To apply this to the motion of two planets, we propose the
following problem :
The planet Venus, as seen from the sun, describes an arc of
1 36'/)er day, and the earth, as seen from the same point, de-
scribes an arc of 59'. Jit what interval of time will these two
bodies come in a line with the sun on the same side ?
Here a=l36' 6=59' rf=360
Therefore, a 6=37'; and as the denominator is
EQUATIONS. 105
minutes, the numerator must be reduced to minutes also ; hence
the equation becomes
360X60 ,
=583.8 days, nearly.
We have not been very minute, as the motions of the planets
are not perfectly uniform, and the actual interval between succes
sive conjunctions is slightly variable. Hence we were not par-
ticular to take the values of a and b to the utmost fraction. A
more rigid result would have been 583.92 days. Half of this
time is the interval that Venus remains a morning and an even-
ing star.
(Art. 63.) This equation, as simple as it may Ep' er.r, is one
practical illustration of the true spirit and utility oi analysis by
algebra.
The principles and relations of time and motion are fixed and
d
invariable, and the equation t=^^ stands conspicuously as
one of these relations.
If t can be determined by observation, as it may be in respect
to the earth and the superior planets, the mean daily motions of
the planets can be determined ; as f?=360, a=59' 08" the mean
motion of the earth, and suppose b the motion of Mars, for ex-
ample, to be unknown.
When unknown, represent it by x.
Then t= - or at txd.
a x
at d
Therefore x= -. .
t
106 ELEMENTS OF ALGEBRA.
SECTION III.
INVOLUTION.
CHAPTER I.
(Art. 64.) Equations, and the resolution of problems producing
equations, do not always result in the first powers of the un-
known terms, but different powers are frequently involved, and
therefore it is necessary to investigate methods of resolving equa-
tions containing higher powers than the first ; and preparatory
to this we must learn involution and evolution of algebraic
quantities.
(Art. 65.) Involution is the method of raising any quantity to
a given p v =h. Evolution is the reverse of involution, and is
the method of Determining what quantity raised to a proposed
power will produce a given quantity.
As in arithmetic, involution is performed by multiplication, and
evolution by the extraction of roots.
The first power is the root or quantity itself;
The second power, commonly called the square, is the quan-
tity multiplied by itself ;
The third power is the product of the second power by the
quantity ;
The fourth power is the third power multiplied into the quan-
tity, &c.
The second power of a is a X# or a z
The third power is a z Xa or a 3
The fourth power is a 3 Xa or a 4
The second power of a 1 is a 4 X 4 or a 8
The third power of a 4 is a 8 Xa 4 or a 12
The nth power of a 4 has the exponent 4 repeated n times,
or a 4 ". Therefore, to raise a simple literal quantity to any
power, multiply its exponent by the index of the required
power.
Raise x to the 5th power. The exponent is 1 understood,
and this 1 multiplied by 5 gives x 5 for the 5th power.
INVOLUTION. 107
Raise x 3 to the 4th power Am. a? 12 .
Raise y 1 to the third power. Ans. y 21 .
Raise x n to the 6th power. Am. x* n .
Raise x n to the mth power. Am. a? 17 *.
Raise ay? to the 3d power. A*is. a*x*.
Raise ab*x 4 to the 2d power. .tfns. 2 6V.
R*.ise c 2 7/ 4 to the 5th power. .tfns. c 10 */ 20 .
(Art. 66.) By the definition of powers the second power is
any quantity multiplied by itself; hence the second power of
ax is a 2 ^ 2 , the second power of the coefficient a, as well as the
other quantity x ; but a may be a numeral, as 6#, and its second
power is 36x*. Hence, to raise any simple quantity to any
power, raise the numeral coefficient, as in arithmetic, to the
required power, and annex the powers of the given literal
quantities
EXAMPLES.
1. Required the 3d power of 3a;r 2 . Am. 27V.
2. Required the 4th power of |?/ 2 . Jins. Ify 8 .
3. Required the 3d power of 2x. Ans. Sx 3 .
4U Required the 4th power of 3x. Jlns. 8 la; 4 .
Observe, that by the rules laid down for multiplication, the
even powers of minus quantities must be plus, and the odd powers
minus.
5. Required the 2d power of - . Ans. - ^-.
6. Required the 6th power of . Ans.
7. Required the 6th power of -j . Ans ^ .
^a? x
(Art. 67.) The powers of compound quantities are raised by
the mere application of the rule for compound multiplication.
(Art. 12.)
108 ELEMENTS OF ALGEBRA.
Let a-\-b be raised to the 2d, 3d, 4th, &c. powers.
a +b
tf+ab
ab+b*
2d power or square, a z -}-2ab
a+b
3d power or cube,
The 4th power, a 4 +4a 3 6-f Ga*b 2 -{-4ab s +b<
a+b
The 5th power,
&c.
By inspecting the result of each product, we may arrive at
general principles, according to which any power of a binomial
may be expressed, without the labor of actual multiplication.
This theorem for abbreviating powers, and its general application
to both powers and roots, first shown by Sir Isaac Newton, has
given it the name of Newton's binomial, or the binomial theorem.
OBSERVATIONS.
Observe the 5th power : a, being the first, is called the leading
term ; and 6, the second, is called the following term. The sum
of the exponents of the two letters in each and all of the terms
amount to the index of the power. In the 5th power, the sura
INVOLUTION. 109
of the exponents ol a and b is 5 ; in the 4th power it is 4 ; in
the 10th power it would be 10, &c. In the 2d power there are
three terms; in the 3d power there are 4 terms; in the 4th
power there are 5 terms ; always one more term than the index
of the power denotes.
The 2d letter does not appear in the first term ; the 1st letter
does not appear in the last term.
The highest power of the leading term is the index of the
given power, and the powers of that letter decrease by one from
term to term. The second letter appears in the 2d term, and its
exponent increases by one from term to term as the exponent of
the other letter decreases.
The 8th power of (+&) is indicated thus: (a-f-6) 8 . When
expanded, its literal part, (according to the preceding observa-
tions) must commence with a 8 , and the sum of the exponents of
every term amount to 8, and they will stand thus : a 8 , a 7 6, 6 6 2
5 6 3 , a 4 b 4 , 3 6 5 , a 2 6 6 , ab\ b 9 .
The coefficients are not so obvious. However, we observe
that the coefficients of the first and last terms must be unity.
The coefficients of the terms next to the first and last are equal,
and the same as the index of the power. The coefficients in-
crease to the middle of the series, and then decrease in the
same manner, and it is manifest that there must be some law of
connection between the exponents and the coefficients.
By inspecting the 5th power of a-f-# we find that the 2d co-
efficient is 5, and the 3d is 10.
The third coefficient is the 2d, multiplied by the exponent of
ihe leading letter, and divided by the exponent of the second
Letter increased by unity.
In the same manner, the fourth coefficient is the third multi-
plied by the exponent of the leading letter, and divided by the
exponent of the second letter increased by unity, and so on from
coefficient to coefficient.
HO
The 4th coefficient is
The 5th is
The last is
Now let us expand
For the 1st term write
For the 2d term write
For the 3d,
ELEMENTS OF ALGEURA.
1 ft S/ Q
=iO
= 1 understood.
I
a 8
Sa'b
28 e & 2
56a 5 6 3
For the 4th,
For the 5th,
Now as the exponents of a and 6 are equal, we have arrived
at the middle of the power, and of course to the highest coeffi-
cient. The coefficients now decrease in the reverse order which
they increased.
Hence the expanded power is
Let the reader observe, that the exponent of 6, increased by
unity is always equal to the number of terms from the beginning,
or from the left of the power. Thus, b z is in the 3d term, &c.
Therefore in finding the coefficients we may divide by the num-
ber of terms already written, in place of the exponents of the
second term increased by unity.
If the binomial (a-\-b) becomes (a+1,) that is, when b be-
comes unity, the 8th power becomes,
Any power of 1 is 1, and 1 as a factor never appears.
If a becomes 1, then the expanded power becomes,
INVOLUTION. Ill
The manner of arriving at these results is to represent the unit
by a letter, and expand the simple literal terms, and afterwards
substitute their values in the result.
(Art. 68.) If we expand (a 6) in place of (a-\-b^ the expo-
nents and coefficients will be precisely the same, but the princi-
ples of multiplication of quantities affected by different signs
will give the minus sign to the second and to every alternate term.
Thus the 6th power of (a b) is
(Art. 69.) This method of readily expanding the powers of a
binomial quantity is one application of the " binomial theorem"
and it was thus by induction and by observations on the result
of particular cases that the theorem was established. Its rigid
demonstration is somewhat difficult, but its application is simple
and useful.
Its most general form may arise from expanding (-{->)".
When n=3, we can readily expand it;
When =4, we can expand it ;
When n= any whole positive number, we can expand it.
Now let us operate with n just as we would with a known
number, and we shall have
a n - z b 2 , &c.
We know not where the series would terminate until we
know the value of n. We are convinced of the truth of the
result when n represents any positive whole number; but let n
be negative or fractional, and we are not so sure of the result.
To extend it to such cases requires deeper investigation and
rigid demonstration, which it would not be proper to go into at
this time. We shall, therefore, content ourselves with some of
its more simple applications.
EXAMPLES.
1. Required the third power of 3o?+2y.
We cannot well expand this by the binomial theorem, because
the terms are not simple literal quantities. But we can assume
3a?=0 and 2y=b. Then
o-f-& and
112 ELEMENTS OF ALGEBRA.
Now to return to the values of a and &, we have,
X 4i/ 2 =36;r/ 2 .
Hence (
2. Required the 4th power of 2a 2 3.
Let #=2a 2 2/=3. Then expand (a? y} 4 , and return tho
values of x and y, and we shall find the result,
16a 8 96a 6 4-216a 4 216 2 +81.
3. Required the cube of (a-{-b -j-c+rf).
As we can operate in this summary manner only on binomial
quantities, we represent a-\-b by x, or assume x=a-\-b, and
Then
Returning the values of x and y, we have
Now we can expand by the binomial, these quantities con
tained in parenthesis.
4. Required the 4th power of 2a-{-3x.
Ans.
5. Expand (x z +3y*) s .
6. Expand (2a z -\-ax)*
7. Expand (# I) 6 .
8. Expand (3* -5) 3 rfns. 27^135^+225^125.
9. Expand (+2) 4 . Ans. a 4 +8a 3 -f-24a 2 +32a+16
10. Expand (1 a) 4 . tins. 1 2a-r-| -- g+Y6
11. Expand (a-r-6+c) 2 .
12. Expand (a2&) 3 .
13. Expand (1 2x) 5 .
EVOLUTION. U3
EVOLUTION.
CHAPTER II.
(Art. 70). Evolution is the converse of involution, or the ex-
ti action of roots, and the main principle is to observe how powers
are formed, to be able to trace the operations back. Thus, to
square , we double its exponent, (Art. 65), and make it a 2 .
Square this and we have a 4 . Cube a 2 and we have a 6 . Take
the 4th power of x and we have x 4 . The nth power of x 3 is
x s> \
Now, if multiplying exponents raises simple literal quantities
to powers, dividing exponents must extract roots. Thus, the
square root of a 4 is a 2 . The cube root of a 2 must be a*. The
cube root of a must have its exponent, (1 understood,) divided by
3, which will make a*'
Therefore roots are properly expressed by fractional expo-
nents, i
The square root of a is a 2 , and the exponents, 5, I, ^, &c. in-
dicate the third, fourth, and fifth roots. The 6th root of y? is
s_
a? 6 ; hence we perceive that the numerators of the exponent in-
dicate the power of the quantity, and the denominator the root
of that power.
(Art. 71.) The square of ax is aV 2 . We square both
factors, and so, for any other power, we raise all the factors to
the required power. Conversely, then, we extract roots by
taking the required root of all the factors. Thus the cube root
of 8x* is 2x.
A root that can not be exactly expressed is called a surd, or
irrational root. The square root of 2 can not be exactly ex-
pressed, hence, it is called a surd. A root merely expressed by
die radical sign (J ), or by a fractional exponent, is called a
radical quantity. The cube root of 9 may be symbolically ex-
pressed thus, ^/9, or (9)3, and this is a radical quantity; but
numerically the root can not be exactly expressed ; it is, there-
fore, a surd. Surd roots may be found approximately.
10
114 ELEMENTS OP ALGEBRA.
The square root of 64a 6 is obviously 8a 3 , and from this and
the preceding examples we draw the following
RULE. For the extraction of the roots of monomials. Ex-
tract the root of the numeral coefficients and divide the exponent
of each letter by the index of the root.
EXAMPLES.
1. What is the square root of 49a 2 # 4 ? Ans. 7ax 2
2. What is the square root of 25c 10 6 2 1 Jins. 5c 5 b.
3. What is the square root of 2Qax ? fins.
In 20, the square factor 4 can be taken out ; the other factor
is 5. The square root of 4 is 2, which is all the root we can
take ; the root of the other factors can only be indicated as in the
answer.
4. What is the square root of 12a 2 1 ns.
5. What is the square root of 144a 2 cV/? Jins. I2ac*xy.
6. What is the square root of 36.r 4 ? fins. db6a?.
(Art. 72.) The square root of algebraic quantities may be
taken with the double sign, as indicating either plus or minus,
for either quantity will give the same square, and we may not
know which of them produced the power. For example, the
square root of 16 may be either -j-4 or 4, for either of them,
when multiplied by itself, will produce 16.
The cube root of a plus quantity is always plus, and the cube
root of a minus quantity is always minus. For -\-2a cubed
gives -|-8a 3 , and 2a cubed gives 8a 3 , and a may represent
any quantity whatever.
i
EXAMPLES. .
1. What is the cube root of 125a 3 ? tins. 5a.
2. What is the cube root of 64# 6 ? tfns. 4x*
3. What is the cube root of 216y ? rfns. 6y 3
4. What is the cube root of 729 A 12 ? dns. 9a*x*
5. What is the cube root of 320 5 ? Jlns. ~
EVOLUTION. 115
6. What is the 4th root of 256V ? ns.
7. What is the 4th root of 16?
8. What is the 4th root of 64x*y 2 ?
N. B. The 1th root is the square root of the square root.
9. What is the 4th root of 20a# ? Ans.
10. What is the square root of 75 ? ns.
75=25X3.
4a 2 # 4
11. Required the square root of -5-2" ^ n
. 32 3 # 5
13. Required the square root of - . <flns.
N. B. Reduce the fraction as much as possible, and then ex-
tract the root.
13. Required the square root of , ^ . *fl.ns. r*
128a 4
14. Required the nth root of -sr .. rfns. -r-.
nn
a 3 2. ~1. -1
15. Required the nth root of =-. Am. a n b" c n
Observing that j =b~ } c~ l
CHAPTER m.
To extract roots of compound quantities.
(Art. 72.) We shall commence this investigation by confining
our attention to square root, and the only principle to guide us is
the law of formation of squares. The square of a-{-b is 2 -}-
2ab-\-b z . Now on the supposition that we do not know the root
'of this quantity to be a-}-b, we are to find it or extract it out of
the square
We know that a 2 , the first term, must have been formed by the
multiplication of a into itself, and the next te-m is 2X#. That
110 ELEMENTS OF ALGEBRA.
is twice the root of the first term into the second term of the root,
Hence if we divide the second term of the square by twice the
root of the first term, we shall obtain b, the second term of the
root, and as b must be multiplied into itself to form a square, we
add b to 2, and have 2-f-/;, which we call a divisor.
OPERATION.
a 2
2a-\-b}2ab+b*
2ab+b 2
We take a for the first term of the root, and subtract its square
( 2 ) from the whole square. We then double a and divide 2ab
by 2a and we find 6, which we place in both the divisor and quo-
tient. Then we multiply 2a-\-b by b and we have 2ab-{-b z , to
subtract from the two remaining terms of the square, and in this
case nothing remains.
Again, let us take rt+6-f-c, and square it. We shall find its
square to be
a 2
2ab+b 2
2ab+b 2
2aci-2bc+c z
By operating as before, we find the first two terms of the root
to be a-}-b, and a remainder of 2ac-\-2bc-{-c z . Double the root
already found, and we have 2a-}-2b for a partial divisor. Divide
the first term of the remainder 2ac by 2a, and we have c for the
third term of the root, which must be added to 2a-\-2b to com-
plete the divisor. Multiply the divisor by the last term of the
root and set the product under the three terms last brought down,
and we have no remainder.
EVOLUTION. 117
Again, let us take a-\-b-\-c to square; but before we square it
let the single letter s=a-}-b.
Then we shall have s-f-c to square, which produces
s 2 -f-2sc+c 2 . To take the square root of this we repeat the first
operation, and thus the root of any quantity can be brought into
a binomial and the rule for a binomial root will answer for a root
containing any number of terms by considering the root already
found, however great, as one term.
Hence the following rule to extract the square root of a com-
pound quantity.
Arrange the terms according to the powers of some letter,
beginning with the highest, and set the square root of the first
term in the quotient.
Subtract the square of the root thus found from the first
term and bring down the next two terms for a dividend.
Divide the first term of the dividend by double of the root
already found, and set the result both in the root and in t/ie
divisor.
Multiply the divisor, thus completed, by the term of the root
last found, and subtract the product from the dividend, and so on.
EXAMPLES.
1. What is the square root of
2a 2 -f26
2a 2 -f-46 2 40 2 86+4
2. What is the square root of 1 4b-\-4b 2 -\-2y iby-{-y*1
Ans. 1 2b -f-y.
8. What is the sq lare root of 4z 4 4or 3 -{-13r J 6#-J-9 ?
Ans. 2x? a?-f3.
4. What is the sa lore root of 4z 4 lGx 3 +24x 2 16#-f4 ?
118 ELEMENTS OF ALGEBRA.
5. What is the square root of I6x* +24x?+S9x*+6Qx-{' 100?
Am.
What is the square root of 4x* 16# 3 +8a: 2 -f-16#-{-4 ?
Ans.
7. What is the square root of
3?+2xy-\-y 2 -\-6xz -\-6yz-\- 9z*l dm.
8. What is the square root of
Ans.
a 2 b"
0. What is the square root of ^ 2-\- -? ?
a b b a
Jlns. T - or - T
b a a b
1O. What is the square root of a.* fcB*
^-T/ or y-x
^j^L- d.TM.cM^.. fW&d, J0 & :
(Art. 73.) Every square root will be equally a root if we change
the sign of all the terms. In the first example, for instance, the
root may be taken a 2 26+2, as well as a z -\-2b 2, for either
one of these quantities, by squaring, will produce the given
square. Also, observe that every square consisting of three
terms only, has a binomial root.
(Art. 74.) Algebraic squares may be taken for formulas, cor-
responding to numeral squares, and their roots may be extracted
in the same way, and by the same rule.
For example, a-f b squared is a z -}-2ab-{-b 2 , and to apply this
to numerals, suppose a =40 and 6=7.
Then the square of 40 is 2 =1600
2ab= 560
6 2 = 49
Therefore, (47) 2 =2209
Now the necessary divisions of this square number, 2209, are
not visible, and the chief difficulty in discovering the root is to
make these separations.
EVOLUTION. 119
The first observation to make is that the square of 10 is 100,
of 100 is 10000, and so on. Hence, the square root of any
square number less than 100 consists of one figure, and of any
square number over 100 and less than 10000 of two figures, and
so on. Every two places in a power demanding one place in
its root.
Hence, to find the number of places or figures in a root, we
must separate the power into periods of two figures, beginning
at the unit's place. For example, let us require the square root
of 22-09. Here are two periods indicating two places in the
root, corresponding to tens and units. The greatest square in 22
is 16, its root is 4, or 4 tens =40. Hence a=40.
22 09(40+7^47
a?= 16 00
2a-K>=80-{-7=87 )609
609
Then 2a=80, which we use as a divisor for 609, and find it
is contained 7 times. The 7 is taken as the value of b, and
2+6, the complete divisor, is 87, which multiplied by 7 gives
the two last terms of the binomial square. 2a6-|-6 2 =560+49
609, and the entire root 40+7=47 is found.
Arithmetically, a may be taken as 4 in place of 40, and 1600
as 16, the place occupied by the 16 makes it 16 hundred, and
the ciphers are superfluous. Also, 2 may be considered 8 in
place of 80, and 8 in 60 (not in 609) is contained 7 times, <fec.
If the square consists of more than two periods, treat it as two,
and obtain the two superior figures of the root, and when obtain-
ed bring down another period to the remainder, and consider the
root already obtained as one quantity, or one figure.
For another example, let the square root of 399424 be ex-
tracted.
39-94-24 || 632
36
123
394
369
1262
2524
2524
120 ELEMENTS OF ALGEBRA.
In this example, if we disregard the local value of the figures,
we have a=6, 2=12, and 12 in 39, 3 times, which gives 6=3
Afterwards we suppose =63, and 2a=126, 126 in 252, 2 times,
or the second value of b =-2. In the same manner, we would
repeat the formula of a binomial square as many times as we
have periods.
EXERCISES FOR PRACTICE.
1. What is the square root of 8836? Jlns. 94.
2. What is the square root of 106929? Jlns. 327.
3. What is the square root of 4782969? Jlns. 2187.
4. What is the square root of 43046721? Jlns. 6561.
5. What is the square root of 387420489? Jlns. 19683.
When there are whole numbers and decimals, point off periods
both ways from the decimal point, and make the decimal places
even, by annexing ciphers when necessary, extending the deci-
mal as far as desired. When there are decimals only, commence
pointing off from the decimal point.
EXAMPLES.
1. What is the square root of 10-4976? Jlns. 3-24.
2. What is the square root of 3271-4207? Jlns. 57-19+.
3. What is the square root of 4795-25731 ? JZns. 69-247+.
4. What is the square root of -0036 ? Ans. -06
5. What is the square root of -00032754? rfns. -01809+.
6. What is the square root of -00103041 ? rfns. -0321.
(Art. 75.) As the square of any quantity is the quantity mul-
tiplied by itself, and the product of T by j (Art. 64.) is p ;
hence to take the square root of a fraction we must extract the
square root of both numerator and denominator.
A fraction may be equal to a square, and the terms, as given,
not square numbers ; such may be reduced to square numbers.
EVOLUTION. Itl
EXAMPLES.
What is the square root of yYj
Observe T y f = gf . Hence the square root is f .
1. What is the square root of T 9 / ? rfns. .
3. What is the square root of J r || ? Ans. J.
3. What is the square root of \\\ ? Ans. f .
4. What is the square root of fUJ? ^w*.
When the given fractions cannot be reduced to square terms,
reduce the value to a decimal, and extract the root, as in the last
article.
CHAPTER IV.
To extract the cube root of compound quantities.
(Art. 76.) We may extract the cube root in a similar manner
as the square root, by dissecting or retracing the combination of
terms in the formation of a binomial cube.
The cube of a+b is a s -\-3a?b-\-3ab 2 -\-b* (Art. 67). Now
to extract the root, it is evident we must take the root of the*
first term (a 3 ), and the next term is 3a 2 6. Three times the square
of the first Utter or terra of the root multiplied by the 2d term
of the root.
Therefore to find this second term of the root we must divide
the second term of the power (3a 2 6) by three times the square
of the root already found (a).
When we can decide the value of 6, we may obtain the com-
plete divisor for the remainder after the cube of the first term is
subtracted, thus :
The remainder is 3a z b-}-3ab 2 -\-b 3
Take out the factor 6, and 3a 2 -f-3a6-j-6 2 is the complete
divisor for the remainder. But this divisor contains 6, the very
term we wish to find by means of the divisor ; hence it must be
found before the divisor can be completed. In distinct algebraic
11
122 ELEMENTS OF ALGEBRA.
quantities there can be no difficulty, as the terms stand separate,
and we find b by dividing simply 3a?b by 3a 2 ; but in numbers
the terms are mingled together, and b can only be found by trial.
Again, the terms 3a 2 -^-3ab-\-b 2 explain the common arithme-
tical rule, as 3a 2 stands in the place of hundreds, it corresponds
with the words : " Multiply the square of the quotient by 300 y "
"and the quotient by 30," (3a,) &c.
By inspecting the various powers of a-\-b, (Art. 67,) we draw
the following general rule for the extraction of roots :
Arrange the terms according to the powers of some letter ;
take the required root of the first term and place it in the quo-
tient : subtract its corresponding power from the first term, and
bring down the second term for a dividend.
Divide this term by twice the root already found for the SQUARE
root, three times the square of it for the CUBE root, four times
the third power for the fourth root, &c., and the quotient will
be the next term of the root. Involve the whole of the root,
thus found, to its proper power, which subtract from the given
quantity, and divide the first term of the remainder by the same
divisor as before: proceed in this manner till the whole root i9
determined.
EXAMPLES.
1. What is the cube root of x-}-Qx 5 4Qx?-\-QQx 64 ?
a*
Divisor 3# 4 ) 6r>= 1st remainder.
Divisor 3# 4 ) I2# 4 =2d remainder.
2. What is the cube root of 27a 3 +108 2 +144a+64 ?
Jlns. 3-f-4.
a- What is the cube root of a 3 Qa z x+I2ax 9 ' 8x* ?
Jlns. a 2x
EVOLUTION. 123
4. What is the cube root of a,' 6 3^ 5 +5x 3 3^1 1
Ans. x z x 1
5. What is the cube root of a 3 Gcfb+lZab 2 Qb 3 ?
Am. a 26.
O I
6. What is the cube root of x*-\-3x-\ h-
* Extract the fourth root of
a 4 +8a 3 -f-24a 2 -f-32a-H 6(a-}-2
a 4
Ans. x-\ .
4a 3 ) 8a 3 , &c.
(Art. 77.) To apply this general rule to the extraction of the
cube root of numbers, we must first observe that the cube of 10
is 1000, of 100 is 1000000, &c.; ten times the root producing
1000 times the power, or one cipher in the root producing 3 in
the power ; hence any cube within 3 places of figures can have
only one in its root, any cube within 6 places can have only two
places in its root, &c. Therefore we must divide off the given
power into periods consisting of three places, commencing at the
unit. If the power contains decimals, commence at the unit
place, and count three places each way, and the number of pe-
riods will indicate the number of figures in the root.'
EXAMPLES.
1. Required the cube root of 12812904.
12-812-904(234
0=2 3 = 8
Divisor 3a 2 =12 )48
12167 = (23) 3
3(23) 2 = 1587) 6459 (4
12812904 =
124 ELEMENTS OF ALGEBRA.
Here 12 is contained in 48, 4 times; but it must be remem-
bered that 12 is only a trial or partial divisor; when completed
it will exceed 12, and of course the next figure of the root can-
not exceed 3.
The first figure in the root was 2. Then we assumed a~2.
Afterwards we found the next figure must be 3. Then we as-
sumed a=23. To have found a succeeding figure, had there
been a remainder, we should have assumed a=234, &c., and
from it obtained a new partial divisor.
2. What is the cube root of 148877? Ans. 53.
3. What is the cube root of 571787? dns. 83.
4. What is the cube root of 1367631 ? rfns. 111.
5. What is the cube root of 2048383 ? rfns. 127.
6. What is the cube root of 16581375? dm. 255.
7. What is the cube root of 44361864 ? Ans. 354.
. What is the cube root of 100544625? Ans. 465.
(Art. 78.) When the power is not complete, and, of course,
its root surd, the methods of direct extraction are all too tedious
to be much used, and several eminent mathematicians have given
more brief and practical methods of approximation.
One of the most useful methods may be investigated as follows :
Suppose a and a-\-c two cube roots, c being very small in
relation to . a 3 and 3 +3a 2 c+3c 2 +c 3 are the cubes of the
supposed roots.
Now if we double the first cube (a 3 ), and add it to the second,
we shall have 3a >+3<?c+3ac>+c>.
If we double the second cube and add it to the first, we shall
have
As c is a very small fraction compared to , the terms con-
taining c 2 and c 3 are very small in relation to the others, and
the relation of these two sums will not be materially changed by
rejecting those terms containing c 2 and c 3 , and the same will
lhen be 3+3 etc
And 3o 3 +6a'c
EVOLUTIOIS 7 125
The ratio of these terms is the same as the ratio* of a-\-c to
a-f2c.
Or the ratio is H ;
But the ratio of the roots a to +> is 1 -\ .
a
Observing again that c is supposed to be very small in rela-
tion to a. the fractional parts of the ratios ; and - are both
a-f-c a
small and very near in value to each other. Hence we have
found an operation on two cubes which are near each other in
magnitude, that will give a proportion very near in proportion to
their roots ; and by knowing the root of one of the cubes, by this
ratio we can find the other.
For example, let it be required to find the cube root of 28, true
to 4 or 5 places of decimals. As we wish to find the cube root
of 28, we may assume that 28 is a cube. 27 is a cube near in
value to 28, and the root of 27 we know to be 3.
Hence a, in our investigation, corresponds to 3 in this exam-
ple, and c is unknown; but the cube of a-\-c is 28, and a 8
is 27.
Then 27 28
2 2
54 56
Add 28 27
Sums 82 : 83 : : 3 : a-f-c very nearly.
Or (a-J-c)= 2 7 4 2 9 =3'03658+, which is the cube root of 28,
true to 5 places of decimals.
By the laws of proportion, which we hope more fully to in-
vestigate in a subsequent part of this work, the above propor-
tion, 82 : 83 : : a : a-}-c, may take this change .
82 : 1 : : a : c
Hence, c= T V c being a correction to the known root,
which, being applied, will give the unknown or sought root.
From what precedes, we may draw the following rule for find-
ing approximate cube roots :
126 ELEMENTS OF ALGEBRA.
RULE. Take the nearest rational cube to the given number,
and call it an assumed cube ; or, assume a root to the given
number and cube it. Double the assumed cube and add the
given number to it ; also, double the given number and add
the assumed cube to it. Then, by proportion, as the first sum
is to the second, so is the known root to the required root. Or
take the difference of these sums, then say, as double of the
assumed cube, added to the number, is to this difference, so is
the assumed root to a correction.
This correction, added to or subtracted from the assumed root,
as the case may require, will give the cube root very nearly.
By repeating the operation with the root last found as an as-
sumed root, we may obtain results to any degree of exactness ;
one operation, however, is generally sufficient.
EXAMPLES.
1. What is the approximate cube root of 120?
tins. 4-93242-K
2. What is the approximate cube root of 8*5 ?
ns. 2-0408-f.
3. What is the approximate cube root of 63 ?
Jlns. 3-97905-}-.
4. What is the approximate cube root of 515 ?
tins. 8-01559-)-.
5. What is the approximate cube root of 16?
The cube root of 8 is 2, and of 27 is 3 ; therefore the cube
root of 16 is between 2 and 3. Suppose it 2-5. The cube of
this root is 15-625, which shows that the cube root of 16 is a
little more than 2-5, and by the rule
31-25
32
16
15-625
47-25
: 47-625
: : 2-5 : to the
required root.
47-25
: -375
:: 2-5 : -01984
Assumed root
2-50000
Correction
01984
Approximate root 2-51984.
EVOLUTION. 127
We give the last as an example to be followed in most cases
where the root is about midway between two integer numbers.
This rule may be used with advantage to extract the root of
perfect cubes, when the powers are very large.
EXAMPLE.
The number 22-069'810'125 is a cube; required its root.
Dividing this cube into periods, we find that the root must
contain 4 figures, and the superior period is 22, and the cube root
of 22 is near 3, and of course th< whole root near 3000; but less
than 3000. Suppose it 2800 and cube this number. The
cube is 21952000000, which being less than the given number,
shows that our assumed root is not large enough.
To apply the rule, it will be sufficient to take six superior
figure^ of the given and assumed cubes. Then by the rule,
220698
2
439040 4413%
220698 219520
659738 : 660916 : : 2800
659738
659738 : 1178 :: 2800
2800
942400
2356
659738)3298400(5
3298690
Assumed root, 2800
Correction, 5
True root, 2805
The resuit of the last proportion is not exactly 5, as will be
seen by inspecting the work ; the slight imperfection arises from
the rule being approximate, not perfect.
When we have cubes, however, we can always decide the unit
figure by inspection, and, in the present example, the unit figure
128 ELEMENTS OF ALGEBRA.
in the cube being 5, the unit figure in the root must be 5, as no
other figure when cubed will give 5 in the place of units.
[For several other abbreviations and expedients in extracting
cube root in numerals, see Robinson's Arithmetic.]
(Art. 79.) To obtain the 4th root, we may extract the square root
of the square root. To obtain the 6th root, we may take the
square root first, and then the cube root of that quantity.
To extract odd roots of high powers in numeral quantities is
very tedious and of no practical utility ; we therefore give no ex-
amples.
(Art. 80.) Some radical quantities may be simplified, and oth-
ers are not susceptible of simplification. For instance, the square
root of 75, written ^/75, is equal to 5 times the square root of
3; that is, ^/75=5^/3. But the square root of 71 can not be
fully expressed except by the sign, thus, ^/71, because 71 con-
tains no square factor.
It is obvious that *J4Q==*J8 :r 5==* i JS *^Ss=S(5)lj but
39 and 41 containing no cube factor, their cube roots can only
be indicated by the radical sign over them.
When simplification is possible, it can be effected by the
following
RULE. Separate the quantity into two factors, one of which
is a perfect power of the required root. Extract the root of
that factor and prefix the result as a coefficient to the other
factor placed under the radical sign.
We give the following examples for practice :
1. Reduce the square root of 75 to lower terms, or reduce
tins, 5^/3.
2. Reduce J98a 2 to lower terms. Jlns. 7 > /2.
3. Reduce jVZx 9 -y to lower terms. Jlns. 2x,j3y.
4. Reduce 3 /54.T 4 to lower terms. Jlns. 3x
5. Reduce 4 ^/T08 to lower terms. Jlns.
6. Reduce Jx? V to lower terms. Jlns. xjx a 2 .
EVOLUTION. 20 129
7. Reduce 3 /32a 3 to lower terms. Ans. 2 3 > /4.
8. Reduce J'ZScfx? to lower terms. Ans. 2axj7a
9. Reduce 74 J to lower terms.
Where terms under the radical are fractional, it is expedient
to reduce the denominator to a power corresponding to the radi-
cal sign ; then by extracting the root there will be no fraction
under the radical.
The above example may be treated thus :
-y K'-J X33=ftV88. Aw.
We divided || into the factors and y ; the first factor is a
square ; the other factor, y , we multiply both numerator and
denominator by 3, to make the denominator a square.
In like manner reduce the following :
10. Reduce 3 7^ to more simple terms. Ans. %*J\Q.
11. Reduce 3 7V to more simple terms. Ans.
12. Reduce to more sim ^ e terms. Ans.
13. Reduce 3 f Ja^-\-a 3 b 2 to more simple terms.
Ans. a
14. Reduce ,J~ to more simple terms. Ans.
(Art. 81.) Radical quantities may be put into one sum, or the
difference of two may be determined, provided the parts essen-
tially radical are the same.
Thus the sum of ^/8 and ,/72 is 8^/2 and their
difference is 4^/2
For ^8=
And 72=
Sum 872
Difference 472"
When radical quantities are not and cannot be reduced to the
130 ELEMENTS OF ALGEBRA.
same quantity under the sign, their sum and difference can only
tie taken by the signs plus and minus.
EXAMPLES.
1. Find the sum and difference of t J\.Qa 2 x and ,j4a 2 x.
Ans. Sum, Qajx ; difference, 2a t jjs.
2. Find the sum and difference of /128 and ^72.
Jlns. Sum, 14^/2 ; difference,
3. Find the sum and difference of 3 ^/135 and 3 N /40.
Am. Sum, 5 3 ,y5 ; difference,
4. Find the sum and difference of V 108 and 9 V 4.
Jins. Sum, 12 3 ^/4; difference, 6 3 > /4.
5. Find the sum and difference of ^/| and J'l.
Ans. Sum, f^2 ; difference, J^/2 U
6. Find the sum and difference of 3 ^/56 and 3 > /189.
fins. Sum, 5 3 ,y7; difference, 3 ^/?.
7. Find the sum and difference of 3Ja z b and
Am. Sum, 14-3a6; difference, 12^
(Art. 82.) We multiply letters together by writing them one
after another, as abxy. If they are numeral quantities, their
product appears as a number ; if two or more of them are nu-
meral, the product of these quantities will appear as a number.
This fundamental principle of multiplication may be applied to
the multiplication of surds. Let it be required to multiply
5^/2 by 3^/7. Here suppose =5, 6=3, x=j2, y=,Jl.
Then the product of 5^/2T by S^/Y is abxy or 15^/2X7=
15^14. Hence, for the multiplication of quantities affected by
the same radical sign, we draw the following
RULE. Multiply the rational parts together for the rational
part of the product, and the radical parts together for the
radical part of the product.
EVOLUTION. 131
EXAMPLES.
1. Required the product of 5^/5 and 3^/8.
Product reduced, tins.
2. Required the product of 4J12 and 3^/27 Am.
3. Required the product of 3,/2~ and 2J~8. .#ns. 24.
4. Required the product of 2yl4 and 3 3 ^/47
Ans. 12 V7.
5. Required the product of 2^5~and 2^/To. Jlns.
(Art. 83.) We multiply different powers of the same quantity
together by simply adding the exponents as, a 3 multiplied by a 3
is a 8 , and this holds when the exponents are fractional, as a 2 into
a 3 ; for the product we simply add J and -J, which make | to
write over a for the product of a 2 into a? which gives cfi.
When the radical quantities are different, as () 2 and 6^, and
the exponents different, are to be multiplied together, or one to be
divided by another, we better lay aside all rules and apply that
powerful engine, an equation. For illustration:
Required the product of a 2 into b^.
The product is P, then we have
Raise both members to the 6th power, then,
Conceiving the second member to be a single quantity, and
taking the 6th root, will give us
EXAMPLES.
1. Required the product of (+&)' and (a-\-b)*.
Jlns.
2. Required the product of JT~ and *JT, rfns. (7 5 )
132 ELEMENTS OF ALGEBRA.
(Art. 84.) If we divide one quantity by another, the product
of the quotient and the divisor must equal the dividend.
Divide the Q Jl2 by ^/2~: the quotient is Q.
Then,
By squaring,
Cubing 8 6=72.
Whence, Q*=9 Q 3 =3
3. Required the product of 2J3 and 3(4)^.
_ _
4. Required the product of 3 ^/15 and ^10.
Am. 6^/225000.
EXAMPLES.
1. Divide 4^/50 by 2</57 Ans. 2JW.
2. Divide 6 yiOO by 3 y~5~. Am. 2 y20i
3. Divide Jl by B 4 J'7.
Let a=7. Then the example is, to divide the a 2 by a a , or
a by a ; quotient a?. As we always subtract the exponents
of like quantities to perform division (Art. 17) ; therefore 7 must
be the required quotient.
4. Divide 6^/54 by 3J2. Ans.
5. Divide (tfb z cl?Y by A -#rcs.
6. Divide (I6tfI2a 2 x)* by 2. ^?is. (4a 3xf.
(Art. 85.) In the course of algebraical investigations, we might
fall on the square root of a minus quantity, as J a, ,J 1,
J b, &c., and it is important that the pupil should readily
understand that such quantities have no real existence ; for no
quantity, either plus or minus, multiplied by itself will give
minus a, minus b, or minus any quantity whatever ; hence there
is no value to J a, &c., and such symbols are said to be im-
possible or imaginary.
EVOLUTION. 133
(Art. 86.) The square and cube root of any quantity, as a,
being expressed by .Ja and 3 Ja, and as by involving the root
we obtain the power, hence the square of ,Ja, is a; and the
cube of I/a, is a. Hence, removing the sign involves to the
corresponding power.
EXAMPLES.
1. What is the square of J3ax 1 Jlns. Sax.
2. What is the cube of V% 2 ? J3ns. 6y 2 .
3. What is the square of Ja 2 x 2 ? JLns. a 2 x 2 .
4. What is the square of /vL_ ? rfns. -
6-f-z 6+0?
5. What is the cube of *Jl+a*. rfns. l+a.
(Art. 87.) When we have two or more quantities, and the
radical sign not extending over the whole, involution will not
remove the radical, but will change it from one term to another.
Thus, Jx-\-a the square will be x-{-2a,Jx-\-a* ; the radical
sign is still present, but not in the first term.
PURE EQUATIONS.
CHAPTER IV.
(Art. 88.) Pure equations in general are those in which a com-
plete power of the unknown quantity is contained, and no special
artifice is requisite to render the power complete.
The unknown quantity may appear in one or in several terms ;
when it appears in several, its exponents will be regular, descend-
ing from a higher to a lower value, or the reverse.
In such cases we must reduce the equation by evolution.
Like roots of equal quantities are equal. (Ax. 8.)
EXAMPLES.
1. Given 3x 2 9=66 to find the value of x.
Solution, 3,r 2 =^75 # 2 =25. Hence, #=5.
134 ELEMENTS OF ALGEBRA.
We place the double sign before 5, as we cannot determine
\vhether25 was produced by the square of +5 or of 5.
In practical problems, the nature of the case will commonly
determine, but in every abstract problem we must take the
double sign.
S5. Given x z 6#-|-9=a 2 to find the value of x.
By evolution, x 3=a. Hence, a?=3#, Jlns.
3. Given x 3 x z -{-ix^=a s b 3 to find x.
Take cube root and x ^ab; and x=ab-\-\, Jlns.
4. Given a; 4 2^+1 = 16 to find the value of x.
Jlns. x=^ or
5. Given x z 4#+4=64 to find the value of x.
Jlns. #=10, or 6.
6. Given x 2 y 2 -}-2xy-}-l=4 : x 2 y 2 and x=2y to find the values
of x andy. Jlns. #==t,/2^ y
7. Given x 2 -\-y 2 =I3 and x 2 2/ 2 =5 to find the values of x
and y. Jlns. #=3 y=2.
(Art. 89.) The unknown quantity of an equation is as likely
to appear under a radical sign as to be involved to a power. In
such cases we free the unknown quantity from radicals by involu-
tion (Art. 86.), having previously transposed all the terms not
under the radical to one side of the equation, the radical being
on the other.
9. Given ,Jx-{-l=a 1 to find the value of x.
By involution, #-{-l=a 2 2a-r-l. Hence, xa 2 2a.
10. Given ,J 1 2 -f- x =3 -\-Jx to find the value of x.
Square both sides, and we have
Drop 9-}- x and 3=6^; or Jx~=s. x=%, Jim.
11. Given Jx 16=8 Jx to find x. Jlns. #=25.
PURE EQUATIONS. 135
X ax J x
12. Given 1= = - to find x.
v# x
Multiply by ^/^ observing that x divided by x gives i ; and
we have x ax=l,
Or (I d)x=l,
Therefore x=- .
1 a
13. Given .= = r to find the value of x.
By clearing of fractions, we have
tf+34/H-168=;r 4-42^+152.
Reducing, 16=8^
By division, 2= Jx, or 4=x.
To call out attention and cultivate tact, we give another solu-
tion. Divide each numerator by its denominator, and we have
Drop unity from both sides, and divide by 8 ; we then have
3 4
Clearing of fractions, 3 t Jx-\-l8=4 i Jx-{-lQ
Dropping equals, 2= ,Jx. Hence, x=4.
14. Given ,J x-{- ,J a-\-x= . = to find x. Jlns. x=\
V a-\-x
2a z
15. Given x-}-Ja z -\-x 2 =j to find x.
Jlns. a?=a,u
. Given a?+a=Va 2 +a?V6 2 +ic 2 to find x.
b*4a 2
Jlns. x=
40
136 ELEMENTS OF ALGEBRA.
Jftx 2 4tjQx 9
17. briven ^ = ^ to find a?.* Jlns.
18. Given V64-f- x 2 -^Sx= to find a:, dns. #
15
19. Given J&+x-}-Jx= - to find a?. rfns. #=4
V 5 +*
20. Given Jx+JxJx Jx=-( --^j- ) to find x.
2\#-rv#/
25
Jin,. *.
Because a 2 6 2 =( a
We infer that 5z 9=(^/5a?-f3) (j5x3)
Therefore, =^5^ 3. The given equation then
--
becomes .^53: 3=1-1^- -- .
2
Now assume
Then y=l+iy. Consequently, 2/=2. Returning to equa
tion (w5), we have ^5a; 3=2
t j5x=5. Therefore, x=5, Ans.
Jaxb 3j~ax2b Qb*
22. Given ^- ==-^ - to find x. Am. x=
Jax+b 3,Jax-{-5b
(Compare 22 with examples 13, and 17.)
23. Given (1+^7^+12) = l+a? to find x. rfns.
21. Given ^!!_ 1 + ^=9, to find x.
* See 2d solution to Equation 13.
PURE EQUATIONS. 137
25. Given a 2 2ax-{-x z =b, to find x. ns. #= Jb
\ 1/3
26. Given =-^ r to find x.
Jlns. x=i.
4
27. Given r =|, to find x. Ans. x=5.
x 2 2#-{-l
28. Given -==l-\-. to find x. Aw. x=
Q/
29. Given x--x 9= to find a?.
30 Given =- to find a?. ^ns. a=4
31. Given = L to find the value of a?.
Multiply the first member, numerator and denominator, by
x-\- ,Jx a), then both members by a, and extract square root.
32. Given ' q l^ =&> to find a-.
Assume a-{-#=2/. Then the equation becomes
'=&. Hence, v=
And
(Art. 90.) To resolve the following examples, requires a de-
gree of tact not to be learned from rules. Quickness of percep-
tion is requisite, as well as sound reasoning. Quickness to per-
ceive the form of binomial squares, and binomial cubes, and a
12
138 ELEMENTS OF ALGEBRA.
readiness to resolve quantities into simple or compound factors,
as the case may require.
1. Given x*-}-2x=9-+- to find the value of x
Multiply by x, and x 3 +2x*=9x+18.
Separate into factors, thus : (a?-f-2)a? 2 =(#-{-2)9.
Divide by the common factor, tf-J-2, and x*=9, or #=3.
Given i a? 2 ,^~ 24 to find the values of x and y.
Add the two equations together, and we have
Extract square root, and x-\-y=6. (.#)
From the first equation we have (x-\-y)x=l.2. (B)
Divide equation (B} by (.#), and a?=2.
This example required perception to recognise the binomial
square, and also to separate into factors.
| Q /
3. Given x z -{-y z = - and xv= - to find the values
xy xy
of x and y.
From the first equation subtract twice the second, and
Therefore, (x #) 3 =1, and x y=l.
Continuing the operation, we shall find #=3, and y=2.
4. Given x 2 y-{-xy 2 =lSO and x s +y*= 189, to find the values
of x and y. J2ns. x=5 or 4 ; y=4 or 5.
To resolve this problem, requires the formation of a cube, or
to resolve quantities into factors.
5. Given x*-\-y*=(x-{-y}xy, and x-}-y=4, to find the
values of x and y. Jlns. x=2] 2/=2.
6. Given x-{-y : x :: 7 : 5, and xy+y*=I26i to find tha
values of x and y. Jlns. #=15, ?/=6
PUKE EQUATIONS. 139
7. Given x y : y :: 4 : 5, and x 2 +4y*=l8], to find the
values of x and y. rfns. #=9, y=5.
8. Given Jx+Jy > Jx Jy :: 4 : 1, and x */=16, to
find the values of x and y. Ans. #=25, i/=9.
9 Given -^+|-f7=9 to find a?. Am x=7.
y o 4
10. Given x+y : x-y :: 3 = 1 ? to find x and
And a? 3 y =56 5
11. Given y?y-}-xy z =30 }
1 i 5 I to find x and
And + ' =
Observe that xy(x-\-y)=x?y-{-xy z .
Clear the 2d equation of fractions, and y-}-x or x-\-y= ~.
Now assume x-\-y=s, and xyp. Then the original equa-
tions become s/?=30
And 6s=5p
Equations which readily give s and j9, and from them we de-
termine x and y.
N. B. When two unknown quantities, as x and y, produce
equations in the form of
x+y=s (1)
And xy=p (2)
such equation can be resolved in the following manner :
Square (1), and x*-}-2xy-{-y 2 =s 2
Subtract 4 times (2) 4xy ==4p
I Diff. is x 2 2xy+y 2 =s 2 4p
By evolution x yJs* 4p (3)
Add equation (1) and (3), and we have,
(4)
Sub. (3) from (1), and 2y=sJs 2 4p (5)
140 ELEMENTS OF ALGEBRA.
To verify equations (4) and (5), add them and divide by 2,
and we have x-\-y=s. Multiply (4) by (5), and divide by 4,
and we have xy=p.
(Art. 91.) No person can become very skilful in algebraic
operations as long as he feels averse to substitution; for judi-
cious substitution stands in the same relation to common algebra,
as algebra stands to arithmetic. The last example is an illustra-
tion of this remark. To acquire the habit of substituting, may
require some extra attention at first, but the power and advantage
gained will a thousand fold repay for all additional exertion.
When two equations involve quantities in the form (x-\-y) and
xy, or xy, and xy, and any product of these quantities, the
equations are easily solved.
The following examples will illustrate :
12. Given x++y= 197 find and
And x 2 -\-xy+y 2 =133$
Put x -\-y =5, and
Then s+p= 19 ()
And s 2 /=133 (E)
Divide (B) by (.#), and we have sp=7, &c.
Ans. #=9 or 4, y=4 or 9.
13. Given x 4 -}-2x z y 2 +y 4 =l2QQ-^4xy(x 2 -{-xy-{-y 2 ) and
x y=4, to find x and y.
Put x*-\-y 2 =s, and xy=p
Then the first equation becomes s 2 =1296 4p(s-\-p)
Multiply and transpose, and s 2 -\-4sp-\-4p 2 =l29Q
Square root +2p=36
But s-t-2/?=a: 2 -{-2^-f-2/ 2 ==36
Therefore x+= 6, or
Rejecting imaginary quantities, we find x=5 or 1, and
?/=l or 5.
X 2 'U Z
14. Given -- =G, and x-{-y-\-xy=U, to find the values
x y
of x and y. Jlns. x=5 or 1, y=l or 5
PURE EQUATIONS. HI
15. Given x*-{-y s =2xy(x-\-y], and xy=l6, to find the
values of x and y. Jlns. #=2^54-2, 7/^2^/5 2.
16. Given x 3 -\-y 3 , and x z y-}-xy*=a, to find the relative
values of x and y. tfns. xy.
1 7. Given x-\-y : x :: 5 : 3, and xy=Q, to find a? and y.
Ans. #=3, 2/=2.
18. Given #-f-?/ : a: :: 7 : 5, and a?7/-l-?/ 2 =126, to find a:
and y. Jlns. a? =15, i/=6.
19. Given ,T 2 -)-?/ 2 =, and xy=b, to find the values of
x and y. Ans. x=.* t Ja-}-2b-{-% l Ja < 2,b.
(Art. A)* Equations in the form of a? 4 2ax 2 -}-a z =b, require
for their complete solution, the square root of an expression in
the form of aJb ; for by extracting the square root of the
equation, we have
Hence x= \/ a
The right hand member of this equation is an expression well
known among mathematicians as
A BINOMIAL SURD.
Expressions in this form may or may not be complete powers ;
and it is very advantageous to extract the root of such as are
complete, for the roots will be smaller, and more simple quantities,
in the form of a'i^/6', or of .Ja'^jW.
Let us now investigate a method of extracting these roots ; and,
for the sake of simplicity, let us square 34-^7.
By the rule- of squaring a binomial, we have 94-6^/7-1-7,
Or, le+G/T;
Conversely, then, the square root of 164-6^/7, is 3+77.
* That the same Articles may number the same in both the School and Col-
lege Edition, we shall designate all additional Articles, in this volume, by
A, B, &c.
142 ELEMENTS OF ALGEBRA.
But when a root consists of two parts, its square consists of tht
sum of the squares of the two parts, and twice the product of
the two parts.
Now we readily perceive that 16 is the sum of the squares of
the two parts expressing the root; and 6^/7 , the part containing
the radical, is twice the product of the two parts.
To find what this root must be, let a? represent one part of the
root, and y the other :
Then aHy=16, (1)
And 2xy=&J7, (2)
Add equations (1) and (2), and extract square root, and we
have
(3)
Subtract equation (2) from (1), and extract square root, and
we have xy=^lQ Gj7~. (4)
Multiply (3) and (4), and we have
a? 2 2/2=^/256 252= > /T=:2 ; (5)
Add (1) and (5), and we have
#=, or
Sub. (5) from (1), and 2# 2 =14, or y=,
Wberie x-\-y, or the square root of 16-|-6^7 is S-f-^/7.
f e shall now be more general.
Take two roots, one in the form of
and one in the form of
Square both, and we shall have 2 2a > /6-f-6,
And a2,Jab-\-b.
In numerals, and, in short, in all cases, the sum of the squares
of the two parts of the root, as ( 2 +6), in the first square, and
(a-f-6), in the second, contains noradical sign ; and the sum of
these rational parts may be represented by c and c', and the
squares represented in the form of
PURE EQUATIONS. 143
c 2a/6,
or of c'2jab.
Hence, generally, if we represent the parts of the roots by y
and y, we shall have x?-}-y Z: = the sum of the rational parts, and
2xy= the term containing the radical.
The signs to x and y must correspond to the sign between
the terms in the power. If that sign is minus, one of the signs
of the root will be minus ; it is indifferent which one.
EXAMPLES.
1. What is the square root of ll-j-6^/2"? Jlns.
2. What is the square root of 7-f-4 A /3~7 Jlns.
3. What is the square root of 7 2^/10?
Jlns. 75 J2 or </2 J&1
4. What is the square root of 94+42^5'? Jlns. 7+3^5".
5. What is the square root of 28+1073? Jlns. 5-f-^
6. What is the square root of
np-}-2m z 2m t Jnp-\-m 2 1
In this example put anp-{-m*, and x and y to represent
the two parts of the root,
Then x 2 +y 2 m 2 +a,
and 2xy = 2m Ja.
Jlns. db( Jnp-{-m z m) .
. What is the square root of bc-+-2bjbc b 2 1
Jlns.
8. What is the sum of
Jlns. 10.
9. What is the sum of ^/l 1+6^/2 and A /7
3+^/5*.
10
Am. S-\-2^5 or 4.
144 ELEMENTS OF ALGEBRA.
In a similar manner we may extract the cube root of a bino-
mial surd, when the expression is a cube ; but the general solu-
tion involves the solution of a cubic equation, and, of course,
must be omitted at this place ? and, being of little practical utility,
we may omit it altogether.
(Art. 92.) Fractional exponents are at first very troublesome
to young algebraists ; but such exponents can always be ban-
ished from pure equations by substitution. For the exponents
of all such equations must be multiples of each other; otherwise
they would not be pure, but complex equations.
To make the proper substitution, place the unknown quanti-
ties having the lowest exponents equal to other simple quantities,
as P and Q ; and let this be a general direction.
EXAMPLES.
1 4. 2
1. Given x 3 -[-y 5 = G, and re 3 -}-;y s =20, to find the values
of x and y
2_ \_
By the above direction, put x 3 =P, and y 5 Q.
Squaring these auxiliaries, or assumed equations,
And x^=P 2 , and y~* = Q\
Now the original equations become
=Q (1)
=2Q (2)
By squaring equation (1), P 2 -f2P#-f-Q 2 =36.
Subtracting equation (2), we have 2PQ=IQ.
Subtracting this last from equation (2), and we have
By extracting square root P Q=2
But by equation (1), P J rQ= 6
Therefore, P=4 or 2, and Q=2 or 4,
fi
JL &VUV W, ~ =4 or 2, and?/ 5 =2 or 4.
Square root ^=2 or (2)* )
' 3 V y=32 or 1024
Cubing gives x=S or (2)* J
PURE EQUATIONS. 145
2. Given xy z +y=2l, and a?y+2/ 2 =333, to find the values
of x and y.
By comparing exponents in the two equations, we perceive
that if we put xy 2 =P, and y=Q, the equations become
P+Q= 21
P*+ $2=333
Solved as the preceding, gives P=18, Q=3.
From which we obtain a?=2, or T ^ , y=3 or 18.
3. Given x z +x?y*= 208
24 to find the values of x and y.
And
2
Assume x*=P, and y* = Q.
By squaring and cubing these assumed auxiliary equations,
we have 4
Seek the common measure (if there be one) between 208 and
1053.
From the above substitution, the given equations become
P*-\-P 2 Q= 208=13.16 (1)
$3-}-$2p=:1053=13.81 (2)
Separate the left hand members into factors, and
\3.lG (3)
13.81 (4)
Divide equation (4) by (3), and we have
Q 2 81 & 9
-^2=. Extracting square root -^=-
9P
Or QT-- Substitute this value of Q in equation (1),
q/33
And / M -f-^-=13.16
4
Or 4P4-9P 2 =13.64
That is, 13P 3 =13.6i
Hence, P 3 =64 or P=4
13
146 ELEMENTS OF ALGEBRA
But 2 =/ w =64. Therefore, #=8
As Q= and P=4, Q9 and #=27.
3 3 3 3_ 1
4. Given # 2 -f-# 4 i/ 4 -H/ 2 = 1009 =a
33 r to find a? and y.
And orM-a^v'-H/ 3 =582193=6
Put x*=P and #* = #
Then ar*=/ and y^=(f
And a- 3 = /* and y z = #*
Our equations then become
/_!_ PQ-L.Q* 1 Equations having no fractional exponents,
p\\piru\Qi t f an( l are of the same form as in Problem
J 12. (Art. 9 1.)
Jim. #=81 or 16, t/=16 or 8].
5. Given x-{-x*y* = l2 }
j j r to find the values of x and y.
And y+x*y*= 4 J
^ns. x=9, y=l
6. Given x-l-*V= |
j j r to find the values of x and y.
And y\-x*y*=zb }
a a* b*
rt jyns. x r^- v= ;-,
a+b 9 a-\-b
333 1
T. Given x 2 +x*y*=a [
3 j- to find the values of x and y
And 1 3
8. Given Jx-\-Jy : Jx Jy :: 4 : I, and a: 1/=16, to
find the values of x and y. As. a?=25, y=9.
9. Given *+y~ 5 tQ find
And ar =13
fins. x=9 or 4, y=4 or 9
PURE EQUATIONS. 147
JO. Given x-\-y : x y :: 3 : 1 } to find the values of
And * 3 */ 3 =56 $ * and y.
fins. #=4, 2/=2.
11. Given x -\-y =35 }
r to find the values of x and y.
And 3+7,3=5 J
CHAPTER V.
Problems producing Pure Equations.
(Art. 93.) We again caution the pupil, to be very careful not to
involve factors, but keep them separate as long as possible, for
greater simplicity and brevity. The solution of one or two of
the following problems will illustrate.
1. It is required to divide the number of 14 into two such parts,
that the quotient of the greater divided by the less, may be to the
quotient of the less divided by the greater, as 16 : 9.
fins. The parts are 8 and 6.
Let x= the greater part. Then 14 x= the less.
x 14 x
Per question, ~- - : -- : : 16:9.
14 x x
AT I ' 1 9X 16(14 X)
Multiply extremes and means, and - - = - - '-
14 x x
Clearing of fractions, we have 9# 2 =16(14 xf
By evolution, 3^=4(14 x) =4. 14 4x
By transposition, 7#=4.14
By division, #=4.2=8, the greater part.
14 "X
Had we actually multiplied - - by 16, in place of indi
rating it, the exact value and form of the factors would have been
lost to view, and the solution mighthave run into an adfected qua-
dratic equation.
The same remark may be applied to many other problems,
and many are put under the head of quadratics that may be re
duced by pure equations.
148 ELEMENTS OF ALGEBRA.
2. Find two numbers, whose difference, multiplied by the
difference of their squares, is 32, and whose sum, multiplied by
the sum of their squares, gives 272.
If we put x= the greater, and y= the less, we shall have
(z-.yX^-yH 32 (i)
And (*+2/)(* 2 -f3/ 2 )=272 (2)
Multiply these factors together, as indicated, and add the equa-
tions together, and divide by 2, and we shall have
o; 3 +y 3 =152 (3)
If we take (1) from (2), after the factors are multiplied, we
shall have 2 t ri/ 2 -f 2arty=240, or xy(x+y} = \'2Q (4)
Three times equation (4) added to equation (3) will give a
cube, &c. A better solution is as follows :
Let x-}-y= the greater number, and x y the less.
Then 2x= their sum, and 2y= their difference.
Also, 4xy is equal the difference of their squares, and
2x 2 -\-2y z = the sum of their squares.
By the conditions, 2yX4xy= 32
And 2#(2or i 4-2*/ 2 )=272
By reduction, xy*= 4
And
By subtraction, x 3 =64 or #=4
Hence, y=l, and the numbers are 5 and 3.
We give these two methods of solution to show how much
depends on skill in taking first assumptions.
3. From two towns, 396 miles asunder, two persons, Ji and Z?,
set out at the same time, and met each other, after traveling as
many days as are equal to the difference of miles they traveled
per day, when it appeared that A had traveled 216 miles. How
many miles did each travel per day? Let x=JPs rate, nnd i/=
fl's rate.
Then x y= the days they traveled before meeting.
By question, (x y)x=21G, and (x y)y=lSQ.
216 180 6 5
vonsequently, -- = - or -=-.
y ' x y x y
PURE EQUATIONS. 14
Therefore, ?/=|x, which substitute in the first equation, and
x 2
eve have (z f#)#=216, or --=216=6X6X6.
By evolution, #=36; therefore y= 30.
4. Two travelers, A and B, set out to meet each other, A
leaving the town C, at the same time that B left D. They traveled
the direct road between C and D ; and on meeting, it appeared
that A had traveled 18 miles more than B, and that A could have
gone 's distance in 1 5| days, but B would have been 28 days
in going ^'s distance. Required the distance between C
and D.
Let x= the number of miles Ji traveled.
Then x 18= the number B traveled.
x 18
=M s daily progress.
15 4
=J9's daily progress.
Sffo
Therefore, , : ,-18 : : f
if 4(x
Divide the denominators by 7, and extract square root, and we
Therefore, #=72 ; and the distance between the two towns
is 126 miles.
5. The difference of two numbers is 4, and their sum multi-
plied by the difference of their second powers, gives 1600.
What are the numbers 1 Jlns. 12 and 8.
6. What two numbers are those whose difference is to the
less as 4 to 3, and their product, multiplied by the less, is
equal to 504? dns. Hand 6.
150 ELEMENTS OF ALGEBRA.
7. A man purchased a field, whose length was to its breadth
as 8 to 5. The number of dollars paid per acre was equal
to the number of rods in the length of the field; and the
number of dollars given for the whole was equal to 13 times
the number of rods round the field. Required the length and
breadth of the field.
Jlns. Length 104 rods, breadth 65 rods.
Put 8,r=the length of the field.
8. There is a stack of hay, whose length is to its breadth as
5 to 4, and whose height is to its breadth as 7 to 8. It is worth
as many cents per cubic foot as it is feet in breadth ; and the
whole is worth at that rate 224 times as many cents as there are
square feet on the bottom. Required the dimensions of the stack.
Put 5x = the length.
Jlns. Length 20 feet, breadth 16 feet; height 14 feet.
9. There is a number, to which if you add 7, and extract the
square root of the sum, and to which if you add 16 and extract
the square root of the sum, the sum of the two roots will be 9.
What is the number ? Jlns. 9.
Put x 2 7= the number.
10. Jl and B carried 100 eggs between them to market, and
each received the same sum. If A had carried as many as
BI he would have received 18 pence for them; and if B had
taken as many as .#, he would have received 8 pence. How
many had each ? Jlns. Jl 40, and B 60.
/
11. The sum of two numbers is 6, and the sum of their cubes
is 72. What are the numbers ? Jlns. 4 and 2.
12. One number is a 2 times as much as another, and the pro-
duct of the two is b 2 . What are the numbers ?
b
Jlns. and ab.
a
13. The sum of two numbers is 100, the difference of
their square roots is 2. What are the numbers?
Jlns. 36 and 64.
PURE EQUATIONS. 151
Put x= the square root of the greater number,
And y= the square root of the less number ; or
Put x-\-y= the square root of the greater, &c.
14. It is required to divide the number 18 into two such parts,
that the squares of those parts may be to each other as 25 to 16
Let x= the greater part. Then 18 x= the less.
By the condition proposed, x*: (18 #) 2 ::25: 16.
Therefore, 16^=25(18 a?) 2
By evolution, 4a?= 5(18 a:)
If we take the plus sign, as we must do by the strict enuncia-
tion of the problem, we find #=10. Then 18 #=8.
And (10) 2 : (8) 2 ::25: 16
If we take the minus sign, we shall find a?=90.
Then 18 x= 18 90= 72.
And (90) 2 : ( 72) 2 ::25: 16; a true proportion, correspond-
ing to the enunciation ; but 18 in this case is not the number
divided, it is the difference between two numbers whose squares
are in proportion of 25 to 16.
15. It is required to divide the number a into two such
parts that the squares of those parts may be in proportion of b
to c.
Let a?= one part, then a x= the other.
By the condition, x 2 : (a x) 2 ::b:c
Therefore, cx 2 =b(a x) 2
By evolution, t jcx= l jb(a x)
Taking the plus sign, x=-~r- and a X-
ajb ajc
Faking the minus sign, x=- rr ^ -r- and a a?= ., *-.
Jb ,Jc
Prob. 14, is a particular case of this general problem, in which
a= 18, 6=25, and c=16; and substituting these values in the re-
sult, we find a?=10, and #=90, as before.
If we take b=c, the two divisions will be equal, each equal
to 5 a, when the plus sign is used ; but when the minus sign is
152 ELEMENTS OF ALGEBRA.
ajb ajb
used, :r= r =i ~^~t a symbol of infinity, as the denomi-
A/^ - V ^
nator is contained in the numerator an infinite number of times.
(Art. 58.) The other part, a x
symbol of infinity; and the two parts,
ajb ajc fl
Jb-Jc (Jb-Jc)
It may appear absurd, that the two parts, both infinite and
having a ratio of equality, (which they must have, if b=C) can
fetill have a difference of a. But this apparent absurdity will
vanish, when we consider that the two parts being infinite in
comparison to our standards of measure, can have a difference
of any finite quantity which may be great, compared with
small standards of measure, but becomes nothing in comparison
with infinite quantities. See (Art. 60.)
Application of the foregoing Problem.
(Art. 94.) It is a well established principle in physics that
light and gravity emanating from any body, diminish in inten-
sity as the square of the distance increases.
Two bodies at a distance from each other, and attracting at a
given point, their intensities of attraction will be to each other
as the masses of the bodies directly and the squares of their
distances inversely. Two lights, at a distance from each other,
illuminating at a given point, will illuminate in proportion to the
magnitudes of the lights directly, and the squares of their dis-
tances inversely.
These principles being admitted
16. Whereabouts on the line between the earth and the moon
will these two bodies attract equally, admitting the mass of the
earth to be 75 times that of the moon, and their distance asunder
30 diameters of the earth?
Represent the mass of the moon by c,
and the mass of the earth by 6,
their distance asunder by a.
PURE EQUATIONS. 153
The distance of the required point from the earth's centre,
represent by x. Then the remaining distance will be (a x).
Now by the principle above cited, x 2 : (a xf ::b:c.
This proportion is the same as appears in the preceding gen-
eral problem ; except that we have here actually made the ap-
plication, and must give the definite values to a, b and c.
ajb ajc
As before, x = rrn j- and a x= y
fl =30, 6=75, c=l.
x= ^ =26&, nearly. Hence, a a?=3.1, nearly.
If we take the second values for the two distances, from the
ajb
neral result, namely, #=- v . a
V^ v c
give the numeral values, we shall have
ajb ajc
general result, namely, #=- v . and a x= . v . , and
V^ v c v v c
*= 3.9, nearly.
These values show that in a line beyond the moon, at a
distance of 3.9 the diameters of the earth, a body would be
attracted as much by the earth as by the moon, and the value
of (a x) being minus, shows that the distance is now counted
the other way from the moon, not as in the first case towards
the earth ; and the real distance, 30, corresponding to a in the
general problem, is now a difference.
We may make very many inquiries concerning the intensity
of attraction on this line, on the same general principle.
For example, we may inquire, whereabouts, on the line be-
tween the earth and moon will the attraction of the earth be 1 6
times the attraction of the moon?
Let x= the distance from the earth.
Then a x the distance from the moon.
The attraction of the earth will be represented by
154 ELEMENTS OF ALGEBRA.
The attraction of the moon at the same point will be
By the
(-*)'
b 16c
3H=?
T> i Jb 4 V C
By evolution, ^_ = -t--^
x a x
Clearing of fractions, ajb l jbx : =4 t jcx.
Using the plus sign, x= ^ . =20.5, nearly.
Using the minus sign, x ,, A , 55.7, nearly, or
25.7 diameters of the earth beyond the moon.
Observe that the 4 which stands as a factor to Jc is the
square root of 16, the number of times the intensity of the
earth's attraction was to exceed that of the moon.
If we propose any other number in the place of 16, its
square root will appear as a factor to /c; we may therefore
inquire at what distance the intensity of the earth's attraction
will be n times that of the moon, and the answer will be from
the earth in a line through the moon,
and a J b
Jb .Jnc
The same application that we have made of this general prob-
lem to the two bodies, the earth and the moon, may be made
to any two bodies in the solar system ; and the same application
we have made to attraction may be made to light, whenever
we can decide the relative intensity of any two lights at any
assumed unity of distance.
(Art. 95.) This problem may be varied in its application to
meet cases where the distances are given, and the compara-
tive intensities of light or attraction are required.
For example, the planet Mars and the moon both transmit
the sun's light to the earth by reflection, and we now inquire
PURE EQUATIONS. 155
the relative intensities of their lights at given distances, and
in given positions.
If the surface of Mars and that of the moon were equal,
they would receive the same light from the sun at equal distance
from that luminary ; but at different distances equal surfaces
vould receive light reciprocally proportional to the squares of
their distances.
The surfaces of globular bodies are in proportion to the squares
of their diameters. Now let M represent the diameter of Mars
and m the diameter of the moon. Also, let R represent the dis-
tance of Mars from the sun, and r the distance of the moon from
the sun.
Then the quantity of light received by Mars may -be expressed
M 2
by -^j ; and the relative quantity received by the moon
m 2
by . But these lights, when reflected to the earth, must be
diminished by the squares of the distances of these two bodies
from the earth. Now if we put D to represent the distance of
Mars from the earth, and d the distance of the moon, we shall
M 2
have ~n2 for the relative illumination by Mars when the whole
m
enlightened face of that planet is towards the earth, and ~^-= for
the light of the full moon.
When the whole illuminated side of Mars is turned towards
the earth, which is the case under consideration, (if we take the
whole diameter of the body,) it is then in opposition to the sun,
and gives us light, we know not how much, as we have no
standard of measure for it ; but we can make a comparative mea-
sure of one by the other, and therefore the light of Mars in this
position may be taken as unity, and in comparison with this let
us call the light of the full moon x.
M 2 m 2
Theretor* *-
156 ELEMENTS OF ALGEBRA.
As the value of a fraction depends only on the relation of the
numerator to the denominator, to find the numeral value of x, it
will be sufficient to seek the relation of m to M> of R to r, and
of D to d.
7)/=4000 miles nearly, and m=2150 ; hence, -r>=^-
Jfl 80
/?=H4000000, and r=95000000 ; or =^
r 95
/)=144000000 95000000=49000000 or ;
a 24
That is, in round numbers, the light of the full moon is twenty-
seven thousand six hundred times the light of Mars, when that
planet is brightest, in its opposition to the sun.
We will add one more example by the way of farther illustra-
tion.
What comparative amount of solar light is reflected to the
earth by Jupiter and Saturn, when those planets are in opposi-
tion to the sun ; the relative diameter of Jupiter being to that
of Saturn as 111 to 83, and the relative distances of the Earth,
Jupiter and Saturn, from the sun, being as 10, 52 and 95, re-
spectively ?
.#iw. Taking the light reflected by Saturn for unity, that by
Jupiter will be expressed by 24. T \\ nearly.
The philosophical student will readily perceive a more ex-
tended application of these principles to computing the relative
light reflected to us by the different planets ; but we have gone to
the utmost limit of propriety, in an elementary work like this.
From Art. 94th to the end of this chapter can hardly be said
to be algebra ; it is natural philosophy, in which the science of
algebra is used ; however, we would offer no apology for thus
giving a glimpse of the utility, the cui bono, and the application
of algebraic science.
QUADRATIC EQUATION. 157
SECTION IV.
QUADRATIC EQUATIONS.
CHAPTER I.
(Art. 96.) Quadratic equations are either simple or compound.
A simple quadratic is that which involves the square of the un-
known quantity only, as or*=; which is one form of pure equa-
tions, such as have been exhibited in the preceding chapter.
Compound quadratics, or, as most authors designate them, ad-
fectf.d quadratics, contain both the square and the first power
of the unknown quantity, and of course cannot be resolved as
simple equations.
All compound quadratic equations, when properly reduced,
may fall under one of the four following forms :
(1) x*+2ax=b
(2) x 2 2ax=b
(3) X s 2ox= -b
(4) x-f 2ax=b
If we take z-f-a and square the sum, we shall have
If we take x a and square, we shall have
If we reject the 3rd terms of these squares, we hare
3~-\-2ax, and x* 2ax
The same expressions that we find in the first members of the
four preceding theoretical equations.
It is therefore obvious that by adding a* to both sides of the
preceding equations, the first members become complete squares.
But in numeral quantities how shall we find the quantity corres-
ponding to cr ? We may obtain a* by the formal process of
taking half the coefficient of the first power of x, or the half of
2 or 2, which is a or a, the square of either being a*.
Hence, when any equation appears in the form of x 2 2x=
rt& ,we may render the first member a complete square, and effect
a solution by the following
158 ELEMENTS OF ALGEBRA.
RULE. Add the square of half the coefficient of the lowest
power of the unknown quantity to the first member to complete
its square; add the same to the second member to preserve the
equality.
Then extract the square root of both members, and ive shall
have equations in the form of
Transposing the known quantity a and the solution is accom-
plished.
In this manner we find the values of x in the four preceding
equations, as follows :
(i) *=.
(2) *=
(3) *=
(4) *=.
When b is greater than a 2 equations (3) and (4) require the
square root of a negative quantity, and there being no roots to
negative quantities, the values of x in such cases are said to be
imaginary.
The double sign is given to the root, as both plus and minus
will give the same power, and this gives rise to two values
of the unknown quantity ; either of which substituted in the
original equation will verify it.
After we reduce an equation to one of the preceding forms,
the solution is only substituting particular values for a and b ;
but in many cases it is more easy to resolve the equation as an
original one, than to refer and substitute from the formula.
(Art. 97.) We may meet with many quadratic equations that
would be very inconvenient to reduce to the form of x 2 -{-2ax=b;
for when reduced to that form 2a and b may both .be
troublesome fractions.
Such equations may be left in the form of
An equation in which the known quantities, c, 6, and c, are all
whole numbers, and at least a and b prime to each other.
QUADRATIC EQUATIONS. 159
We now desire to find some method of making the first mem-
ber of this equation a square, without making fractions. We
therefore cannot divide by 0, because b is not divisible by a,
the two letters being prime to each other by hypothesis. But
the first term of a binomial square is always a square. There-
fore, if we desire the first member of our equation to be convert-
ed into a binomial square, we must render the first term a
square, and we can accomplish that by multiplying every term
by a.
The equation then becomes
a 2 x 2 -}-bax=ca
Put y=ax. Then y z +by=ca
Complete the square by the preceding rule, and we have
.b 2 .b 2
y 4-%-f j=H-j
We are sure the first member is a square ; but one of the terms
is fractional, a condition we wished to avoid ; but the denomina-
tor of the fraction is 4, a square, and a square multiplied by a
square produces a square.
Therefore, multiply by 4, and we have the equation
4y 2 +4by+b 2 =4ca-{-b 2
An equation in which the first member is a binomial square and
not fractional.
If we return the values of y and y z this last equation becomes
4a 2 x 2 +4abx-\-b 2 =4ac-}-b 2
Compare this with the primitive equation
ax z -}-bx= c.
We multiplied this equation first by 0, then by 4, and in ad-
dition to this we find b 2 on both sides of the rectified equation, b
being the coefficient of the first power of the unknown quantity.
From this it is obvious that to convert the expression ax 2 -\-bx
into a binomial square, we may use the following
RULE 2. Multiply by four times the coefficient of x 2 , and
add the square of the coefficient of x.
To preserve equality, both sides of an equation must be mill-
160 ELEMENTS OF ALGEBRA.
ti plied by the same factors, and the same additions to both sides.
We operate on the first member of an adfected equation to
make it a square, we operate on the second member to preserve
equality.
(Art. 98.) For the following method of avoiding fractions in
completing the square, the author is indebted to the late Professor
T. J. Matthews, of Ohio.
Resume the general equation ax 2 -\-bx=c
u , u* bit
Assume x=- Ihen aar= and bx=
a a a
The general equation becomes + =c
a a
Or u 2 +bu=ac
Now when b is even, we can complete the square by the first
rule without making a fraction. In such cases this transforma-
tion is very advantageous.
When b is not even, multiply the general equation by 2, and
the coefficient of x becomes even, and we have
2c (1)
Assume #= - Then 2ax 2 = and 2bx=
2a 2a 2a
With these terms, equation (1) becomes
u 2 2bu
2a + W
Or u*-\-2bu=4ac
Complete the square by the first rule, and we have
An equation essentially the same as that obtained by completing
the square by the rule under (Art. 97.) ; for we perceive the sec-
ond member is the same as would result from that rule ; hence
this method has no superior advantage except when b is even, in
the first instance.
(Art. 99.) The foregoing rules are all that are usually given
for the resolution of quadratic equations ; but there are some
QUADRATIC EQUATIONS. 161
intricate cases in practice that we may meet with, where
neither of the preceding rules appear practical or convenient.
To master these with skill and dexterity, we must return to a
more general and comprehensive knowledge of binomial squares,
x* -\-2ax-\- a 2 is a simple and complete binomial square. Let
us strictly examine it, and we shall perceive,
1st. That it consists of three terms ;
2d. Two of its terms, i\\e first and the third, are squares;
3d. The middle term is twice the product of the square
roots of the first and last term.
Now let us suppose the third term, a\ to be lost, and we have
only x z -\-2ax. We know these two terms cannot make a square,
as a binomial square must consist of three terms.*
We know also that the last term must be a square.
Let it be represented by t*.
Then, by hypothesis, x z ^-2ax-^t* is a complete binomial
square.
It being so, 2xt2ax, by the third observation above.
Therefore, t=a and t*=a 2
Thus a z is brought back.
! Again, 4a 2 -}-4ab are the first and second terms of a bi-
nomial square ; what is the 3rd term ?
Let t 2 represent the third term.
Then 4 2 -f 4ab-}-t 2 is a binomial square.
Hence, 4at4ab or t=b and / 2 =6 2
That is, t 2 represented the 3d term, and b 2 is the identical 3d
lerm, and 4a 2 -\-4ab-\-b 2 is the actual binomial square whose root
Is 2a+b.
2. 36?/ 2 -}-36?/ are the first and 2d terms of a binomial square,
what is the 3d term ? JLns. 9.
3. -\ \-9 are the 2d and 3d terms of a square, what is the
x !
first? Ans. ?
x 2 -
* In binomial surds two terras may make a square, and this may condemn
the technicality here assumed ; but it is nothing against the spirit of this ar-
ticle.
14
162 ELEMENTS OF ALGEBRA.
4Qx 2
4. -- 49 are the 1st and 2d terms of a binomial square,
49
what is the 3d 1 Jlns. .
tt
5. 9?/ 2 Qy are the 1st and 2d terms of a binomial square,
what is the 3d? rfns. 1.
6. tfx*-\-bx are the 1st and 2d terms of a binomial square,
b 2
what is the 3d 'I Jlns.
4a
(
7. 8 la; 2 3 are the 1st and 3d terms of a binomial square,
what is the 2d or middle term ? fins. 18.
8. 2/ 2 Sx^y are the 1st and 2d terms of a binomial square,
what is the 3d ? rfns. I6x.
9. -- ?r+36 are the 2d and 3d terms of a binomial square,
1 7
or 2
what is the 1st ? Ans. - -.
36 1
77 2
10. ^--}-36 are the 1st and 3d terms of a binomial square,
301
what is the middle term ? rfns.
iy
11. If #-}--- are the 2d and 3d terms of a binomial square,
lo
what is the 1st term ? rfns. 4x*.
Qx*
12. The 1st term of a binomial square is the 2d term is
u
:12, what is the 3d term ? Ans. -^-.
(Art. 100.) Adfected quadratic equations, after being reduced
to the form of x?-\-2axb, can be resolved without any formality
of completing the square, by the following substitution :
QUADRATIC EQUATIONS. 163
Assume xy o,
Then a^=?/ 2 2a?/+ 2
And 2ax= -\-2ay-2a*
By addition, x*+2ax=y z a z b
Hence, y= l Jb-\-a z
And x= a-i l Jb-^-a 2 j the same result as
may be found in equation (1), (Art. 96.)
RULE FOR SUBSTITUTION. Assume the value of the unknown
quantity equal to another unknown, annexed to half the coeffi-
cient of the inferior power with a contrary sign.
(Art. 101.) For further illustration of the nature of quadratic
equations, we shall work and discuss the following equation :
Given ^+4^=60, to find x.
Completing the square, (Rule 1st.) 2 +4aH-4=64.
Extracting square root, a?-f-2=rb8.
Hence, x=6 or x= 10.
That is, either plus 6, or minus 10, substituted for x in the
given equation, will verify it.
For 6H-4X 6=60. Also, ( 10) 2 4X 10=60
If x=6 then x 6=0
Ifa:= 10 then z+10=0
Multiply these equations together, and we have
x 6
x +10
Product, #M-4aN 60=0
Transpose, and 0^+42?= 60, the original equation.
Thus we perceive, that a quadratic equation maybe considered
as the product of two simple equations, and these values of x in
the simple equations are said to be roots of the quadratic, and this
view of the subject gives the rationale of the unknown quantity
having two values.
164 ELEMENTS OF ALGEBRA.
In equations where but one value can be found, we infer that
the other value is the same, and the two roots equal, or one of
them a cipher.
EXAMPLES FOR PRACTICE.
1. Given x 2 6# 7=33, to find x. Jlns. #=10 or 4.
2. Given x 2 20#= 96, to find x. Jlns. 12 or 8.
3. Given x*-{-6x-{-l=92, to find x. Jlns. 7 or -13.
4. Given y 2 +12y=589, to find y. Jlns. 19 or 31.
5. Given t/ 2 6y-|- 10=65, to find y. Jlns. 1 1 or 5.
6. Given a?-}-12;c-}-2=110, to find x. Jlns. 6 or 18.
7. Given x 2 14z=51, to find x. Jlns. 17 or 3.
8. Given a? 2 -f6a:-|-6=9, to find x. Jlns. 32,y3.
9. Given ar4-8a?=12,to findtf. Jlns..
10. Given a?-{-l2x=lQ, to find x. Jlns.
The reader will observe that the preceding examples are in,
or can be immediately reduced to the form of # 2 2a#=6, and of
course their solution is comparatively easy. The following are
mostly in the form of ax*-}-bx=c.
11. Given 5#*-j-4:e=204, to find x.
According to (Art. 98,) put x-. Then 5# 2 = and
4w w 2 4w
4x , and the equation becomes _-f ^-=204.
o o 5
Clearing of fractions, w 2 +4tt=1020.
Completing the square and extracting the root, we have,
w-f-2=rb32, or w=30 or 34
But x=~. Therefore, x 6 or , Jlns.
5 5
12. Given 5^+4#=273, to find x. Am. 7 or 7 4 S
13. Given 7a? 2 20z=32, to find x. Jlns. 4 or f .
14. Given 25# 2 20#= 3, to find x. Jlns. | or ^.
15. Given 21 x 2 292#= 500, to find x.
ins. Ilifor2
QUADRATIC EQUATIONS. 165
16. Given 2x 2 5#=117, to find x.
Here, as 5 or b of the general equation is not even, we must
multiply the whole equation by 2, to apply the above principle ;
or we may take Rule 2. (Art. 97.)
Multiply by 8, and add 5 2 or 25 to both members.
Then 16^40a:4-25=961
Square root, 4x 5=31. Hence, x =9 or 65.
(Art. 102.) It should be observed that all quadratic equations
can be reduced to the form of x z .2ax=b t or, as most authors
give it, x z px=q; but when the terms would become fractional
by such reduction, we prefer the form ax*bx=c, for the sake
of practical convenience, as mentioned in (Art. 97.)
(Art. 103.) It is not essential that the unknown quantity
should be involved literally to its first and second powers ; it is
only essential that one index should be double that of the other.
In such cases the equations can be resolved as quadratics. For
example, a? 6 4^=621 is an impure equation of the sixth
degree, yet with a view to its solution, it may be called a quad-
ratic. For we can assume y=x 3 ; then y 2 =x 6 , and the equa-
tion becomes y z 4i/=621, a quadratic in relation to y, giving
y=27, or 23.
Therefore, #"=27 or 23
And a?=3 or V 23.
There are other values of x; but it would be improper to seek
for them now; such inquiries belong to the higher order of
equations.
3
For another example, take x 3 x 2 =56, to find the values
of x.
Here we perceive one exponent of x is double that of the other,
it is therefore essentially a quadratic.
Such cases can be made clear by assuming the lowest power
of the unknown quantity equal to any simple letter. In the
3
present case assume y=x 2 ; then y 2 =# 3 , and the equation is
166 ELEMENTS OF ALGEBRA.
By Rule 3, 4y*4y+ 1=225
By evolution, 2y 1=15
Hence, y=& r 7
3 3
And by returning to the assumption y=x^ we find # 2 8,
or a? 4 =2. Hence, a?=4; or, by taking the minus value of y,
(Art. 104.) When a compound quantity appears under differ-
ent powers or fractional exponents, one exponent being double
that of the other, we may put the quantity equal to a single letter,
and make its quadratic form apparent and simple. For example,
suppose the values of x were required in the equation
Assume J2
Then by involution, 2# 2 -}-3#-f 9=?/ 2 (A]
And the equation becomes y 2 5y=6 (J9)
Which equation gives y=& or 1. These values of y, sub-
stituted fory in the equation (fi give 2,r 2 -j-3x-j-9=36
Or
From the first of tl^e we find a?=3 or
From the las* we find a?=|( 3dby 55,) imaginary quan-
tities.
EXAMPLES.
I. Given (#+12)*+ (a>H2)* = 6 to find the values of x.
rfns. x=4 or 6U
nd tlie values of or
Ans. xb*a or 81 b' 1 a.*
* It is proper to remark, that in many instances it would be difficult to verify
the equation by taking the second values of x, as by squaring, the minus quan-
tity becomes plus, and in returning the values, there is no method but trial to
decide whether we shall take a plus or a minus root. Hence, these second
answers are sometimes called roots of solution. In many instances hereafter,
we shall give the rational and positive root only.
Given (aH-</)~*-f 2&(#-f-a)* =3b 2 , to find the values of ar
QUADRATIC EQUATIONa 107
3. Given 9;r+4+2,/9a'+4=15, to find the values of x.
Jlns. a?=f J.
4. Given (10+z)* (10+ar)*=2, to find x
Jlns. a?=6.
5. Given (# 5) 3 3(tf 5)2=40, to find x.
Ans. o?=9.
6. Given 2(1 -far a? 2 ) (l-f a; ^)^+^=0, to find a%
7. Given s+16 3(tf+16f=10, to find a;. Am. x=Q.
8. Given So.- 2 ' 1 2o? n =8, to find a*. #ns. o?=%/2.
9. Given 3^+^=756, to find x. Jlns. a?=243.
Q -I rt
1O. Given - -- r^ 1 "!" -- to ^ n( ^ ^ ^ n5 - a7 = 3 or 1 -
11. Given 4a?+=39, to find x. Am. a?=729.
12. Given ^2^+6(^2^+5)^ = 11, to find x.
Jlns. x=l
it 2 12ic
13. Given - T^ == 32, to find the value of #..
rfns. a?=152 or 76.
If much difficulty is found in resolving this 13th example, the
pupil can observe the 9th example, (Art. 99).
14. Given 8 la^+ 17+^=99, to find the values of x. ]
Jlns. x=l, or 1, or |.
Observe that the 1st and 3d terms of the first number are
squares, see (Art. 99.)
1 R41 2^2
15. Given 81^+17+-,-=-^-+ +15, to find x. J
X XT X
rfns. x=2 or 1J.
16. Given 25# 2 +6+ '--, to find the values of x.~-
Jlns. a? 2, or 2, or
15
168 ELEMENTS OF ALGEBRA.
IT. Given -^--j--|=6, to find the values of x.
Ans. x=7 or 11J.
(Art. 105.) Equations of the third, fourth, and higher degrees,
can be resolved as quadratics, provided we can find a compound
quantity in the given equation involved to its first and second
power, with known coefficients.
To determine in any particular case, whether such a com-
pound quantity is involved in the equation, we must transpose
all the terms to the first member, and if the highest power of
the unknown quantity is not even, multiply every term of the
equation by the unknown letter to make it even, and then extract
the square root, to two or three terms, as the case may require ;
and if we find a remainder to be any multiple or any aliquot
part of the terms of the root, a reduction to the quadratic form
is effected ; otherwise it is impossible, and the equation cannot
be resolved as a quadratic.
For example, reduce the following equation to the quadratic
form, if it be possible.
1. Given x 4 8a# 3 -f 8aV-f32a 3 a>- 9a 4 =0, to find the values
of x by quadratics.
OPERATION.
a- 4 Sax 3 +8aV+32a 3 *--9a 4 ==0
This remainder can be put into this form :
Now we observe the original equation can be written thus :
9a 4 =0
By putting # 2 4ax=y we have
y 2 8a 2 y=9a* a quadratic.
Completing the square?/ 2 8a 2 y-j-16a 4 =25a 4
By evolution y 4a 2 =5a 2
Hence y=9a 2 2
QUADRATIC EQUATIONS. 109
Or x 2 4ax=Qa 2 or a 2
Completing the squares 2 4ax-\-4a 2 =l3a 2 or 3a 2
By evolution x 2a=a l jl3 or aj3
Hence x may have the four following 1 values (2-{->/iy)
(2aaJ~T3), (2-f aj3), (2aaj3). Either of which being
substituted in the original equation will verify it.
2. Reduce a7 3 -f-2ax 2 +5a 2 ^-f 4 3 =0 to a quadratic.
As ihe highest power of x is not ct'en, we must multiply by x
to make it even. Then
x 4 +2ax 3 +5aV-f-4tt 3 .r =0
By extracting two terms of the square root, and observing the
remainder, the part that will not come into the root, we find that
Divide by (x 2 -{-ax) and x z -\-ax-}-4a 2 =Q a quadratic.
3. Given x*+2x*7x 2 8#-hl2=0, to find the values of x.
This equation may be put in the following form :
Ans. x=l or 2, or 2 or 3.
4. Given x 3 8^ 2 -fl9 x 12=0, to find the values of x.
Am. x=l or 3 or 4.
5. Given a: 4 10^ 3 -|-35a; 2 50#-{-24=0, to find the values
of a?. JJns. x=l, 2, 3 or 4.
6. Given x 4 2x 3 +a?=132, to find the values of x.
Jjns. x=4 or 3.
7. Given i/ 4 2n/ 3 -f(c 2 2)t/ 2 -f 2cy=C", to find the values
f y-
(Art. 106.) The object of this article is to point out a few lit-
tle artifices in resolving quadratics, which apply in particular
cases only, but which at times may save much labor. It is there-
fore proper that they should be presented, though some minds
prefer uniformity to facility.
15
J70 ELEMENTS OF ALGEBRA,
For example, take equation (J5) (Art. 104.)
1. y* -5y=6 Put 2a=5
Then y* 2ay=2a+l.
Add a 2 to both sides to complete the square, (Rule 1 .)
And i/ 2 2ay-f-a 2 = 2 +2a-}-l
By evolution, y fl=(a-f-l.)
Hence, y=2a-\-l=G or 1
2. Given y* 7i/=8, to find y. Jlns. y=8 or 1.
3. Given a: 2 -j-ll#=26, to find the values of x.
Assume 2a=ll ; then 4a-J-4=26.
Now put these values in place of the numerals, and complete
the square, and o; 2 -{-2rt#-J-a 2 =a 2 -|-4a+4.
By evolution, x -\-a=(a-\-2] Hence, x=Z or 13.
4. Given x 2 17^=60, to find the values of x.
Assume 2a=17; then 6a-f-9=60
And ** 2a?+a 2 =a 2 +6a-{-9.
By evolution, x a(a-\-3.) Hence, rr=20 or 3.
5. Given a^-f-19#=92, to find the values of x.
Assume 2a=19 ; then 8+ 16=92
Putting these values and completing the square we have
a?-}-2ax + 2 =a 2 +8a4- 1 6
ar-{-a=(a-r-4) or x=4 or 23.
Observe that in the preceding equations we invariably put the
coefficient of the first power of the unknown quantity equal 2a.
Then if we find the absolute term in the second member of the
equation equal to 2a-{- 1
or 4a-f 4
or 6a-f- 9
or 8a-}-16
Or, in general, m2a-\-m z . That is, any multiplier of 2 plus
the square of the same multiplier equal to the second member,
then the equation can be resolved in this manner ; for in fact one
QUADRATIC EQUATIONS. 171
of the roots of the equation is this multiplier of 2a, and the other
root is (2a-j-77i), m being the multiplier, and it may represent
any number, integral or fractional ; but there is no utility in ope-
rating by this method unless m is an integer, and not very large
To present a case where m is fractional, we give tie following
equation: x z 9#=y, to find the value of x.
Put 20=9; then Ix2a+J= l 4 9 an( ^ ^ ie e q uat i n becomes
Therefore, x a=(fl-H). Hence, x= or 2a+s=9|.
(Art. 107.) When the roots of the equation are irrational or
surd, of course this method of operation will not apply ; but we
can readily determine whether the roots will be surd or not. For
example, take the equation # 2 -r"13#=40.
Put 2a=13; then 4a-|-4=30 And 60+9=48
From this, we observe that one of the roots of the equation
lies between 2 and 3.
(Art. 108.) When the roots of an equation are irrational or
surd, no artifice will avail us, and we must conform to set rules ;
but when the roots are small integers, we can frequently find
some method to avoid high numeral quantities ; but special artifi-
ces can only be taught by examples, not by precept. The follow-
ing are given as examples :
1. Given r j -f-9984#=l 60000, to find the values of x.
Observe that 9984=1000016
Put 2a=10000; then 320=160000
These substitutions transform the equation to
Completing the square by (Rule 1) and
a*+(2 16)a?-r-(a 8) 2 =a 2 -f-16a-f-64
By evolution, x-\-(a 8)=db(a-f-8)
Hence, #=16 or 2a= 10000.
2. Given # 2 +45#=9000, to find the values of x.
If we put 20=45, the multiplier and its square, requisite to
172 ELEMENTS OF ALGEBRA.
produce 9000, is so large that it is not obvious, and of course
there will be no advantage in adopting this method ; at the same
time, we wish to avoid the high numerals we must encounter by
any set rule of solution;
We observe that 45 X 200=9000. Put a=45
Then x?+ax=2QOa
Complete the square by (Rule 2,) and
(13?+ 2 =a 2 +8000
By evolution, 2#+a=ya(a+800)= > /45X845
Multiply one of the factors, under the radical, by 5, and divide
the other by 5, and the equivalent factors will be 225 X 169, both
squares. Taking their root, resuming the value of a, and the
equation becomes
2;r+3.15=13.J5
Drop 3.15 from both sides
And 2a?=10.15 or #=75, rfns.
3. Given IQx 2 225#=225, to find the values of x.
This equation is found in many of the popular works on
algebra, and in several of them the common method of resolving
it may be seen.
Observe that 225=15X15.
Put a=15; then a+l = 16, and the equation becomes
Completing the square by (Rule 2), and
4(+])V 4(+l)a 2 a: +a 4 =a 4
By evolution, 2(a+l)a? 2 =<z 2 +2a
Transpose a 2 and divide by 2, and we have
Divide by (0+1) and x=a=\ 5, Jlns
We give one more example of the utility of representing nume-
rals, or numeral factors, by letters, in reducing the following
equation :
QUADRATIC EQUATIONS. 173
18 , 81 rr 2 r 2 65
4. Given --[ ^ -
By examining the numerals, we find 9, and several multiples
of 9. Therefore, let a=9, and using a in the place of 9 the
equation becomes
2a , a 2 y?_x* 65
"P" 1 ^~ : : ~~&T~
Clearing of fractions, we have
16 2 -f8a 2 ;r 8x*=x 4 65^
Transposing all to one side, and arranging the terms according
to the powers of .T, we have
16a=0
; 16a 2
Or 2 (x 2 4-8^+16)
Therefore, by (Art. 105,) the equation becomes
Or (ar+4)V= 2 (a:+4) 2
By division, ic 2 = 2
And a; ==9,
The preceding examples may be of service in reducing some
of the following
EXAMPLES.
1. Given a; 2 -}- 11 a? =80, to find a?. ^ns. a:=5, car 16
.
2. Given 5a? -- -=2a?-f - - , to find a?.
# 3 *4
rfns. ar=4, or 1
3. Given - ^+^ ~~"fi"' to fill(1 ^ *^ n5 ' a?=2
174 ELEMENTS OF ALGEBRA.
72a? 250
4. Given 7x-{-- ^-=50, to find x. Ans. #=2, or -
10 3x 21
/6 V /6 \
5. Given ( h*/)-H hy)=30, to find tne values of
Ans. y=3 or 2, or
_4
6. Given # 3 -|-7# 3 =44, to find the values of x.
Am. #=8 or (
. Given ^-j-ll-j-^/i/ 2 -!-! 1+2=44, to find the values of y.
fins. 2/=5 or
it -1-7
it -- --#
8. Given 14-}-2.r -- =#4- - , to find the values of x.
Ans. #=28 or 9.
9. Given 3z? Qx 4=80, to find the values of x.
Am. x=7 or 4.
10. Given =T, to find x. Ans. x=4.
11. Given -{-x -2=24 3x, to find the values
x 3
of x. Ans. x=G or 5.
12. Given --- : = , to find the values of x.
x x* 9
A)is. a? 3 or y' f .
-^ _ l Ox 2 4-1
13. Given - ^ r-jr- =x 3, to find the values of x.
XT \)X~\~ J
Ans. x=l or 28.
14. Given ma? 2mxjn=nx* mn, to find x.
a Jmn
Ans. x=
15. Given # 4 H -=34a?+16, to find the values of x.
Z
Exms. Art. 99.) Ans ,r=2, or 2, or 8, or 5.
QUADRATIC EQUATIONS. 175
16. Given ^(J- V 3 =^, to find the values of *.
N.B. Put -1-=. fins. x=S or 9.
17. Given y* ty 2 =y4-8y-f-12, to find the values of y.
(See Art. 99.) Jlns. y=3 or 2.
o
18. Given x 1=2+ , to find the values of a:.
Jx
Am. x=4 or 1
19. Given (-^~ -Y=# 2, to find the values of x
\zJx*9/
tfns. x=5 or 3.
CHAPTER II.
Quadratic Equations, containing two or more unknown
quantities.
(Art. 108.) We have thus far, in quadratics, considered equa-
tions involving only one unknown quantity; but we are now
fully prepared to carry our investigations farther.
Two equations, essentially quadratic, involving two unknown
quantities, depend for their solution on a resulting equation of the
fourth degree.
This principle may be shown in the following manner :
Two equations, essentially quadratic, and in the most gencrul
form, involving two unknown quantities, may be represented
thus :
=0,
We do not represent the first terms with a coefficient, as any
coefficient may be reduced to unity by division ; and a, b, c, &c.,
and a', b' t &c. may represent the result of such division ; and,
of course, may be of any value, whole or fractional, positive or
negative.
176 ELEMENTS OF ALGEBRA.
Arranging the terms, in the above equations, in reference to
we have
a*+(m/+c)x+hf+dy+e =0, (1)
>=Q. (2)
13 y subtracting (2) from (1), we have
'}f+(dd>)y+e e'=0;
Therefore
Tliis expression for a?, substituted in either equation (1) or (2)
will give a final equation, involving only one unknown quan
tity, y.
But to effect this substitution would lead to a very complicated
result ; and as our object is only to show the degree to which
the resulting equation will rise, we may observe that the ex-
pression for the value of x is in the form of r -. This
ry+s
put in either of the equations (1) or (2), its square, or the ex-
pression for ar 2 , will be of the fourth degree ; and no term can
contain y of a higher degree than the fourth.
Therefore, in general, the resolution of two equations of the
second degree, involving two unknown quantities depends upon
that of an equation of the fourth degree involving one unknown
quantity.
(Art. 109.) Two or more equations, involving two or more
unknown quantities, can be resolved by quadratics, when they
fall under one of the following cases :
1st. One of the equations only may be quadratic; the other
must be simple, or capable of being reduced to a simple form.
2d. The equations must be similar in form, or the unknown
quantities similarly involved or combined in a similar manner,
as they combine in regular powers ; or,
3d. The equations must be homogeneous ; that is, the expo-
nents of the unknown quantities must make the same sum in
every term.
In the first and second oases, we eliminate one quantity in
QUADRATIC EQUATIONS. J77
one equation and substitute its value in the other, or perform an
equivalent operation, by rules already explained.
In the third case, we throw in a factor to one unknown quan-
tity, to make it equal to the other, or assume it to be so ; but
these principles can only be explained by
EXAMPLES.
ax* -\-bxy -\- cy z =e,
These are homogeneous equations, for the exponents of the
unknown quantities make the same sum 2, in every term. In
such cases, assume x=vyi then the equations become
__
or =
a'v*+b'v+c'
An equation involving the 1st and 2d powers of v, and of
course, a quadratic.
The solution of this equation will give v. Having v, we
have y 2 and y, and from vy we obtain x.
For a particular example, we give the following :
1. Given 4 ^-- 2; /= 12 I to find the values of x and y.
Put x=vy ; then the equations become
4v 2 y 2 2vy 2 =12, or y*=-
u 2 v)2
And
f* Q
Whence, = - =-. Dividing by 2, then clearing of
Zv v 2-\-3v
fractions, we have
v=8t> 2 4.v or 8u 2
12
178 ELEMENTS OF ALGEBRA.
This last equation gives v=2 or |.
8 8
Omitting the negative value y*= -r- =-=l.
-- o
Therefore, 7/=l, and x=vy=2.
2. Given 2x 3y=l, and 2x*-}-xy 5y z =2Q, to find the
values of x and y.
These equations correspond to the first observation, one of
them only being quadratic, the other simple ; and the solution
is effected by finding the value of x in the first equation. Sub-
stituting that value - in the 2d, and reducing, we have
2?/ 2 -{-77/=39, which gives 2/=3. Hence, x=5.
3. Given & 2 -\-y z x #=78, and xy+x-\-y=39, to find
the values of x and y.
In these equations x and y are similarly involved, not equally
involved ; nor are the equations homogeneous. In cases of this
kind, as we have before remarked, a solution by a quadratic can
be effected, but no general or definite rule of operation can be
laid down ; the hitherto acquired skill of the learner, and his
power of comparison to discern the similarity, will do more than
any formal rules.
To resolve this example, we multiply the 2d equation by 2,
and add the product to the first ; we then have
l5G, or
Put x-}-y=s ;' then s 2 -f-s=156, a quadratic, which gives .9,
or x-{-y=l2. This value of or-J-y, taken in the second equa-
tion, gives xy=27. From this sum and product of x and y,
we find a?=9 or 3, and y=3 or 9.
Again, after we multiply the second equation by 2, if we sub-
tract it from the first, we shall have
or (xy) 2 (a?+y)=0
or (# y) 2 =3Xl2=36
or x yQ
But x+y=l$
Hence, 2#=18 or 6, and #=9 or 3, as before.
QUADRATIC EQUATIONS. 179
(Art. 110.) There are some equations to which the foregoing
observations do not immediately apply, or not until after reduc-
tions and changes take place. The following is one of them.
4. Given y > to find the values of x and y.
Here neither of the equations is simple, nor are both letters
similarly involved, nor are the equations homogeneous ; yet we
can find a solution by a quadratic, because the two equations
have a common compound factor, which taken away, will bring
the equations far within the limits or condition laid down ; and
this remark will apply to all problems that can be resolved by
quadratics not seemingly within the limits of the three con-
ditions.
12
From the first of these equations, we have y
From the second, - ... y= .
12 18
Hence, ~- = Divide the denominators by (#+1)
X"-\~X X + 1
find the numerators by 6, and we have
2 3
~ == Z5 ITTT a Q ua ^ratic equation.
Clearing of fractions, and 2# 2 2x-{-2=3x
or 2.x 2 5#= -2.
(Rule 2.) I6x 2 .#+25=2516=9.
We write Jl to represent the second term. It is immaterial
what its numeral value may be, as it always disappears in taking
the root.
By evolution, 4# 5= 3
Hence, o?=2 or 5.
12 12 12 12X4
* r
The following is of a similar character :
180 ELEMENTS OF ALGEBRA.
5 Given $ **"" 2/ 2 ~ (*+#) = 8 ? to find the values of 3,
W$ andy.
Divide the first equation by (x-\-y) and x y 1 -- (.#).
x-{-y
oo
Divide the second by (x-\-y) and (x 2/) 2 =-- (B).
Put x-\-y=s, and transpose minus 1, in equation (.#), and
=* + l. By squaring, (^-2/) 2 =^4-y-M.
00
Equation (5) gives (x y) 2 =
5
Therefore, V- 6 +l=??.
S' 2 5 5
Clearing of fractions, and transposing 32s, we have
64 -16s-|-s 2 =0
By evolution, 8 s=0 or s=8. That is, a:+y=8, which
value, put in equation (^j, gives x y=%.
Whence, x=5 and 2/=3.
MISCELLANEOUS EXAMPLES.
1. Given x=2y 2 and 5(0? 2/)=5, to find the values of x
and y. Jlns. x=l8 or 12^.
T/= 3 or 2j|.
2. Given 2^4-y=22, and a?i/+2i/ 2 =120, to find the values
of x and ?/. Jlns. x=8, y=G.
3. Given x+y : x y : : 13 : 5, and #-|-7/ 2 =25, to find
the values of x and y. Jlns. x=9, and 2/=4.
4. Given J^+S/ 3 ^ 8 ^ ? to find the values of # and y.
+y =12 5 ^5. a?=8 or 4, =4 or 8
5. Given z*-{-2xy-{-y 2 =l2Q 2x 2y, and xy / 2 =8, to
find the values of x and y.
x=Q or 9, or i
v=4 or 1, or 3
6. Given
n
QUADRATIC EQUATIONS.
=56
=56 ^
=60 S to
va * ues
or 10.
to find the values of x
and .
8. Given
C
s 4 2 i 3 ?/
/ = 68 ?
?/== i60 S to
e va ^ ues
Jlns. ic=
In the first four examples, one of the equations is simple ; in
the 5th and 6th, x and y are similarly involved ; and the Oth,
7th and 8th are homogeneous.
(Art. 111.) When we have fractional exponents, we can re-
move them, as explained in (Art. 92.) ; but in some cases it may
not be important to do so.
EXAMPLES.
1. Given x*-}-y 3 '=3x and x*-\-y*=x, to find the values
of x and y.
Put a^=P; then x=P* and x^=P*
And^=#; then y=Q? and y%=Q*
Now the primitive equations become
P3+Q*=3P*, and P+Q=P*
From the 1st, # 2 =(3 P)P*
From the 2d, Q =(P1)P
By squaring, Q 2 =(P 1) 2 P 2
Put the two values of Q 2 equal to each other, rejecting or
dividing by the common factor P 2 , and we have
182 ELEMENTS OF ALGEBRA.
or P 2 2/M-l=3P
or P*P=2
Hence, P =2 or 1, and #=4 or 1, and y=8.
2. Given #^+^+2#^+2y^=23, and #V 3 "=36, to find
the values of x and y. Jlns. #=27 or 8, y=8 or 27.
3. Given #"-i-i/ 5 =8, and # 3 -f-7/^+# 3 i/ 3 "=259, to find the
values of # and y.
; #=125 or 27,
\ y= 27 or 125.
4. Given x-}-y-\-x+y=2Q 1 and #=8, to find the
values of x and y. Jlns. #=8, i/=32.
# 4# 5 33
5. Given h f == -j- an( l # ^2/ =5 to ^ nt ^ ^ ie values of X
and y. *fl.ns. a?=9, y=4
to
6. Given \ to ^ n( e va l ues of ^ an(1 2/
*=2|, y=16
7. Given #(y a + lJ-^# a + i;4-2;r 2 ?/ 2 =55, and #y a +
t^ v t >a =30, to find the values of # and y. ( #=4 or 9
< 3/=9 or 4
?. 3. l
8. Given x 3 y z =2y z and 8# 3 1/^ = 14, to find the value*
of # and y. ^ m C #=2744 or 8
^y=9604 or 4
(Art. 1 12.) No additional principles, to those already given,
are requisite for the solution of problems containing three or
more unknown quantities in quadratics. As in simple equations,
we must have as many independent equations as unknown
quantities.
As auxiliary to the solution of certain problems, particularly in
geometrical progression, we give the following problem :
QUADRATIC EQUATIONS. 183
Given x+y=s, and xy=p, to find the values of x 2 -j-y 2 ,
x'-J-y 5 ? x 4 +?/ 4 , and x 5 -J-y 5 , expressed in terms of s and p.
Squaring the first, x*-{-2xy-{-y 2 =s*
Subtract twice the second, 2xy 2p
1st result, x*+y 2 =s* 2p (A)
Again,
= P s
Subtract
2d result,
Square (A], and
Subtract
x*+y*=s* Bps
x 4 -\-2x 2 y z +y 4 =s* 4s*p+4p z
2x*y 2 = 2p z
3d result,
Multiply (A) by (#) and
x 4 +y*=s 4 4s z p-}-2p z
Subtract
x 2 y 2 (x-\-y) =
sp z
4th result,
(C)
x 5 +y 5 =s 5 5s 3 p-}-5sp 2 (D)
CHAPTER III.
Questions producing Quadratic Equations.
(Art. 113.) The method of proceeding to reduce the question
into equations, is the same as in simple equations ; and, in fact,
many problems which result in a quadratic may be brought out
by simple equations, by foresight and skill in notation. Others
again are so essentially quadratic, that no expedient can change
their form.
EXAMPLES.
1. A person bought a number of sheep for $240. If there
had been 8 more, they would have cost him $1 a-piece less.
What was the cost of a sheep, and how many did he purchase ?
240
Let x= the number of sheep j then =cost of one.
184 ELEMENTS OF ALGEBRA.
If he had x-\-S sheep, _=:cost of one.
240 240 ,
By the question, - : -- 1-1
x 3?-p8
Clearing of fractions, 240or-)-1920=240;r-{-a: 2 -f-8;c
Or
Resolving gives #=40 or 48 ; but a minus number will not
apply to sheep ; the other value only will apply to the problem
as enunciated.
This question can be brought into a simple equation thus : Let
x 4= the number of sheep, then 8 more would be expressed
by #4-4, and the equation would be
, ' +1. Put =240.
aH-4
Then " = +1
x 4 x -|-4
Clearing of fractions, ax-}-4aax 4a-\-x* 16.
Transposing, # 2 =8a-l-16=8(a+2)=16X 121
Extracting square root, #=4 X 1 1=44. Hence, a: 4=40, the
number of sheep. Divide 240 by 40, and we have $6 for the
price of one sheep.
(Art. 114.) In resolving problems, if the second member is
negative after completing the square, it indicates some impossi-
bility in the conditions from which the equation is derived, or an
error in forming the equation, and in such cases the values of the
unknown quantity are both imaginary.
2. For example, let it be required to divide 20 into two such
parts that their product shall be 140.
Let r one part, then 20 x= the other.
By conditions, 20# ^=140
Or, x* 20#=--140
Completing the square, x 2 20#-}-100= 40
By evolution, x 10=^=2^ 10
Or, x =10
QUADRATIC EQUATIONS. 185
This result shows an impossibility ; there are no such parts
of 20 as here expressed. It is impossible to divide 20 into two
such parts that their product shall be over 100, the product of
10 by 10, and so on with any other number. The product of
two parts is the greatest possible, when the parts are equal.
3. Find two numbers, such that the sum of their squares being
subtracted from three times their product, 11 will remain; and
the difference of their squares being subtracted from twice their
product, the remainder will be 14.
Let x= the greater number, and i/=the less.
By the conditions, '3xy x 2 t/ 2 =ll
And 2xy # 2 -f-i/ 2 =14
These are homogeneous equations; therefore, put x=vy ;
Then 3w/ 2 ify 2 y 2 =l 1 (./?)
And 2vy*vy+y*=l4 (B)
Conceive (./?) divided by (B) and the fraction reduced, we have
3f v* 1^11
l 14
Clearing of fractions and reducing, we find
3u 2 20u = 25.
5
A solution gives one value of v, -
3
Put this value in equation (.#), and we have
5y 2 v 2 = ! 1
9 f
Multiply by 9. and 457/ 2 25^ 2 9y 2 =llX9,
Or, lli/ 2 =llX9,
y z = 9 or y=3. Hence, x5.
A. A company dining at a house of entertainment, had to pay
$3.50 ; but before the bill was presented two of them went away ;
in consequence of which, those who remained had to pay each
20 cents more than if all had been present. How many persons
dined ? Jlns. 7.
16
186 ELEMENTS OF ALGEBRA.
5. There is a certain number, which being subtracted from
22, and the remainder multiplied by the number, the product will
be 117. What is the number ? Jlns. 13 or 9.
6. In a certain number of hours a man traveled 36 miles, but
if he had traveled one mile more per hour, he would have taken
3 hours less than he did to perform his journey. How many
miles did he travel per hour? Ans, 3 miles.
7. A person dies, leaving children and a fortune of $46,800,
which, by the will, is to be divided equally among them ; but it
happens that immediately after the death of the father, two of the
children also die ; and if, in consequence of this, each remaining
child receive $1950 more than he or she was entitled to by the
will, how many children were there? Jlns. 8 children.
8. A gentleman bought a number of pieces of cloth for 675
dollars, which he sold again at 48 dollars by the piece, and gain-
ed by the bargain as much as one piece cost him. What was
the number of pieces I Jlns. 15.
This problem produces one of the equations in (Art. 10.)
9. A merchant sends for a piece of goods and pays a certain
sum for it, besides 4 per cent, for carriage ; he sells it for $390,
and thus gains as much per cent, on the cost and carnage as the
12th part of the purchase money amounted to. For how much
did he buy it? Ans. $300.
10. Divide the number 60 into two such parts that their pro-
duce shall be 704. Jins. 44 and 16.
11. A merchant sold a piece of cloth for $39, and gained
as much per cent, as it cost him. What did he pay for it?
Jlns. $30.
12. Jl and B distributed 1200 dollars each, among a certain
number of persons. Jl relieved 40 persons more than B, and
B gave to each individual 5 dollars more than Jl. How many
were relieved by A and B? Jlns. 120 by Jl, and 80 by B.
This problem can be brought into a pure equation, in like man-
ner as (Problem 1.)
QUADRATIC EQUATIONS. 187
13. A vintner sold 7 dozen of sherry and 12 dozen of claret
for 50, and finds that he has sold 3 dozen more of sherry for
10 than he has of claret for 6. Required the price of each?
Ans. Sherry, 2 per dozen; claret, 3.
14. Ji set out from C towards J9, and traveled 7 miles a
day. After he had gone 32 miles, B set out from D towards
C, and went every day ^ of the whole journey ; and after
he had traveled as many days as he went miles in a day, he met
Jl. Required the distance from C to 2).
J$ns. 76 or 152 miles ; both numbers will answer the con-
dition.
15. A farmer received $24 for a certain quantity of wheat,
and an equal sum at a price 25 cents less by the bushel for a
quantity of barley, which exceeded the quantity of wheat by 16
bushels. How many bushels were there of each ?
Jlns. 32 bushels of wheat, and 48 of barley.
16. A and B hired a pasture, into which Jl put 4 horses, and
B as many as cost him 18 shillings a week ; afterwards B put
in two additional horses, and found that he must pay 20 shillings
a week. At what rate was the pasture hired?
*flns. B had six horses in the pasture at first, and the price
of the whole pasture was 30 shillings per week.
I 1 ?. Find those two numbers whose sum, product, and dif-
ference of their squares, are all equal to each other.
Ans. i36, and
18. If a certain number be divided by the product of its
two digits, the quotient will be 2, and if 27 be added to the num-
ber, the digits will be inverted. What is the number?
JJns. 36.
19. It is required to find three numbers, whose sum is 33,
such that the difference of the first and second shall exceed the
difference of the second and third by 6, and the sum of whose
squares is 441. fins. 4, 13, and 16.
188 ELEMENTS OF ALGEBRA.
2O. Find those two numeral quantities whose sum, product,
and sum of their squares, are all equal to each other.
Jlns. No such numeral quantities exist. In a strictly algebraic
sense, the quantities are
21. What two numbers are those whose product is 24, and
whose sum added to the sum of their squares is 62?
fins. 4 and G.
22. It is required to find two numbers, such that if their pro-
duct be added to their sum it shall make 47, and if their sum be
taken from the sum of their squares, the remainder shall be 62?
*ftns. 7 and 5.
23. The sum of two numbers is 27, and the sum of their
cubes 5103. What are their numbers? J%ns. 12 and 15
24. The sum of two numbers is 9, and the sum of their fourth
powers 2417. What are the numbers? Jlns. 7 and 2.
25. The product of two numbers multiplied by the sum of
their squares, is 1248, and the difference of their squares is 20
What are the numbers? Jlns. 6 and 4.
Let x-}-y=ihe greater, and x y=the less.
26. Two men are employed to do a piece of work, which
they can finish in 12 days. In how many days could each do
the \vork alone, provided it would take one 10 days longer than
the other? Ans. 20 and 30 days.
27. The joint stock of two partners, Jl and B, was $1000.
.tf's money was in trade 9 months, and J5's 6 months ; when
they shared stock and gain, Ji received $1,140 and B $640.
What was each man's stock?
Jlns. JFs stock was $600 ; B's $400.
28. A speculator from market, going out to buy cattle, met
with four droves. In the second were 4 more than 4 times the
square root of one half the number in the first. The third con
tained three times as many as the first and second. The fourth
was one half the number in the third and 10 more, and the whole
ARITHMETICAL PROGRESSION. 199
number in the four droves was 1121. How many weie in each
drove? Ans. 1st, 162 ; 2d, 40 ; 3d, 606 ; 4th, 313
29. Divide the number 20 into two such parts, that the pro-
duct of their squares shall be 9216. Jlns. 12 and 8.
30. Divide the number a into two such parts that the product
of their squares shall be b.
rfns. Greater part -f-
Less part |- \(a*-4 jlty*.
31. Find two numbers, such that their product shall be equal
to the difference of their squares, and the sum of their squares
shall be equal to the difference of their cubes.
dns. ^5" and <l(5,/5)
SECTION V.
ARITHMETICAL PROGRESSION.
CHAPTER I.
A sertes of numbers or quantities, increasing or decreasing by
the same difference, from term to term, is called arithmetical pro-
gression.
Thus, 2, 4, 6, 8, 10, 12, &c., is an increasing or ascending
arithmetical series, having a common difference of 2 ; and 20,
17, 14, 11, 8, <fec., is a decreasing series, having a common dif-
ference of 3.
(Art. 1 15.) We can more readily investigate the properties of
an arithmetical series from literal than from numeral terms. Thus
let a represent the first term of a series, and d the common dif-
ference.
190 ELEMENTS OF ALGEBRA.
a,(a-{-d),(a J r2d),(a-\-3d) 9 (a-\-'ld), &c., represent an ascend
ing series ; and
a, (a t?),(a 2d),(a 3d), (a 4cQ, &c., represent a descend-
ing series.
Observe that the coefficient of d, in any term is equal to the
number of the preceding terms.
The first term exists without the common difference. All
other terms consist of the first term and the common difference
multiplied by one less than the number of terms from the first.
Wherever the series is supposed to terminate, is the last term,
and if such term be designated by Z, and the number of terms
by n, the last term must be a-\-(n l)d, or a (n l)d, accord-
ing as the series may be ascending or descending, which we draw
from inspection.
Hence, L=a-(nl)d (A)
(Art. 116.) It is manifest that the sum of the terms will be the
same, in whatever order they are written.
Take, for instance, the series 3, 5, 7, 9, 11,
And the same inverted, 11, 9, 7, 5, 3.
The sums of the terms will be 14, 14, 14, 14, 14.
Take the series #, a-f- rf, a-{-2d, a-}-3d, a-\-4d t
Inverted, -j-4c/, a-{-3d, a-\-2d, a-}- d, a
Sums will be 2a+4rf, 2a+4d, 2a+4d, 2a-\-4d, 2a+4d.
Here we discover the important property, that, in an arithmeti-
cal progression, the sum of the extremes is equal to the sum of
any other two terms equally distant from the extremes. Also,
that twice the sum of any series is equal to the extremes, or
first and last term repeated as many times as the series contains
terms.
Hence, if S represents the sum of a series, and n the num-
ber of terms, a the first term, and L the last term, we shall
have 2S=n(a+L)
Or S=?(a+L) (B)
The two equations (.#) and (B] contain five quantities, a, d,
ARITHMETICAL PROGRESSION. 191
Z, n, and S; any three of them being given, the other two can
be determined.
Two independent equations are sufficient to determine two un-
known quantities, (Art. 45,) and it is immaterial which two are
unknown if the other three are given.
By examining the two equations
Z=a4-(rc l)rf (JS)
S=%(a+L) (B)
We perceive that the value of any letter, L for example, can be
drawn from equation (B) as well as from (A).
It can also be drawn from either of the equations after n or a
is eliminated from them. Hence, the value of L may take/ot/r
different forms. The same may be said of the other letters,
and there being five quantities or letters and four different
forms to each, the subject of arithmetical progression may in-
clude twenty different equations. But we prefer to make no
display with these equations, believing they would add dark-
ness rather than light, as they are all essentially included in the
two equations, (A) and (B), and these can be remembered literal-
ly and philosophically, and the entire subject more surely under-
stood.
These two equations are sufficient for problems relating to
arithmetical series, and we may use them without modification
by putting in the given values just as they stand, and afterwards
reducing them as numeral equations.
EXAMPLES.
1 The sum of an arithmetical series is 1455, the first term
5, and the number of terms 30. What is the common difference!
Ans. 3.
Here =1455, a=5, n=W. L and d are sought.
Equation (B) 1455=(5-{-Z)15. Reduced Z=92
Equation (.#) 92=5-]-29<?. Reduced d=3, fins.
192 ELEMENTS OF ALGEBRA.
2. The sum of an arithmetical series is 567, the first term 7,
and the common difference 2. What is the number of terms?
Am. 21
Here 5=567, a=7, d=2. L and n are sought.
Equation (A) Z=7-f-2n 2=5+2n
Equation (B) 567=(7+5-f-2n)~=6tt-|-n 2
Or n 2 -j-6/i-f-9=576
n-}-3=24, or n=2l,Ans.
3. Find seven arithmetical means between 1 and 49.
Observe that the series must consist of 9 terms.
Hence, o=l, Z=49, n=Q.
Ans. 7, 13, 19,25, 31, 37, 43.
4. The first term of an arithmetical series is 1, the sum of the
terms 280, the number of terms 32. What is the common dif-
ference, and the last term? Ans. d=z, Z=16j.
5. Insert three arithmetical means between | and 5*
Ans. The means are ~, T 5 2, \\
6. Find nine arithmetical means between 9 and 109.
Ans. d--=lO.
"7. What debt can be discharged in a year by paying 1 cent
the first day, 3 cents the second, 5 cents the third, and so on, in-
creasing the payment each day by 2 cents?
Ans. 1332 dollars 25 cents.
8. A footman travels the first day 20 miles, 23 the second, 20
the third, and so on, increasing the distance each day 3 miles.
How many days must he travel at this rate to go 438 miles?
Ans. 12.
9. What is the sum of n terms of the progression of 1,2, 3,
1O. The sum of the terms of an arithmetical series is 950,
the common difference is 3, and the number of terms 25. What
is the first term? Ans. 2.
ARITHMETICAL PROGRESSION. 193
11. A man bought a certain number of acres of land, paying
for the first, $5 ; for the second, $| ; and so on. When he came
to settle -he had to pay $3775. How many acres did he pur-
chase, and what did it average per acre ?
Ans. 150 acres at $25 per acre.
Problems in Arithmetical Progression to which the preceding
formulas, (A) and (B), do not immediately apply.
(Art. 117.) When three quantities are in arithmetical progres-
sion, it is evident that the middle one must be the exact mean
of the three, otherwise it would not be arithmetical progression ;
therefore the sum of the extremes must be double of the mean.
Take, for example, any three consecutive terms of a series, as
and we perceive by inspection that the sum of the extremes is
double the mean.
When there are four terms, the sum of the extremes is equal
to the sum of the means, by (Art. 116.)
To facilitate the solution of problems, when three terms are
in question, let them be represented by (x y), x, (x-\-y], y being
the common difference.
When four numbers are in question, let them be represented
by (x 3y), (x y), (aj+y), (x-\-3y)i 2y being the common dif-
ference.
So in general for any other number, assume such terms that
the common difference will disappear by addition.
1. There are five numbers in arithmetical progression, the
sum of these numbers is 65, and the sum of their squares is
1005. What are the numbers ?
Let x= the middle term, and y the common difference. Then
X 2y, xy, x, x-\-y, x-\-2y, will represent the numbers,
and their sum will be 5#=65, or #=13. Also, the sum of
their squares will be
5# 2 -f-10 < j/ 2 =1005 or ic 2 +2/ 2 =201.
But # 2 =169; therefore, 2?/ 2 ^32, i/ 2 =16 or y=i.
Hence, the numbers are 13 8=5, 9, 13, 17 and 21
17
194 ELEMENTS OF ALGEBRA.
2. There are three numbers in arithmetical progression, their
sum is 18, and the sum of their squares 158. What are those
numbers? Jlns. 1, 6 and 11
3. It is required to find four numbers in arithmetical progres-
sion, the common difference of which shall be 4, and their coiv
tinued product 176985. Jlns. 15, 19, 23 and 27
4. There are four numbers in arithmetical progression, the
sum of the extremes is 8, and the product of the means 15
What are the numbers? Jlna. 1, 3, 5, 7.
*5. A person travels from a certain place, goes 1 mile the first
day, 2 the second, 3 the third, and so on ; and in six days after,
another sets out from the same place to overtake him, and travels
uniformly 15 miles a day. How many days must elapse after
the second starts before they come together?
Jlns. 3 days and 14 days.
Reconcile these two answers.
6. A man borrowed $60 ; what sum shall he pay daily to can-
cel the debt, principal and interest, in 60 days; interest at 10 per
cent, for 12 months, of 30 days each?
Jlns. $1 and f ^ of a cent.
7. There are four numbers in arithmetical progression, the
sum of the squares of the extremes is 50, the sum of the squares-
of means is 34 ; what are the numbers? J$ns. 1, 3, 5, 7
. The sum of four numbers in arithmetical progression is
24, their continued product is 945. What are the numbers ''
Jlns. 3, 5, 7, 9.
9. A certain number consists of three digits, which are in
arithmetical progression, and the number divided by the sum of
its digits is equal to 26 ; but if 198 be added to the number its-
digits will be inverted What is the number 1 fins. 234
GEOMETRICAL PROGRESSION. 195
CHAPTER II.
GEOMETRICAL PROGRESSION.
(Art. 118.) When numbers or quantities differ from each
othfr by a constant multiplier in regular succession, they consti-
tute a geometrical series, and if the multiplier be greater than
unity, the series is ascending ; if it be less than unity, the series
is descending.
Thus, 2 : 6 : 18 : 54 : 162 : 486, is an ascending series, the
multiplier, called the ratio, being three ; and 81 : 27 : 9 : 3 : 1 :
l : i, &c., is a descending series, the multiplier or ratio being |.
Hence, a : ar : ar* : r 3 : ar 4 : ar 5 : w 6 : &c., may represent
any geometrical series, and if r be greater than 1, the series is
ascending, if less than 1, it is descending.
(Art. 119.) Observe that the first power of r stands in the
2d term, the 2d power in the 3d term, the third power in the
4th term, and thus universally the power of the ratio in any
term is one less than the number of the term.
The, first term is a factor in every term. Hence the 10th
term of this general series is ar 9 . The 17th term would be ar 16 .
The rath term would be ar n ~ l .
Therefore, if n represent the number of terms in any series,
and L the last term, then L=ai**~ l (1)
(Art. 120.) If we represent the sum of any geometrical series
by 5, we have
5 =a-j-ar+cfr 2 -}-r 3 4- &c. . . ar n - 2 +ar n ~ l .
Multiply this equation by r, and we have
r 5 =ar-}-tfr 2 -f ar 3 -f- &c. ar n ~ l -\-ar n .
Subtract the upper from the lower, and observe thai
Zr=ar n ; then (r l)s=Lr a.
Therefore, 8 =- ( 2 )
As these two equations are fundamental, and cover the whole
subject of geometrical progression, let them be brought together
for critical inspection.
196 ELEMENTS OF ALGEBRA.
(1), S= (2).
These two equations furnish the rules given for the operations
in common arithmetic.
Here we perceive five quantities, 0, ?*, n, L and S, and any
three of them being given in any problem, the other two can be
determined from the equations.
To these equations we may apply the same remarks as were
made to the two equations in arithmetical progression (Art. 116.)
Equation (2), put in words, gives the following rule for the
sum of a geometrical series ;
RULE. Multiply the last term by the ratio, .and from the
vroduct subtract the first term, and divide the remainder by the
ratio less one.
EXAMPLES FOR THE APPLICATION OF EQUATIONS
(1) AND (2).
1. Required the sum of 9 terms of the series, 1, 2, 4, 8, 16,
&c. Ans. 511.
3. Required the 8th term of the progression, 2, 6, 18, 54,
&c. Ans. 4374.
3. What is the sum of ten terms of the series 1, f, J, &c. ?
* WAV
4. Required two geometrical means between 24 and 192.
N. B. When the two means are found, the series will consist
of four terms ; the first term 24 and the last term 192.
By equation (1) L=ar n ~\
Here a 24, Z=192, n=4, and the equation becomes
192=24r 3 or r=2.
Hence, 48 and 96 are the means required.
5. Required 7 geometrical means between 3 and 768.
tins. 6, 12, 24, 48, 96, 192,
6. The first term of a geometrical series is 5, the last term
1215, and the number of terms 6. Wliat is the ratio 1 fins. 3.
GEOMETRICAL PROGRESSION. J97
7". A man purchased a house, giving $1 for the fiist door, $2
for the second, $4 for the third, and so on, there being 10 doors.
What did the house cost him 1 tins. $1023.
(Art. 121.) By Equation (2), and the Rule subsequently given,
we perceive that the sum of a series depends on the first and last
terms and the ratio, and not on the number of terms ; and
whether the terms be many or few, there is no variation in the
rule. Hence, we may require the sum of any descending series,
as 1, a i, \, &c., to infinity, provided we determine the LAST
term. Now we perceive the magnitude of the terms decrease
as the series advances ; the hundredth term would be e'xtremely
small, the thousandth term very much less, and the infinite term
nothing ; not too small to be noted, as some tell us, but absolutely
nothing.
Hence, in any decreasing series, when the number of terms
is conceived to be infinite, the last term, Z, becomes
0, and Equation (2) becomes s=--
By .change of signs s=- -
This gives the following rule for the sum of a decreasing infi-
nite series :
RULE. Divide the first term by the difference between unity
and the ratio.
EXAMPLES.
1. Find the value of 1, , T 9 , &c., to infinity.
o=l, r=\. JLns. 4
2. Find the exact value of the decimal .3333, &c., to infinity.
rfns. 5.
This may be expressed thus : T \+ T f , <fec. Hence.
3. Find the value of .323232, &c., to infinity.
=T 3 2 o r= T <nr 2 oo ; therefore r= T - . Ans. |
4. Find the value of .777, &c., to infinity. *ftns. .
J98 ELEMENTS OF ALGEBRA.
5. Find the value of f : 1 : J : / 3 , &c. to infinity.
tins. 4 '
6. Find tlie value of 5 : f : , &c. to infinity. Jlns. 7|
7. Find the value of the series 1, ^, &c., to infinity ? Am. |.
8. What is the value of the decimal .71333, &c., to infinity?
On*. Hi-
O. What is the value of the decimal .212121, &c., to infinity?
*>. A-
(Art. 122.) If we observe the general series, (Art. 11 8.) a: an
ar 2 : ar 3 : ar 4 , &c., we shall find, by taking three consecutive
terms anywhere along in the series, that the product of the ex-
tremes will equal the square of the mean. Hence, to find a
geometrical mean between two numbers, we must multiply them
together, and take the square root. If we take four consecutive
terms, theproduct of the extremes will be equal to the product
of the means.
(Art. 123.) This last property belongs equally to geometrical
proportion, as well as to a geometrical series, and the learner must
be careful not to confound proportion with a series.
a: ar : : b : br, is a geometrical proportion, not a continued
series. The ratio is the same in the two couplets, but the mag-
nitudes a and b, to which the ratio is applied, may be very dif-
ferent.
We may suppose a : ar two consecutive terms of one series,
and b : br any two consecutive terms of another series having
the same ratio as the first series, and being brought together they
form a geometrical proportion. Hence, the equality of the ra-
tio constitutes proportion.
To facilitate the solution of some difficult problems in geomet-
rical progression, it is desirable, if possible, to express several
terms by two letters only, and have them stand symmetrically.
Three terms maybe expressed by x : ,Jxy : y, or by x z : xy :
y\ as the product of the extremes is here evidently equal to the
square of the mean.
To express four terms with x and y symmetrically, we at first
GEOMETRICAL PROGRESSION. 199
write P : x \ : y : Q. The firstthree being in geometrical progres-
y& y*
sion, gives Py=x 2 orP= . In the same manner, we find $= -
s? y 2
And taking these values of P and Q we have : x:: y: to ren-
y x
resent four numbers symmetrically with two letters.
Taking three numbers as above, and placing them between P
and Q, thus, P : x z : xy : y 2 : , we have five numbers ; and
3?
by reducing P and Q into functions of x and ?/, we have: #* :
V 3
xy : y z : y , for five terms symmetrically expressed.
X
3? z 2 y 2 y*
Six numbers thus, : : x : : y : : ^
t/ 2 y xx*
Sometimes we may more advantageously express unknown
numbers in geometrical proportion by x, xy, xy 2 , <fec. ; x being
the first term, and y the ratio.
HARMONICAL PROPORTION.
(Art. 124.) When three magnitudes, a, ft, c, have the relation
of a : c : : a b : b c ; that is, the first is to the third as the dif-
ference between the first and second is to the difference between
the second and third, the quantities a, ft, c, are said to be in har-
monical proportion.
(Art. 125.) Four magnitudes are in harmonical proportion
when the first is to the fourth as the difference between the first
and second Is to the difference between the third and fourth.
Thus, , ft, c, d, are in harmonical proportion when a : d : :
a ft : c d, or when a : d : : b a : d c.
An harmonical mean between two numbers is equal to twice
their product divided by their sum. For a : x : b representing
three numbers in harmonical proportion, we have by the definition,
(Art. 124.) a : ft : : a x : xb.
Therefore, ax ab=ab bx or x= r-y.
200 ELEMENTS OF AUJEHKA.
1. Find the harmonica! mean between 6 and 12. JJns. 8.
2. Find the third harmonical number to 234 and 144.
rfns. 104.
3. Find the fourth harmonical proportion to the numbers 24 :
16 : 4. Am. 3,
4U There are four numbers in harmonical proportion, the firs'
is 16, the third 3, the fourth 2. The second is lost ; find it.
Ans. 8
PROBLEMS IN GEOMETRICAL PROGRESSION, AND
HARMONICAL PROPORTION.
1. The sum of three numbers in geometrical progression is
26, and the sum of their squares 364. What are the numbers ?
Let the numbers be represented by x : ,Jxy : y.
Then x+Jxy+y = 26=a (1)
And x 2 -!- ^4-2/2=364=6 (2)
Transpose, ,Jxy in Equation (I) and square, we have
a*+2xy+if=(fi2ajxy+xy (3)
Or x*-\- xy-{-y 2 =a~2a l Jxy (4)
The left hand members of Equations (2) and (4) are equal ;
hence,
a 2 2ajxy=b.
Therefore, 7^==6 (5)
This gives the second term of the progression, and now from
equations (1) and (5) we find #=2, y=I8, and the numbers
are 2, 6, 18.
2. The sum of four numbers in geometrical progression is 15,
(), and the sum of their squares 85 (6). What are the
numbers ?
Let the numbers be represented as in (Art. 123.)
Then
GEOMETRICAL PROGRESSION. 201
Assume x-{-y=s
xy=p
Then by (Art. 112.)
X' "'"*
And
y
And x*-\-y*=s 8 3sp
Transposing (x-\-y) in Equation (1), and (x 2 -\-y 2 ) in Equation
(2), we have
=as (3) and +=b--#+2 p (4)
y *x y 2 x z
Square (3) and transpose 2xy or 2p and
|'+| 4 =(-*) 2 -2/> (5)
The left hand members of equations (4) and (5) are equal, there
fore, (ar-8)*2p=b s 2 +2p
Or a 2 2as-{-2s z 4p=b (6)
Clear equation (3) of fractions, and x 3J ry 3 ap ps.
s 3
That is, 6- 3 3span ps or = : - (7}
a+2s
Put this value of p in equation (6) and reduce, we have,
Or as*+bs=:-
2
Taking the given values of a and b we have
15s 2 -|-856'=70X15
Or 3* 2 -{-17s=210, an equation which gives s=6
Put the values of a and s in equation (7), and p=S
That is, x-\-y=Q, and xy=8, from which we find #=2, am!
=4 ; therefore the required numbers are
1,2,4 and 8, Ans
3. The arithmetical mean of two numbers exceeds the geo
202 ELEMENTS OF ALGEBRA.
metrical mean by 13, and the geometrical mean exceeds the har-
monical mean by 12. What are the numbers?
Let x and y represent the numbers.
Then | (#+?/)= the arithmetical mean, > Jxi/= the geome-
trical mean, (Art. 112.) and r^-= the harmonical mean.
v+y
Let o=12;
Then, by the question, i(x-\-y) = i Jxy -\-a-\- 1 (1)
(2)
By our customary substitution, these equations become
5*=V/>+-fl (3)
And <//>=T+ a ( 4 )
o
Take the ralue of s from equation (3) and put it into equation
(4), dividing me numerator and the denominator by 2, and we
have . p .
Clearing of fractions, we shall have
Drop equals, and Jp=(a-\-l}a (6)
Put this value of Jp in equation (3) and we have
Or 5=2(0+ 1) 2 (7)
For the sake of brevity, put (a-j-l)=6; squaring equation
(6) and restoring the values of s and p in equations (6) and (7),
and we have xy=a z b* (Jl)
x+y=2b* (JS)
Square (S) and
Subtract 4
/ a\
times ^
And x z 2^-l-i/ 8 =4& 2 (6 2 2 )=46 2 (6+)(6 a) (C)
GEOMETRICAL PROGRESSION. 803
As =12 and b=l3, b-\- a=25, and b a=l.
Therefore, (C) becomes (a? 2/) 2 =4& 2 X25X 1.
By evolution, a? y=2bX5
Equation () #-{-i/=2 2
By addition 2#
Or x= 6 2 -f- 5&=(6-f 5)6=18X13=234
By subtraction, 2y=2b 2 Wb
y= b 2 56=(6 5)6= 8X13=104.
A more brief solution is the following :
Let x y and x-\-y represent the numbers.
Then x= the arithmetical mean, >Jx 2 y z = the geometri-
cal mean, (Art. 112), and = the harmonical mean. By
the question,
x~ I3 = jx 2 y 2 (1), and - ^-4-12 = 7^ y 2 (2)
The right hand members of equations (1) and (2) being the
y2 yZ
same, therefore, ^--{-I2=x 13.
x
By reduction, y 2 =25x.
Put this value of y z in equation (1), and by squaring
# 2 26z-l-(13) 2 =z 2 25*, or a;=(13) 2 =169.
Hence, i/=65, and the numbers are 104 and 234.
4. Divide the number 210 into three parts, so that the last
shall exceed the first by 90, and the parts be in geometrical pro-
gression, dns. 30, 60, and 120.
5. The sum of four numbers in geometrical progression is
30 ; and the last term divided by the sum of the mean terms is
1|. What are the numbers ? Ans. 2, 4, 8, and 16.
6. The sum of the first and third of four numbers in geo-
metrical progression is 1 48, and the sum of the second and
fourth is 888. What are the numbers ?
Ans. 4, 24, 144, and 864.
204 ELEMENTS OF ALGEBRA.
7". It is required to find three numbers in geometrical progres-
sion, such that their sum shall be 14, and the sum of their squares
84. Jlns. 2, 4, and 8.
8. There are four numbers in geometrical progression, the
second of which is less than the fourth by 24 ; and the sum of
the extremes is to the sum of the means, as 7 to 3. What are
the numbers ? Jlns. 1, 3, 9 and 27.
9. The sum of four numbers in geometrical progression is
equal to the common ratio -}-! and the first term is y 1 ^. What
are the numbers ? fins. T 1 ^, T \, T \, T .
10. The sum of three numbers in harmonical proportion is
26, and the product of the first and third is 72. What are the
numbers ? Jlns. 12, 8, and 6
11. The continued product of three numbers in geometrical
progression is 216, and the sum of the squares of the extremes
is 328. What are the numbers ? tins. 2, 6, 18.
12. The sum of three numbers in geometrical progression is
13, and the sum of the extremes being multiplied by the mean,
the product is 30. What are the numbers ?
1, 3, and 9
13. There are three numbers in harmonical proportion, the
sum of the first and third is 18, and the product of the three is
576. What are the numbers ? Jlns. 6, 8, 12.
14. There are three numbers in geometrical progression, the
difference of whose difference is 6, and their sum 42, What
are the numbers? Jlns. 6, 12,24.
15. There are three numbers in harmonical proportion, the
difference of whose difference is 2, and three times the product
of the first and third is 216. What are the numbers ?
Jlns. 6, 8, and 12.
16. Divide 120. dollars between four persons, in such a
way, that their shares may be in arithmetical progression ; and
if the second and third each receive 12 dollars less, and tho
GEOMETRICAL PROPORTION. 205
fourth 24 dollars more, the shares would then be in geometri-
cal progression. Required each share.
Jlns. Their shares were 3, 21, 39, and 57, respectively.
17. There are three numbers in geometrical progression,
whose sum is 31, and the sum of the first and last is 26. What
are the numbers ? Jlns. 1, 5, and 25.
18 The sum of six numbers in geometrical progression is
189, and the sum of the second and fifth is 54. What are the
numbers ? rfns. 3, 6, 12, 24, 48 and 96.
19. The sum of six numbers in geometrical progression is
189, and the sum of the two means is 36. What are the num-
bers? Am. 3, 6, 12, 24, 48 and 96.
CHAPTER III.
PROPORTION.
(Art. 176.) We have given the definition of geometrical pro-
portion in (Art. 41.) and demonstrated the most essential prop-
erty, the equality of the products between extremes and means.
We now propose to extend our investigations a little farther.
Proportion can only exist between magnitudes of the same
kind, and the number of times and parts of a time, that one
measures another, is called the ratio. Ratio is always a num-
ber, and not a quantity.
(Theorem 1.) If two magnitudes have the same ratio as
two others, the first two as numerator and denominator may
form one member of an equation ; and the other two magnitudes
as numerator and denominator will form the other member.
Let A and B represent the first two magnitudes and r their ratio.
Also C and D the other two magnitudes, and r their ratio.
Then,^=r and =r Therefore, (Ax. 7) =
(Theorem 2.) Magnitudes which are proportional to tlie
same proportionals, are proportional to each other.
Suppose a : b : : P : Q \ Then we are to prove that
and c: d:: P: Q V a:b::c:d
and x : y : : P : Q ) and a : b : : x : y, &c.
200 ELEMENTS OF ALGEBRA.
b Q
From the first proportion, -=73
From the second, = ~
Hence, (Ax. 7) = or a : b : : c : d
In the same manner we prove a : b : : x : y
And c : d : : x : y
(Theorem 3.) If four magnitudes constitute a proportion,
the first will be to the sum of the first and second, as the
third is to the sum of the third and fourth.
By hypothesis, a : b : : c : d ;^then we are to prove that
a : a-\-b : : c : c-\-d.
By the given proportion, =-. Add unity to both members,
then reducing to the form of a fraction we have = .
a c
Throwing this equation into its equivalent proportional form, we
have ' a : a+b : : c : c+d.
N. B. In place of adding unity, subtract it, and we shall find
that a : a-b : : c : c-d.
or a : b a : : c : d c.
(Theorem 4.) If four magnitudes be proportional, the sum
of the first and second is to their difference, as the sum of
the third and fourth is to their difference.
Admitting that a : b : : c : d, we are to prove that
a-\-b : a b : : c-{-d : c d
X
From the same hypothesis, (Theorem 3.) gives
a : a-\-b : : c : c-\-d
And a : a b : : c : c d
Changing the means, (which will not affect the product of- the
extremes and means, and of course will not destroy proportion-
ality,) and we have,
GEOMETRICAL PROPORTION. 207
arc:: -}-& : c-\-d
a i c : : a b : c d
Now by (Theorem 2.) a-{-b : c-\-d : : a b : c d
Changing the means, a-\-b : a b : : c-\-d : c d
(Theorem 5.) If four magnitudes be proportional, like
powers or rootsof the same, will be proportional.
Admitting a : b : : c : d, we are to show that
n n n n
a n : b n : : c n : er, and a : b : : c : d
By the hypothesis, ~j- = -7' Raising both members of this
equation to the nth power, and
a n c n
Changing this to the proportion a n : b n : : c n : d n
Resuming again the equation T=-y and taking the nth root
of each member, we have j- =7-. Converting this equation
r~**~ j"~
b d
into its equivalent proportion, we have
JL _L i i
n , n n ,~n~
a : b : : c id.
Now by the first part of this theorem, we have
m jrn m m
a n : b" : : c n : d" , m representing any
power whatever, and n representing any root.
(Theorem 6.) If four magnitudes be proportional, also
four others, their compound, or product of term by term, will
form a proportion.
208 ELEMENTS OF ALGEBRA.
Admitting that a : b : : c . d
And x : y : : m : n
We are to show that, ax : by : : me : nd
From the first proportion, T = y
From the second, -=
y n
Multiply these equations, member by member, and
ax_mc
by nd
Or ax : by : : me : nd.
The same would be true in any number of proportions.
(Theorem 7.) Taking the same hypothesis as in (Theorem
6.) we propose to show, that a proportion may be formed by di-
viding one proportion by the other, term by term.
By hypothesis, a : b : : c : d
And x : y : : m : n
Multiply extremes and means, ad=bc (1)
And nx=my (2)
Divide (1) by (2), and -X -=- X -
x n m y
Convert these four terms, which make two equal products, into
a proportion, and we shall have
a ^ b c f d
x ' y m ' n
By comparing this with the given proportions, we find it com-
posed of the quotients of the several terms of the first propor-
tion divided by the corresponding term of the second.
(Theorem 8.) If four magnitudes be proportional, we may
multiply the first couplet or the second couplet, the antecedents
or the consequents, or divide them by the same factor, and the
results will be proportional in every case.
GEOMETRICAL PROPORTION. 209
Suppose a : b : : c : d
Multiply ex. and means, and ad=bc (I)
Multiply this eq. by ?n, and .... madmbc
Now, in this last equation, ma may be considered as a single
term or factor, or md may be so considered. So, in the second
member, we may take mb as one factor, or me. Hence we may
convert this equation into a proportion in four different ways.
Thus, as . . ma : mb : : c : d
or as a : b :: me : md
or as ma : b : : me : d
or as a : mb :: c : md
If we resume the original equation (1), and divide it by any
number, m, in place of multiplying it, we can have, by the same
course of reasoning,
a
b
: : c
: d
m
m
a
: b
c
d
m
m
a
: b
c
: d
m
m
b
d
a
: : c
__
m
771
The following examples are intended to illustrate the practi-
cal utility of the foregoing theorems :
EXAMPLES.
! Find two numbers, the greater of which shall be to the
less, as their sum to 42 ; and as their difference is to 6.
Let a?=the greater, y=the less.
18
210 ELEMENTS OF ALGEBRA.
( x : y : : x-\-y :42 (1)
fhen.pcr quest, j x . * . . ^ : 6 fo
(Theorem 2.) x-\-y : 42 : : # y : 6
Changing the means x-}-y ' x y : : 42 : 6
(Theorem 4.) 2x : 2y : : 48 : 36
(Art. 42.) x : y : : 4 : 3
(Theorem 2.) 4:3:: xy : 6
And 4:3:: x+y : 42
From these last proportions, x y= 8
And x+y=5Q. Hence, x=32, i/=24.
2. Divide the number 14 into two such parts, that the
quotient of the greater divided by the less shall be to the quo
dent of the less divided by the greater, as 16 to 9.
Let #== the greater part, and 14 a?=the less.
x 14 x
By the conditions, - : -- : : 16:9
14 x x
Multiplying terms, x 2 : (14 a?) 2 : : 16:9
Extracting root, x : (14 x) : : 4:3 (Theor. 5.;
Adding terms, x : 14 : : 4:7
Dividing terms, x : 2 : : 4 : 1
Therefore, x=S.
3. There are three numbers in geometrical progression whose
sum is 52, and the sum of the extremes is to the mean as 10 to
3. What are the numbers ? Ans. 4, 12, and 36
Let x, xy, xy z represent the numbers.
Then, by the conditions, x-\-xy-\-xy z =52 (I)
And xy z -i-x : xy : : 10 : 3
(Art. 42.) yM-1 : y : : 10 : 3
Double 2d and 4th, ?/ 2 +l : 2y : : 10 : 6
Adding and sub. terms, 2 -f2?/4-l : y 2 2y-\-l : : 16 : 4
Extracting square root, 2/+1 : y 1 : : 4 : 2
Adding and sub. terms, y : 1 : : 3 : 1 Hence, y=3.
52 52 52
From equation (1), *=__=__- =-=4.
GEOMETRICAL PROPORTION. *jll
4. The product of two numbers is 35, the difference of their
cubes, is to the cube of their difference as 109 to 4. What arc
the numbers ? Ans. 7 and 5
Let x and y represent the numbers.
Then, by the conditions, xy=35, and x 2 y* : (x y)* : : 109:4
Divide by (xy) (Art. 42.) and x 2 +xy-\-y* : (xy)* : : 109:4
Expanding, and # 2 -f xy -\-y z : x 2 2xy-{-y z : : 109:4
(Theorem 3.) 3xy : (z-^) ? ' : : 105:4
ttuiSxy, we know from the first equation, is equal to 105.
Therefore, (x y) 2 =4, or x y=2.
We can obtain a very good solution of this problem by putting
x-\-y= the greater, and x y= the less of the two numbers.
5. What two numbers are those, whose difference is to their
sum as 2 to 9, and whose sum is to their product as 18 to 77?
Jlns. 1 1 and 7
6. Two numbers have such a relation to each other, that if 4
be added to each, they will be in proportion as 3 to 4 ; and if 4
be subtracted from each, they will be to each other as 1 to 4
What are the numbers ? Ans. 5 and 8
7. Divide the number 16 into two such parts that their pro-
duct shall be to the sum of their squares as 15 to 34.
. 10, and 6.
8. In a mixture of rum and brandy, the difference between
the quantities of each, is to the quantity of brandy, as 100 is to
the number of gallons of rum; and the same difference is to the
quantity of rum, as 4 to the number of gallons of brandy.
How many gallons are there of each ?
tins. 25 of rum, and 5 of brandy.
9 There are two numbers whose product is 320 ; and the
difference of their cubes, is to the cube of their difference, as 61
to 1. What are the numbers? Am. 20 and 16.
1O. Divide 60 into two such parts, that their product shall be
to the sum of their squares as 2 to 5. Ans. 40 and 20
21-2 ELEMENTS OF ALGEBRA.
11. There are two numbers which are to each other as 3 to
2. If 6 be added to the greater and subtracted from tie less,
the sum and the remainder will be to each other, as 3 to 1.
What are the numbers ? Jlns. 24 and 1 6.
12. There are two numbers, which are to each other, as 16
co 9, and 24 is a mean proportional between them. What are
the numbers 1 Ans. 32 and 18.
13. The sum of two numbers is to their difference as 4 to 1,
and the sum of their squares is to the greater as 102 to 5.
What are the numbers 1 Jlns. 15 and 9.
14. If the number 20 be divided into two parts, which are to
each other in the duplicate ratio of 3 to 1, what number is a
mean proportional between those parts ?
Jlns. 18 and 2 are the parts, and 6 is the mean proportion
between them.
15. There are two numbers in proportion of 3 to 2 ; and if
6 be added to the greater, and subtracted from the less, the results
will be as 3 to 1. What are the numbers ? Jins. 24 and 16.
16. There are three numbers in geometrical progression, the
product of the first and second, is to the product of the second
and third, as the first is to twice the second ; and the sum of
the first and third is 300. What are the numbers ?
Jlns. 60, 120, and 240.
17. The sum of the cubes of two numbers, is to the differ-
ence of their cubes, as 559 to 127 ; and the square of the first,
multiplied by the second, is equal to 294. What are the num-
bers ? rfns. 7 and 6.
18. There are two numbers, the cube of the first is to the
square of the second, as 3 to 1 ; and the cube of the second is
to the square of the first as 96 to 1. What are the numbers ?
tins. 12 and 24.
BINOMIAL THEOREM. 213
SECTION VI.
CHAPTER I.
INVESTIGATION AND GENERAL APPLICATION OF
THE BINOMIAL THEOREM.
(Art. 127.) It may seem natural to continue right on to the
higher order of equations, but in the resolution of some cases in
cubics, we require the aid of the binomial theorem ; it is there-
fore requisite to investigate that subject now.
The just celebrity of this theorem, and its great utility in the
higher branches of analysis, induce the author to give a general
demonstration : and the pupil cannot be urged too strongly to
give it special attention.
In (Art. 67.) we have expanded a binomial to several powers
by actual multiplication, and in that case, derived a law for form-
ing exponents and coefficients when the power was a whole
positive number ; but the great value and importance of the
theorem arises from the fact that the general law drawn from that
case is equally true, when the exponent is fractional or negative,
and therefore it enables us to extract roots, as well as to ex-
pand powers.
(Art. 128.) Preparatory to our investigation, we must prove
the truth of the following theorem :
If there be two series arising from different modes of ex-
panding the same, or equal quantities, with a varying quan-
tity having regular powers in each series ; then the coefficients
of the same powers of the varying quantity in the two series
are equal.
For example, suppose
Jl+Bx+Cy?+nx*, &c. =a-\-bx+cx z +dx*, &c.
This equation is true by hypothesis, through all values of a?.
It is true then, when #=0. Make this supposition, and <ft=a
Now let these equal values be taken away, and the remainder
divided by x. Then again, suppose #=0, and we shall find
fi=b. In the same manner we find Cc, D=d, E=e, &c.
214 ELEMENTS OF ALGEBRA.
(Art. 129.) A binomial in the form of a-\-x may be put in
the form of a X f 1 -j- - ) ; for we have only to perform
the multiplication here indicated to obtain a-\-x. Hence
(x\ m
1 + - ) > it will be sufficient to mul-
tiply every term of the expanded series by a m for the expansion
of (-j-.r) OT , but as every power or root of 1 is t, the first term
/ x \ m
of the expansion of ( l-f~ J is 1, and this multiplied by a m
must give a m for the first term of the expansion of (a-\-x) m , what-
ever m may be, positive or negative, whole or fractional.
99
As we may put x in place of -, we perceive that any bino-
mial may be reduced to the form of (1-j-o?), which, for greater
facility, we shall operate upon.
(Art. 130.) Let it be required to expand (l-f-a;) m , when m is
a positive whole number. By actual multiplication, it can be
shown, as in (Art. 67.) that the first term will be 1, and the
second term mx. For if m=2, then
(l+a?) m =(l+a?) 8 =l+2ar, &c.
If m=3, (l-hr) ra =l+3x, &c.
And in general, (l-\-x) m =l+mx-\-tfx z ,+JBx*, &c.
The exponent of x increasing by unity every term, and .#,
It, C, &c., unknown coefficients, which have some law of de-
pendence on the exponent m, which it is the object of this investi-
gation to discover.
(Art. 131.) Now if m is supposed to be a fraction, or if m=-,
the expansion of (l-}-x) m will be a root in place of a power, and
we must expand (l-\-x) r . _i
For example, let us suppose r=2, then r
BINOMIAL THEOREM. 21
and to examine the form the series would take, let us actually
undertake to extract the square root of (l-\-x) by the common
rule.
OPERATION.
2-r-ar _i^
Thus we perceive that in case of square root, the first term of
the series must be unity, and the coefficient of the second term
is the index of the binomial, and the powers of x increase by
unity from term to term.
We should find the same laws to govern the form of the series,
if we attempted to extract cube, or any other root; but, to be
general and scientific, we must return to the literal expression
Now as any root of 1 is 1, the first term of this root must be
1, and the second term will have some coefficient to x. Let
that coefficient be represented by p ; and as the powers of r will
increase by unity every term, we shall have
\ &c.
Take the r power of both members, and we shall have
l+x^^+px+dx 2 , &c.) r
As r is a whole number, we can expand this second mem-
ber by multiplication ; that is, by (Art. 130.), the second mem-
ber must take the following form
l+*=l-|-rjoa:-M'* 2 , &c.
Drop 1, and divide by x, and we have
Ky (Art. 128.) l=rp or p=- ;
210 ELEMENTS OF ALGEBRA.
That is, the coefficient represented by p must be equal to tho
index of the binomial.
Therefore (\-\-x) r = \ + \.x-{- Ay?-\- Bx* -f, &c. ; the same
general form as when the exponent was considered a whole
number.
(Art. 132.) If we take m=- we have to expand the root of
a power. The first term must be 1, and the second term will
contain a?, with some coefficient, and the coefficients of x will
rise higher and higher every term.
n
That is, (l+x)^ = l+7?^+^^ 2 , &c. Take the r power of
both members, and (l-}-x) n =(l-\-px t &c.) r .
Expanding both members, as in (Art. 130),
l+nx-{-ax z , &c.=l+ rp.r-f-.tftf 2 , &c.
Now, by (Art. 128), 71=773 or p=^.
n
Therefore, (1-J-J?) r =l+ x+Jlx*+E3?, &c. ; the first two
terms following the same law, relative to the exponents, as in the
former cases. Now let us suppose m negative. Then
(1-Hr) 1 * will become (l-\-x)~ m = (Art. 18.)
Or by (Art. 130.) - * 2
; l+maj-f-^a: 2 , &c.
By actual division, l+mx+rfx*, &c.) 1 (J mx-\-Jl'3f, &c.
mx 3?
mx m 2 x 2
That is, (\-\-x)~ m =lmx J rJl l yr 1 which shows that the same
general law governs the coefficient of the second term, as in the
former cases.
Hence it appears that whether the exponer* m of a binomial
BINOMIAL THEOREM. 217
be positive or negative, whole or fractional, th same general
form of expression must be preserved.
That is, in all cases (l+ar) m =l+roar+-foM- Btf, &c.
(Art 133.) For clearness of conception, let the pupil bear in
mind that the coefficients of an expanded binomial quantity
depend not at all on the magnitudes of the quantities themselves,
but on the exponent. Thus, (a-\-b) to the 5th, or to any other
power, the coefficients will be the same, whether a and b are
great or small quantities, or whatever be their relation to each
other.
(Art. 134.) The equation (l+x^^l+mx+rftf+Ba*, &c.,
is supposed to be true, therefore it must be true, if we square
both members. But we have only a portion of one member.
We have, however, as much as we please to assume, and suffi-
cient to determine the leading terms of its square, which is all
that we desire. Square both members, and (l-}-2a:-{-# 8 ) m =
(l+mH-^-l-^-f-Ck 4 , &c.) 2 .
By expanding the second member, and arranging the terms
according to the powers of tf, we shall have
Now if we assume yZx-}-^, the first member of this equa-
tion becomes (l-j-t/) m . If we expand this binomial into a series.
it must have the same coefficients as the expansion of (l-f-#) OT ,
because the coefficients depend entirely on the exponent m.
(Art. 133.)
Therefore, l+*=l+rm/-Mi 2 -r- 8 -f &c -
Put the values of y, T/*, y 3 , &c., in this equation, and arrange
the terms according to the powers of #, and we have
SB
* t <fcc. (2)
The left hand members of equations (1) and (2), are identical,
19
218 ELEMENTS OF ALGEBRA.
and the coefficients of like powers in the second member must
be equal. (Art. 128.)
Therefore, 4^-l-m=m 2 +2^, or
And 8
mu r r> - - - /o\
Therefore B=~~- ^=m. - -- - (3)
o . o
By putting the coefficients of the fourth powers of x equal,
we have
To obtain the value of C from this equation, in the requisite
form, is somewhat difficult.
We must make free use of the preceding values of Jl and B,
which are alone sufficient ; but, to facilitate the operation, we
shall make use of the following auxiliary.
Assume P=m 2, then P-\-im 1,
m,P-\-m m 1 - , .
and o = m .-=rf (a)
Also, by inspecting equation (3), we perceive that
35=^P, and 2m=^^ (b)
3
By putting the values of 125 and 2m5 in the primitive equa-
tion, and dividing every term by .#, and in the second member
taking the value of Jl from equation (), and we shall have
_i_^ p ,,
_____ :__ __
Multiply by 6, and substitute the value of
24F-}- 6 =24?7i 42, because P=m 2, and we have
84O
__-|-24m 42=4m/M-3mP-r-3ra.
Jl
Collecting terms, and dividing by 7, gives
BINOMINAL THEOREM. 2-1-9
But 3m 6=3P, which substitute and transpose, and
ll:=mP_3P=P(m 3)
yl
Or C^ f ( m ~ 3 )=m m ~ 1 m ~~ m ~ 3
~~12 2 ~~3~" 4
Therefore the development of (l-\-x) m must be
2 2
234
whatever be the value of m, positive or negative, whole or
fractional, and thus far the law of development has been demon-
strated, and we infer, and can only infer, that this law will
continue true, whatever be the number of terms. Hence, the
demonstration is complete, only so far as we extend it.
This series will terminate when (771) is a whole positive
uumber, but in all other cases it will be infinite.
Few pupils, who pay attention, find difficulty in compre-
hending the preceding method of demonstration, and for that
reason we preferred it ; but to the following demonstration, no
objection as to completeness or perfection can be found.
SECOND DEMONSTRATION.
Assume (l+ x )*==l+Ax+x a + Cx*+2)x*-}- &c. (1)
Then (\J ry y=\J r Ay+By* + Cy*-\-Dy* + & c > W
By subtracting (2) from (1) and dividing the result by
(ff y) we obtain
x y xy xy xy
+ <&c. (3)
Place (l+z)=P, and(l+y) = . (a)
Then xy=PQ. Whence (3) becomes
^U c
y
If we divide (x 2 y 2 ) by (x y) the quotient will term-
inate with the second term, and (x 3 y 3 ) divided by (x y) the
quotient will terminate with its third term, and so on.
220 ELEMENTS OF ALGEEBRA.
Then if n is a whole number, the division of the first member
as indicated will terminate with the n th term of the quotient.
Now, divide m the last equation as indicated, and after-
ward suppose xy. It then follows that P= Q, and we have
p*-i_^p*-a g_|_p- 3_|_ < c> to n terms. And if P= Q the
whole quotient must be nP D ~ l for the first member.
Whence, nP*- l =A+2Bx-{-3Cx s -{-4I>x 3 +5Ex* &c.
But P = l+x.
Multiply these two equations, member by member, and
we shall have,
But we perceive, by inspecting equation (1), that
By equating the right hand members of these two equa-
tions, and observing that the coefficients of the like powers of
x, must be equal, we shall have
or =
or C
or D=
<KC.
Whence A=n
2
n 1 n2
_-
n 1 n 2 n 3
234
and so on, the law of continuation being obvious from the
equations, to whatever term it may be extended.
Thus the demonstration is complete when n is a whole posi-
tive number.
Now let n be an irreducible fraction, as ~,
BINOMIAL THEOREM. 221
1 ,.x
Then (l+*) t= =:
And
j. i^
Now assume ( 1 -f-#) l =P and ( 1 +y ) l =
8
8
Then, (J+*) 1 =P
And l+a=
Whence zy=P i (>* (c)
Now if we subtract (b) from (), and divide the remainder
by x y, we shall have
Now divide numerator and denominator of the first member
by (P Q), and it will become
, fcc.
Now, because 5 and J are whole numbers, the division in both
numerator and denominator will be complete, and if we sup-
pose x=y, it follows that P= Q, and the results of the several
divisions will be
+, &c. (e)
But JP* =!+.
Multiplying these equations, member by member, will
produce
, 3
But by inspecting equation (a), and observing that (l-J-#)r
222 ELEMENTS OF ALGEBRA.
s
-P =-+-.Ax+-Bx*-\--Cx*+, Ac. (g)
t t t ' t t
Equating (/) and (g), and operating as before, we find
A=f-
3 * 2 3~'
&c. &c.
The same law of development as before, and this completes the
demonstration for all positive fractional exponents.
Now, let the exponent be a negative fraction,
Then, (l~}-3;)-^=l+Ax-{-x a -\-Cx 3 -^, &c. (a')
*
g j
Now assume, (l-fz)-7= />" Then, (l+#)-T= p~ 4
Or, Pt = i_|^.
In like manner, Q l =
By subtraction, ^P l Q l x y
Subtracting (V) from (a') and dividing by (x y), as in the
other cases, we shall obtain
Ac.
The first member of this equation is
1 1
-
BINOMIAL THEOREM. 223
But _=-
t
When the division is performed, and P supposed equal to Q.
hence, * P~ 8 " 1 =A+2
But, P i =l+x.
Whence, * P~ 8 " 1 =A+2x+3 Cx 2 +, &c.
Therefore, *-P * =
A
-, <fec.
But, - P =-JLLAa;-ix* ,
t t t t
Whence, A= 8 - J5= l.lil!, &c.
t t 2
and thus the demonstration is shown to be complete, under
all possible circumstances.
We may use either of the three following forms :*
Ac. (1)
^+,&c.(2)
2 a 2 2 3 a
2 3 a
&c. (3)
_ 1 ; !!Z^a-3 ;z .3 +> &c .
APPLICATION".
I. Required the square root of (l-|-a?).
Apply formula (1), making w*=-|- ; then the development
2.4 2.4.6 2.4.6.8
The law of the series, in almost every case, will become
apparent, after expanding three or four terms, provided we
keep the factors separate.
*In practical examples we may be required to expand (1 x) as well as
(1-t-tf), and the exponent may be negative as well as positive. But in all
cases the products in the second member must conform to the rules of mul-
tiplication.
224 ELEMENTS OF ALBKBRA.
The above will be the coefficients for any binomial square
root (Art. 133 ); hence the square root of 2 is actually expressed
in the preceding series, if we make a?=l.
Then (1-f 1) 5 ==1-K_-L, & c . The square root of 3 w
expressed by the same series, when we make #=2, &c.
2. Required the cube root of (a-\-b) or its equal, af H V
Formula (2) gives
This expresses the cube root of any binomial quantity, or any
quantity that we can put into a binomial form, by giving the
proper values to a and b. For example, required the cube root
of 10, or its equal, 8+2. Here a=8, 6=2. Then ^=2,
and b -=$.
a
1 2 25
Therefore, 2(1+-^ - z +^ Q ' Q ^ &c.) is the cube root
of 10, and so for any other number.
S. Expand - -r into a series, or its equal, (a b)~ l .
1 b .b*b*
b* 6 4 . b*
4. Expand (a 2 -h&*) into a series, or its equal, M-| 2
Expand d(c*-}-x?) * into a series.
d a?
BINOMIAL THEOREM. 225
6. Expand (a 2 #*)* into a series.
Ans.
f
2 5 a 2 2 7 a 4
7. Expand
8. Find the cube root of
a 3 +b
9. Find the cube root of 31.
31=27+4=27/^1+^
277-
, .
. 1 - + -- - , &c
3a 3 9a 6 8 la 9
Place-i=a
27/ 27
Ans. 3l+_-. +.-- 3 +,&c. =3.156066
V 327 36 (27)^369 (27) 3 /
1O. Find the seventh root of 2245, or any other number, by
the binomial theorem.
N. B. Find by trial the greatest seventh power, less than the
given number, (2245), and divide the given number by it.
Thus, 3 7 =2187
2187 2187 2187
Ans. sf 1+1. Jl? __ A YJLY+&0. ^=3. 01 13574
\ ' 7 2187 7.14 \2187/ /
11. Find the cube root of 9, by the binomial series.
9=8(1+|). Cube root,
Ans. 2.080084.
19.' Find the cube root of 7, by the binomial series.
7=(8 1)=8(1 ). Cube root, 2(1 J)*
Ans. 1.912933.
13. Expand (x -y)" 1 , into a series.
n 1
Ans. x* tt B
226 ELEMENTS OF ALGEBRA.
When n = 1, this series becomes
V^JJ/ 3 , y 3 ,
SF?FiTF^, &c., Ac.
14. Find an approximate cube root of 100, by the binomial
series.
N. B. The nearest cube number to 100, is 125.
/ 25 \
Therefore, we place 100=125 25=125^1 ~\%).
Or, 100=125(l). Whence (100)^=5(1 $)%.
Ans. 4.64159, nearly.
Thus, by a little artifice, this series can be used to extract
the cube (or any other root), of any number.
15. Expand - into a series.
a+cx
N. B. It is much easier to expand all such expressions as
this, by actual division, than by the binomial theorem, and
we take the example, merely to show that it can be expanded
by the binomial theorem.
The above expression is the same as 1 ( ~r c ) x t
a-\-cx
Place -J-c=w, then we are required to expand
/ M, nix/* , cx\
mx(a+cx) n or, _(!+ )
a \ a/
because n= 1.
Because n= 1, this series becomes
IJL-i, Ac.
a 5
this multiplied by , and the product subtracted from 1, pro-
a
, mx , mcx 2 me 2 x 3 . mc 3 x* mc*x 5 ., u
duces 1 _ -4- ___ - -- -t- --- - , &c. the result
a a 2 a 2 a 4 a 5
sought, when (b-\-c) is written in place of m.
INFINITE SERIES. 227
16. Expand ' , or its equal 14- _ , or its equal
x 1 x 1
-^.
1 x
Ans. 1 2* 2s 3 2* 3 2# 4 , &c., &o.
17 Expand z(l a?)-" 3 into a series.
Ans. *+3:r 2 +6z 3 -f-10z 4 -f.
18. Expand (64+1)*, or its equal, 8(1+^)*, into a series.
N. B. This is the same as demanding the cube root of 65.
CHAPTER II.
OF INFINITE SERIES.
(Art 136.) An infinite series is a continued rank, or progres-
sion of quantities in regular order, in respect both to magnitudes
and signs, and they usually arise from the division of one quantity
by another.
The roots of imperfect powers, as shown by the examples in
the last article, produce one class of infinite series. Some of
the examples under (Art. 121.) show the geometrical infinite
series.
Examples in common division may produce infinite series for
quotients ; or, in other words, we may say the division is con-
tinuous. Thus, 10 divided by 3, and carried out in decimals,
gives 3.3333, &c., without end, and the sum of such a series is
3|. (Art. 121.)
(Art. 137.) Two series may appear very different, which arise
from the same source ; thus 1, divided by 1-f-a, gives, as we may
bee, by actual division, as follows:
l-f-a)l (1 a-j-a 2 a 3 -f-a 4 , <fcc. without end.
228 ELEMENTS OF ALGEBRA.
Also, +!)! ( -+- j, &c. without end.
' \ a a* a* a*
a
These two quotients appear very different, and in respect to
blngle terms are so ; but in these divisions there is always a re-
mainder, and either quotient is incomplete without the remain-
der for a numerator and the divisor for a denominator, and when
these are taken into consideration the two quotients will be
equal.
We may clearly illustrate this by the following example :
Divide 3 by l-f-2, the quotient is manifestly 1 ; but suppose them
literal quantities, and the division would appear thus :
(3 6+12, &c.
3+6
6
612
12
12+24
24
Again, divide the same, having the 2 stand first.
|+f,&c.
Now let us take either quotient, with the real value of its re-
mainder, and we shall have the same result.
Thus, 3+12=15; and 6, and the remainder 24 divided
INFINITE SERIES. 229
by 3, gives 8, which make 14 ; hence, the whole quotient
is 1.
Again, H-t=V, and |i=l.
Hence, y |=f =1 the proper quotient.
If we more closely examine the terms of these quotients, we
shall discover that one is diverging, the other converging, and
by the same ratio 2, and in general this is all a series can show,
the degree of convergency.
(Art. 138.) We convert quantities into series by extracting
the roots of imperfect powers, as by the binomial, and by actual
division, thus :
1. Convert -. into an infinite series.
a-\-x
Thus, +*, (l_+_ + , & c .
2. Convert - into an infinite series.
a x
Observe that these two examples are the same, except the
signs of x : when that sign is plus the signs in the series will be
alternately plus and minus ; when minus, all will be plus.
3. What series will arise from - ?
1 x
Jlns. 1+2*+ 2*M-2# 3 , &c.
Observe that in this case the series commences with 2x. The
unit is a proper quotient, and the series arises alone from r --
1 z,
the remainder after the quotient 1 is obtained.
230 ELEMENTS OF ALQEBRA.
4. What series will arise from -r- r ^ ?
,,,2 ^4
_ f___^
a a 4
Observe, in this example, the term x, in the numerator does
not find a place in the operation ; it will be always in the re-
mainder : therefore,^-; - will give the same series.
2 -- z
5. What series will arise from dividing 1 by 1 a-}- a 2 , or
from
- - -, -? rfns. 1-l-a a 3 a 4 -}-a 6 +a 7 a 9 a 10 ,
1 a-j-cr
In this example, observe that the signs are not alternately plus
and minus, but two terms in succession plus, then two minus ;
this arises from there being two terms in place of one after the
minus sign in the divisor.
6. What series will arise from y~~ ?
fins. a-j-ar-r-ar 2 -|-ar 3 -l-ar 4 , &c.
Observe that this is the regular geometrical series, as appears
in (Art. 118.)
7. What series will arise from - - ?
rfns. 1-f-l-f-l-fl, &c
That is - is 1 repeated an infinite number of times, or infinity,
a re/ U corresponding to observations under (Art. 60.)
SUMMATION OF SERIES. 231
8. What series will arise from the fraction ?
If =&, this series will be -, repeated an infinite number of
times.
This series can also be expanded by the binomial theorem,
for . j=h(a 6) . This observation is applicable to seve-
ral other examples.
(Art. 139.) A fraction of a complex nature, or having com-
j /
pound terms, such as 2 , may give rise to an infinite
1 &XdX
series, but there will be no obvious ratio between the terms.
Some general relation, however, will exist between any one term
and several preceding terms, which is called the scale of rela-
tion, and such a series is called a recurring series. Thus
the preceding fraction, by actual division, gives l-}-x-{-5x c *-\r
ISa^-f-^la^+lSltf 5 , &c., a recurring series, which, when carried
to infinity, will be equal to the fraction from which it is derived.
l+2a?
Expand -^ into a series
CHAPTER III.
SUMMATION OF SERIES.
(Art, 140.) We have partially treated of this subject in geo-
metrical progression, in (Art. 121) ; the investigation is now
more general and comprehensive, and the object in some respects
different. There we required the actual sum of a given number
of terms, or the sum of a converging infinite series. Here the
series may not be in the strictest sense geometrical, and we may
not require the sum of the series, but what terms or fractional
quantities will produce a series of a given convergent]/.
232 ELEMENTS OF ALGEBRA.
The object then, is the converse of the last chapter ; and for
every geometrical series, our rule will be drawn from the sixth
example in that chapter ; that is, - - , a being the first term
of any series, and r the ratio. We find the ratio by dividing
any term by its preceding term.
Hence, to find what fraction may have produced any geomet-
rical series, we have the following rule:
RULE. Divide the first term of the series by the alge-
braic difference between unity and the ratio.
EXAMPLES.
1. What fraction will produce the series 2, 4, 8, 16, &c?
2
Here a=2, and r=2 ; therefore, - - Jlns
1 Z
2. What fraction will produce the series 3 9-J-27 81, &c.?
Here 0=3, and r= 3; then r=3,
Hence, -fL-
3. What fraction will produce the series ^, T 7 , &c.
Here = T V and r= T V; therefore,
10 3 31
4. What fraction will produce the series,
_+__ 3 , & c . ? [See example 1, (Art. 138.)]
x 1 a
Here 0=1, and r= - ; then - = = ,
a .x a-i-x
a
5. What fraction will produce the series
&c. ? [See example 3, (Art. 138.) and the observation in con-
nection.]
SUMMATION OF SERIES. 233
2#
Here a=2x, r=x\ then - - , will give all after the first
2iX 1 l %
term: therefore, 1+- -- =- - , flns.
, w . e d db . db 3 (/6 s
6. What fraction will produce the series - H F -- 4
c er c* c*
&c. ? AM.
What fraction will produce the series 7+- r^ i rr~ '
'b ax'
8. What fraction will produce the series
2ab
3 ,- ' FT'
a a 2 a 3 a-J-o
9. What fraction will produce the series
!-}- a 3 a 4 4-a 6 +a 7 a 9 a 10 , &c. ?
See example 5, (Art. 138.)
Put l+a=6; then a 3 -J-a 4 =a 3 6, and a e + 7 =a 6 6, and the
scries becomes b a?b-\-a*b , <fec.
b 1-fa 1
Hence, the fraction required is = 3=^-: 5 =- : 5.
1-j-a 3 1+a 3 1 +a 2
10. What fraction will produce the series a?+a^-f-a^, &c. ?
Ana. ..
1 x
If a?s=l f this expression becomes j f~o' a S 7 m bol of in
fmity
11. What fraction will produce the series 1-f-f+JL, &c. ?
1 5
53
Hence, the sum of this series, carried to infinity, is 2.
20
234 ELEMENTS OF ALGEBRA.
In the same manner, we may resolve every question in (Art.
tin.)
12. What is the sum oi the series, or what fraction will pro-
duce the series 1 a-\-a? 4 -f-a 6 a'-\-cP , &c. ?
Jlns.
13. What is the value of the infinite series
RECURRING SERIES.
(Art. 141.) We have explained recurring series in (Art. 139.)
and it is evident that we cannot find their equivalent fractions
by the operation which belongs to the geometrical order, as no
common relation exists between the single terms. The fraction
|_l_2<>'
--- 2> by actual division, gives the series l-|~3#-f-4#M-7a: 8
L^~~3C' i 3C
-\-llx* -\-\8x*, &c., without termination ; or, in other words, the
division would continue to infinity.
Now, having a few of these terms, it is desirable to find a
method of deducing the fraction.
There is no such thing as deducing the fraction, or in fact no
fraction could exist corresponding to the given series, unless
order or a law of dependence exists among the terms ; therefore
some order must exist, but that order is not apparent.
Let the given series be represented by
d+B+C+D+E+F, &c.
Two or three of these terms must be given, and then each sue
oeeding term may depend on two or three or more of its pre-
ceding terms.
In cases where the terms depend on two preceding terms we
may have
C=mBx-}-nJlx*
J)=mCx-\-nBx*
RECURRING SERIES. 236
In cases where the terms, or law of progression depend on
three preceding terms we may have
fi=m
E=mDx+n
F=m
&c.=&c.
The reason of the regular powers of x coming in as factors,
will be perfectly obvious, by inspecting any series.
The values of m, n and r express the unknown relation, or law
that governs the progression, and are called the scale of relation
We shall show how to obtain the values of these quantities in a
subsequent article.
(Art. 142.) Let us suppose the series of equations (1), to be
extended indefinitely, or, as we may express it, to infinity, and
add them together, representing the entire sum of Jl-\-B-\-
C-\-D, <fec., to infinity, by S ; then the first member of the
resulting equation must be (S J? B), and the other member
is equally obvious, giving
S A
A+BmJlx
Hence, S=-l -- r (a)
1 mx no?
In the same manner, from equations (2), we may find
s _J+B+G-(Jl+B)mx--Jlna* ,
1 mx nx* rx* * '
(Art. 143.) The/orw of a series does not depend on the
value of x, and any series is true for all values of x. Equations
(1) then, will be true, if we make 2=1.
Making this supposition, and taking the first two equations of
the series (1), we have
C^mB+nA ) (c)
And D=mC+nB \
In these equations, ,/?, B, (7, 7J, are known, and m and n un-
known ; but two unknown quantities can be determined from two
equations ; hence m and n can be determined
236 ELEMENTS OF ALGEBRA.
When the scale of relation depends upon three terms, wt
take three of the equations (2), making x=l, and we determine
m, n, and r, as in simple equations.
EXAMPLES.
1. What fraction will produce the series
To determine the scale of relation, we have w#=l, .#=3,
=4, and D=7. Then from equations (c), we have
n+3m=4
3n+4m=7
These equations give m=l, and 72= 1.
Now to apply the general equation (a), we have Jj=\ t 23=3x.
A+B-mAx 1+3*-* 1+2*
1 hen o= - 5- = -; -- ?=; -- a t *
1 mx nx* 1 x ar 1 x or
2. What fractional quantity will produce the series
l + 6z+12arM-48z 3 -f 120;r 4 , &c., to infinity?
Here .#=1, B=Qx, m=l, n=6.
Hence, by applying equation (a), we find - - j for the
1 ~^x o* ,
expression required.
3. What quantity will produce the series
l_L3;r-f Sa^-fTar'-i-Qtf 4 , to infinity ? l-|-ar
4. What quantity will produce the infinite series
1-f 4^+6^+11^+ 28^ 4 +63^, <fec. ? in which m=2, n= 1,
r=3.
[Apply equation (6).] ^ns.
5. What fraction will produce the series
, &c.?
REVERSION OF A SERIES.
6. What fraction will produce the series
, &c.
287
REVERSION OF A SERIES-
(Art. B.) To revert a series is to express the value of the un-
known quantity in it, (which appears in the several terms under
regular powers,) by means of another series involving the powers
of some other quantity.
Thus, let x and y represent two indeterminate quantities, and
the value of y be expressed by a series involving the regular
powers of x. That is
&c.
To revert this, is to obtain the simple value of ar, by means
of another series containing only the known quantities, a, b, c,
d, &c. and the powers of y.
To accomplish this,asswme
*, &c.*
Substitute the assumed value of x for x, x*, a? 8 , <fcc., in the pro-
posed series, and transposing y, we shall have
1
y-\-aB
+3 cA*B
&c.
* By examining previous authors, we have found none that explain the
rationale of such assumptions : but these are points on which the learner
requires the greatest profusion of light. We explain thus : the term x must
have some value, positive, negative, or zero, and the series Ay-\-By 2 -\- Cy*,
fcc., can be positive and of any magnitude. It can be negative, and of any
magnitude, by giving the coefficients A, B and C, negative values. It can bo
zero by making ,y=0. Therefore it can express the value of or, whatever that
may be ; and because the powers of y are regular, the substitution of this
value of x for the several powers of x, in the primitive series, will convert
that series into regular powers of y, which was the object to be accomplished.
238
ELEMENTS OF ALGEBRA.
As every term contains y, \ve can reduce the equation by
dividing by y ; and afterwards, in consideration that the equation
must be true for all values of y : making y=0, we shall have
I 1=0 or
.*-!
a
Continuing the same operation and mode of reasoning as in (Art.
128.), and we shall find, in succession, that
n
a
2b z ac
a 1
= &c.
Substituting these values of .#, B, C, &c., in the assumed
equation and we have the value of x in terms of , b, c, &c.,
and the powers of y as required, or a complete reversion of the
series.
EXAMPLES.
1. Revert the series i/=x-}-x' 2 -}-j: 3 , &c.
Here a=l 6=1 e=l, &c.
Assume x=^.y-f-% 2 +Oj/ 3 -[-&c.
Result, ^= +i/ 3 i/ 4 + 5 &c.
ft. Revert the series x=y J - -\-- - &c.
234
Here a I b=% c=| J= | &c.
Assume y^Ax+Btf+Ctf+Dx*-^ &;c., and find
the values of rf, B, C, &c., from the formulas (F),
Result, ^+l. + _+.-. & c
LOGARITHMS.
3. Revert the series y=rc+3a? 2 +5a >3 +7a? 4 +9a ;5 , &c.
Result x=
239
In case the given series is of the form of xay-\- by*-}-cy 5 ,
&c., the powers of y varying by 2, the equations (F] will not
apply, and we must assume y=J2x-\-J3x 3 -\-Cx 5 , &c., and sub-
stitute as before, and we shall find
(0)
a 10
In common cases, after the coefficients, as far as D, are deter-
mined, the law of continuation will become apparent, especially
if the factors are kept separate.
CHAPTER IV.
EXPONENTIAL EQUATIONS AND LOGARITHMS,
(Art. 144.) We have thus far used exponents only as known
quantities ; but an exponent, as well as any other quantity, may
be variable and unknown, and we may have an equation in the
form of a*=b.
This is called an exponential equation, and the value of x can
only be determined by successive approximations, or by making
use of a table of logarithms already determined.
(Art. 145.) Logarithms are exponents. A given constant
number may be conceived to be raised to all possible powers, and
thus produce all possible numbers ; the exponents of such pow-
ers are logarithms, each corresponding to the number produced.
Thus, in the equation a*=6, x is the logarithm of the number
b ; and to every variation of x, there will be a corresponding
240 ELEMENTS OF ALGEBRA.
variation to b a is constant, and is called the base of the sys
tern, and differs only in different systems.
The constant a cannot be 1, for every power of 1 is 1, and the
variation of x in that case would give no variation to b ; hence,
the base of a system cannot be unity ; in the common system it
is 10.
In the equation 10 Z =2, x is, in value, a small fraction, and
is the logarithm of the abstract number 2.
In the equation a x =b, if we suppose x=l, the equation becomes
a l =a; that is, the logarithm of the base of any system is unity.
If we suppose 37=0, the equation becomes a=l ; hence, the
logarithm of 1 is 0, in every system of logarithms.
(Art. 146.) The logarithms of two or more numbers added
together give the logarithm of the product of those numbers,
and conversely the difference of two logarithms gives the lo-
garithm of the quotient of one number divided by the other.
For we may have the equations z =&, a v =b', and a*=b".
Multiply these equations together, and as we multiply powers
ly adding the exponents the product will be
Hence, by the definition of logarithms, x-\-y-\-z is the loga-
rithm for the number represented by the product bb'b". Again.
divide the first equation by the second, and we have a x ~ y =j- i ;
and from these results we find that by means of a table of loga-
rithms multiplication may be practically performed by addition,
and division by subtraction, and in this consists the great utility
of logarithms.
(Art. 147.) In the equation a x =b, take a=10, and x succes-
sively equal to 0, 1, 2, 3, 4, &c.
Then 10=1, 10^10, lO^lOO, 10 3 =1000, &c.
Therefore, for the numbers
1, 10, 100, 1000, 10000, 100000, &c., we have for corres-
ponding logarithms
0, 1, 2, 3, 4, 5, <fcc.
LOGARITHMS. 241
Here it may be observed that the numbers increase in geometri-
cal progression, and their logarithms in arithmetical progression.
Hence the number which is the geometrical mean between two
given numbers must have the arithmetical mean of their lo-
garithms, for its logarithm.
On this principle we may approximate to the logarithm of any
proposed number. For example, we propose to find the /a-
garithm of 2.
This number is between 1 and 10, and the geometrical mean
between these two numbers, (Art. 122.), is 3.16227766. The
arithmetical mean between and 1 is 0.5; therefore, the number
3.16227766 has 0.5 for its logarithm.
Now the proposed number 2 is between 1 and 3.162, &c.,and
the geometrical mean between these two numbers is 1.778279,
and the arithmetical mean between 0. and 0.5 is 0.25 ; therefore,
the logarithm of 1.778279 is 0.25.
Now the proposed number 2 lies between 1.778279, and
3.16227766, and the geometrical mean between them will fall
near 2, a little over, and its logarithm will be 0.375. Continuing
the approximations, we shall at length find the logarithm of 2 to
be 0.301030, and in the same manner we may approximate to
the logarithm of any other number, but the operation would be
very tedious.
(Art. 148.) We may take a reverse operation, and propose a
logarithm to find its corresponding number; thus, in the general
equation a z =n; x may be assumed, and the corresponding value
of n computed.
Thus suppose ar= T 3 ff ; then (10) T =n, or 10 3 =n 10 .
Hence, n=J 1000=1. 9952623 15.
That is, the number 1.9952, <fcc., (nearly 2) has 0.3 for its
logarithm. In the same way we may compute the numbers cor-
responding to the logarithms 0.4, 0.5, 0.6, 0.7, &c.
(Art. 149.) We may take another method of operation to find
the logarithm of a number.
21
242 ELEMENTS OF ALGEBRA.
Let the logarithm of 3 be required.
The equation is 10 Z =3, the object is to find x.
It is obvious that x must be a fraction, for 10 1= =10 ; there
fore,
Let #= . Then 10 Z '=3, or 10=3*. Here, we perceive,
that x' must be a little more than 2. Ma.te x'=2-{- p Then
Hit x 1
3 2 X3 =10; or 3 = .
7
Or ( ~) =3.
Here we find by trial that x" is between 10 and 1 1 ; take it 10,
and a?'=2-f-y^ ; hence, a?=!,j=0. 476, for a rough approxima-
tion for the logarithm of 3. A more exact computation gives
0.4771213; but all these operations are exceedingly tedious, and
to avoid them, mathematicians have devised a more expeditious
method by means of a converging series ; which we shall inves-
tigate in a subsequent article.
(Art. 150.) It is only necessary to calculate directly the lo-
garithms of prime numbers, as the logarithms of all others may
be derived from these. Thus, if we would find the logarithm
of 4, we have only to double that of 2; for, taking the equation
(10) z =2, and squaring both members we have, (10) 2z =4 ; or
taking the same equation, and cubing both members, we have
(10) 3z =8 ; which shows that twice the logarithm of 2 is the
logarithm of 4, and three times the logarithm of 2 is the loga-
rithm of 8 ; and, in short, the sum of two or more logarithms
corresponds to the logarithm of the product of their numbers,
(Art. 142.), and the difference of two logarithms corresponds to
the logarithm of the quotient, which is produced from dividing
one number by the other.
Thus, the logarithm of j log. a log. b. Tne abbrevia-
tion, (log. a), is the symbol for the logarithm of a.
LOGARITHMS. 243
(Art. 151.) As the logarithm of 1 is 0, 10 is 1, 100 is 2, &c.,
we may observe that the whole number in the logarithm is one
less than the number of places in the number.
The whole number in a logarithm is called its characteristic,
and is not given in the tables, as it is easily supplied. For ex-
ample, the integral part of the logarithm of the number 67430
must be 4, as the number has 5 places. The same figures will
have the same decimal part for the logarithm when a portion of
them become decimal.
Thus, 67430 logarithm 4.82885
6743.0 3.82885
674.30 2.82885
67.430 1.82885
6.7430 0.82885
.67430 1.82885
For every division by 10 of the number, we must diminish
the characteristic of the logarithm by unity.
The decimal part of a logarithm is always positive ; the index
or characteristic becomes negative, when the number becomes
less than unity.
By reference to (Art. 18.), we find that =10-', = =
10~ 2 , &c. That is : fractions may be considered as numbers
with negative exponents, and logarithms are exponents ; there-
fore the logarithm of , or .1, is 1 ; of , or .01, is 2,
10 100
&c. If any addition is made to .01 the logarithm must be more
than 2 ; but, for convenience, we still let the index remain
2, and make the decimal part plus. Hence the index alone
must be considered as minus.
Negative numbers have no logarithms; for, in fact, as tot
have before observed, there are no such numbers.
(Art. C.) We now design to show a practical method of com-
puting logarithms. The methods hitherto touched upon were
only designed to be explanatory of the nature of logarithms ; but,
to calculate a table, by either of those methods, would exhaust the
244 ELEMENTS OF ALGEBRA.
patience of the most indefatigable. To arrive at easy practical
results requires the clearest theoretical knowledge, and we must
therefore frequently call the attention of students to first prin-
ciples.
The fundamental equation is a*=b, in which a is the con
slant base and x is the logarithm of the number b ; and b may
be of any magnitude or in any form, if it be essentially positive.
Now we may take a=l+c and 6=1 -j
Then the fundamental equation becomes (l-f-c) z =(l-| J (1)
Here ar=log. (l+i)=log.(^i)=log.(/i+l)-]og./i (2)
Raise both members of equation (1) to the nth power, then
we shall have
Expand both members into a series by the Binomial Theorem,
Then l+nxc+nx. ^^c^+nx . ^^ . ^=^c 3 4-
I
nx 1 nx 2 nx 3 4 1 . n 1 1 .
nX 2 8 4-'+' &C " = l+n -j +n -2-p* +
n 1 n 2 1 , n 1 n 2 n 3 1
n.
a 3
Drop unity, from both members, and divide by n, then we have
<, nx 1 , . nx 1 nx 2 . nx 1 nx 2 nx 3 .
c+- a -c'+- . ^-c'+- 2 - . -jr- f-S
\ In 1 1 , n 1 n 2 I n 1 n2 n 3 1
"*" ) ~~ p 2""'/" 1 2~'~3~y~ h ~2~'~3~~T~'^
+, &c.
This equation is true for all values of n. It is, therefore, true
LOGARITHMS. 945
when =0. Making this supposition and the above equation
reduces to
c 2 c 3 c 4 c 5 \ 1 1,1 1.1
- 2 - +rr +5 ^rrw^w^W'^' w
As c remains a constant quantity for all variations of x and /),
the series in the vinculum may be represented by the symbol AI
Then xM=- -- - 2 -f -L - L+J_ & c . (3)
jo 2/> 2 3/ 4jo 4 5/
Take the value of x from equation (2), and substitute it in
equation (3),
Then [\ og .(p+l)-l os .p-]M= l ~;L+;L-+, &c. (4)
Again, we may have the fundamental equation
( I - Y in which y is the logarithm of ( I -- Y the same as
of(l+l).
W as
Or 2/=log.(- ) = log.(j9 1) lg P" (5)
Operating on the equation (l-j-c) 1/ =l , the same as w*
did on equation (1), we shall find
Subtract equation (6) from equation (4), and we obtain
/I 1 1 1 \
[1og.(;;-fl) log.(p 1)]M=2( ^~5~i~^"K~iH ^ & Ct ) P;
Dividing by M, and considering that log/^- - )=log.(p-f-l).
log. ( p 1), the equation can take this form
246 ELEMENTS OF ALGEBRA.
As p may be any positive number, greater than 1, make
" -=1-J . Then j0=2*4-l, and equation (8) becomes
log \~~F L ) =
+
3(2H-1)* ' 5(2z-r-l) 5 '
By this last equation we perceive that the logarithm of (z-f-1)
will become known when the log. of z is known, and some
value assigned for the constant M. Baron Napier, the first dis-
coverer of logarithms, gave M the arbitrary value of unity, for
the sake of convenience.
Then, as in every system of logarithms, the logarithm of 1 is
0, make z=l, in equation (9), and we shall have
h^Lrh &cA=.O.C9314718.
)=
This is called the Napierian logarithm of 2, because its magni-
tude depends on Napier's base, or on the particular value of M
being unity.
Having now the Napierian logarithm of 2, equation (9) will
give us that of 3. Double the log. of 2 will give the logarithm
of 4. Then, with the log. of 4, equation (9) will give the loga-
rithm of 5, and the log. of 5 added to the log. of 2, will give the
logarithm of 10.
Thus the Napierian logarithm of 10 has been found to great
exactness, and is 2.302585093.
The Napierian logarithms are not convenient for arithmetical
computation, and Mr. Briggs converted them into the common
logarithms, of which the base is made equal to 10.
To convert logarithms from one system into another, we
may proceed as follows :
Let e represent the Napierian base, and a the base of the
common system, and N any number.
LOGARITHMS. 247
Also, let 3: represent the logarithm of JV, corresponding to
the base , and y the logarithm of N 9 corresponding to the
base c.
Then a x =N t
and & ~N.
Now, by inspecting these equations, it is apparent that if the
base a is greater than the base e, the log. x will be less than the
log. y.
These equations give a T e y .
Taking the logarithms of both members, observing that x and
y are logarithms already, we have
a? log. a=y \og.e.
This equation is true, whether we consider the logarithms
taken on the one base or on the other. Conceive them taken on
the common base, then
log. 0=1, and x=y\Q<r*e (10)
x
or log.e=-.
y
In this equation x and y must be logarithms of the same num-
ber, and therefore if we take a?=l, which is the logarithm of
10, in the common system, y must be 2.302585093, as pre-
viously determined.
Hence log. f = 2 L- =0.434394482. . . . (11)
This last decimal is called the modulus of the common sys-
tem ; for by equation (10) we perceive that it is the constant
multiplier to convert Napierian or hyperbolic logarithms into
common logarithms.
But equation (9) gives Napierian logarithms when M=\\
therefore the same equation will give the common logarithms by
causing M to disappear, and putting in this decimal as a factor.
Equation (9) becomes the following formula for computing
common logarithms :
ELEMENTS OF ALG'KBKA
log.(z-r-l) log.*=
0.868588961^
(-L_+ ^+^-1--,+, Ac.) (F)
To apply this formula, assume z= 10. Then
log. *=1, and 2z-f-l=21.
21
441
0.86858896
0.0413613791 = 0.041361379
93792 3 = 31264
2125 = 42
.041392685=sum of series.
log.lO= 1.0
log.(*+l) = 1.041 392685 =log.ll.
If we make 2=99, then (z-H) = 100, and Iog.(*-H)=s2,
and 224-1=199.
199
39601
0.86858896
436477-H =0.00436477
11-A_3= 4
0.00436481 =sum of series.
Hence 2.00000log.99-=0.00436481
By transposition log.99=l. 99563519
Subtract log.l l=1.04139269=log.l 1
log. 9=0.95424234
log.9=!og. 3=0.47712 117=log.3.
Thus we may compute logarithms with great accuracy ami
rapidity, using the formula for the prime numbers only.
By equation (11) we perceive that the logarithm of the Na-
pierian base is 0.434294482 ; and this logarithm corresponds to
the number 2.7182818, which must be the base itself.
We may also determine this base directly :
In the fundamental equation (1), the base is represented by
(1+c). In equation (A), c must be taken of such a value as
LOGARITHMS.
249
c 8 c 3 c 4
shall make the series c +- --f, &c., equal to 1. But to
determine what that value shall be, in the first place, put
c 2 , c 3 c 4
^-sf+s-T &c '
Now by reverting the series (Art. B.), we find that
But, by hypothesis, the series involving c equals unity that is,
y=l. Therefore
By taking 12 terms of this series, we find (l-j-c)=2.7182818,
the same as before.
TABLE I. LOGARITHMS FROM J TO 100.
N.
Log.
N.
Log.
N.
Log.
N.
Log.
1
000000
26
1 414973
51
1 707570
76
1 880814
2
301030
27
1 431364
62
1 716003
77
1 886491
3
477121
28
1 447158
53
1 724276
78
1 892095
4
602060
29
1 462398
54
1 732394
79
1 897627
5
698970
30
1 477121
55
1 740363
80
1 903090
6
778151
31
491362
56
1 748188
81
1 908485
7
845098
32
505150
57
1 755875
82
1 913814
8
903090
33
518514
58
1 763428
83
1 919078
9
954243
34
531479
59
1 770852
84
1 924279
10
1 000000
35
544068
60
1 778151
85
1 929419
11
1 041393
36
556303
61
1 785330
86
1 934498
12
1 079181
37
568202
62
1 792392
87
1 939519
13
113943
38
579784
63
1 799341
88
1 944483
14
146128
39
591065
64
1 806180
89
1 949390
15
176091
40
602060
65
1 812913
90
1 954243
16
204120
41
612784
66
1 819544
91
1 959041
17
230449
42
623249
67
1 826075
92
1 963788
18
255273
43
633468
68
1 832509
93
1 968483
19
278754
44
643453
69
1 838849
94
1 973128
20
301030
45
653213
70
1 845098
95
1 977724
21
322219
46
662578
71
1 851258
96
1 982271
22
342423
47
672098
72
1 857333
97
1 986772
23
361728
48
681241
73
1 863323
98
1 991226
24
380211
49
690196
74
1 869232
99
1 995635
25
397940
50
698970
75
1 875061
100
2 000000
250
ELEMENTS OF ALGEBRA.
TABLE II. LOGARITHMS OF LEADING NUMBERS WITHOUT INDICES.
1
2
3
4
6
6
7
8
9
N. 10Q. N. 101.
N. 1(W ! N. 10^. N. 104.
N. 105.
N. U)6.
N. 107.
N. 108.! N. 109.
000000
000434
000867
001301
001734
002166
002598
003029
003461
003891
0043-21
004750
005881
006609
006038
006466
006894
007321
007748
008174
003600
009026
009451
009876
010300
010724
011147
011570
011993
012415
012837
013259
013680
014100
014521
014940
015360
015779
016197
016616
017033
017451
017868
018284
018700
019116
019532
019947
020361
020776
021189
021603
022016
022428
022841
023252
023664
024075
024486
024896
025306
025715
026125
026653
026942
027350
027757
028164
028571
J028978
029384
029789
030195
030600
031004
031408
031812
032216
032619
1033021
033424
033826
034227
034628
035029
035430
035830
036230
036629
037028
0374-26
037825
038223
038620
039017
039414
039811
040207
040602
040998
(Art. 15la.) The preceding tables are sufficient to give us
the logarithms of any number whatever, provided we under-
stand the theory of logarithms, and are able to use a few prac-
tical artifices.
If to this knowledge, and these artifices, we add the formula,
we can with ease find the log. of any number, however large,
or however small, to any required degree of accuracy.*
We can, in fact, find the logarithm of any number, very
nearly, by Table I.
ILLUSTRATIONS.
1. We require the logarithm of 520; but 520 is not in the
table, 52 and 10 are.
Log. of 52 is
Log. of 10 is
Log. of 520, therefore, is
2 The logarithm of 5.2 is,
And of .52 is
1.716003.
1.000000.
2.716003.
0.716003,
1.716003, (fee., <fec.
3. What is the logarithm of 146 ?
146=73X2. Whence log. 73, 1.863323
log. 2, 301030
Log of 146 is 2.164353
*It is very bad policy to give a learner the use of voluminous tables ;
such tables are made to supply the place of theory, and of thought. Con-
tracted tables, like these, are the best educators.
LOGARITHMS. 251
4. What is the logarithm of 3244 ?
3240=60 54. 3250=65 50.
Log. 60, 1.778151 Log. 65, 1.812913
Log. 54, 1.732394 Log. 50, 1.698970
Log. 3240, is 3.510545. Log. 3250, is 3.511883.
(Art. 1515.) Here we have the logarithm of 3240, and of
3250 ; and the number 3244 is intermediate, therefore its loga-
rithm must be intermediate between 3.510545 and 3.511883.
The difference between the two extreme numbers is 10, and
the difference of their logarithms is .001338.
Therefore, by proportion, 10 : .001338 : : 4.
That is -^ths of the difference of the known logarithms is
the correction to add to the less logarithm to obtain the loga-
rithm sought.
Whence to log. 3240 3.510545
Add .0001338X4 635
Log. of 3244 is 3.511080
This example shows that Table I., and the true theory of
logarithms, are sufficient to find the logarithm of any number
whatever.
ANOTHER METHOD.
(Art. 151c.) When the number is large, as in the present
example, the first term of the series will be the increase of the
logarithm corresponding to one ; four times this would be the
increase corresponding to four, and so on.
The logarithm of 3240 is 3.510545, as before found, and
here z=3240, and 22+1=6481,
.86858896
Whence, T- - =0.000134
Whence to losr. 3.540545
Add 0.000134X4 536
Log. 3244 3.511081
Ar.d the two methods agree the first may be clearest to a
learner, but the second is more brief and scientific.
252 ELEMENTS OF ALGEBRA.
(Art. 151d.) Table II. requires some explanation. The
numbers are at the top and side, and only the decimal part of
the logarithm is contained in the table. The operator, in each
particular case, will supply the index, according to the theory
of logarithms.
EXAMPLES.
1. Find the logarithm of 1013. Ans. 3.005609.
We find 101 at the top, and 3 at the side. Under the former,
and opposite the latter, we find the decimal .005609, and this
is the decimal part of the logarithms of all numbers expressed
by the figures 1013, as they here stand.
Thus, it is the decimal logarithm of 1013, 101.3, 10.13, 1.013,
or of the decimals .1013, .01013, &c.
It is also the decimal logarithm of 10130, 101300, <fcc., but
this was sufficiently illustrated in (Art. 151).
2. Find the logarithm of 1 .075. Ans. 0.031408.
3. Find the logarithm of 10940. Ans. 4.039017.
Here, under 109, at the top, and opposite 4, at the side, we
find the decimal, and we take 4 for the index, because there
are five places of figures, in whole numbers.
4. Find the logarithm of 101 .5. Ans. 2.031408.
5. Find the logarithm o/. 001043. Ans. 3.018284.
(Art. 1510.) Table II. will be very useful in solving prob-
lems in compound interest, and annuities ; but its greatest
utility is in solving the following problems :
6. Find the logarithm of 5&1521. Ans. 5.753977.
Divide by the two superior figures (in this example it is 56),
and the quotient is 10134^|.
The divisor, consisting of two figures, its logarithm can
always be found in Table I.
And the four superior figures of the quotient (any quotient)
will be found in Table II.
The quotient in this example may be written thus,
10130+41!-.
The logarithm of 10130 (the decimal part of it), can be
taken directly out of Table II. It is .005609 ; and the next
LOGARITHMS. 253
greater (the number below) is .006038 ; the difference between
the two is .000429, and the correction for 4\% must be 4
times one tenth of .000429, which is .000180.
Whence log. 10130 - 4.005609 Table II.
Correction for 4}|, - - .000180
Log. of quotient, 10134^, 4.005789
Log. of 56, - - 1.748188 Table I.
Log. of 567521, 5.753977.
Thus we may find the logarithm of any number by the use of
these two tables. We give one more example.
7. Find the logarithm of 365. 25638 the days and parts of
a day in a siderial year.
(N. B. As three figures are whole numbers, the index must
be two. Divide by 36, and taking five figures in the quotient
as whole numbers, we shall have 10146.01 for a quotient.)
This quotient may be written in the following form :
10140+6.01.
The Log. of 36, Table I., - 0.556303
Log. of 10140, Table II., - 0.006038
6.01
Correction for -y^- of diff.* .000257
Sum +2.=log. of 365.25638, 2.562598
(Art. 151/.) To find the number corresponding to a given
logarithm we take the converse operation.
For example, we take the logarithm just found, and demand
its corresponding number.
Log. (omit index), 0.562598
Next less in Table I. 36, - .556303
.006295
Next less in Table II. 10140, .006038
.000257
*The difference in the table is the difference between .006038 and the
next greater .006466, which is .000428, one-tenth of this .000042.8, multi-
ply by (6.01) and the product is .000257.
254 ELEMENTS OF ALGEBRA.
Divide this difference by the tabular difference, which is
.000428, and we obtain 6, and a very small remainder.
Whence, to 10140, add 6, and we have 10146, which being
multiplied by 36, produces 365.256, for the number sought.
We take three of the figures for whole numbers, because
the index of the given logarithm is 2. Had it been 3, we
should have taken four figures for whole numbers.
1. What number corresponds to the logarithm 0.317879 ?
Ans. 2.079118.
From the given log. 0.317879
Sub. next less. Table I., 2, 0.301030
0.016849
Next less. Table II., 1039, 0.016616
0.000233
Divide 233 by 417 (tab. diff.), and we obtain 559, nearly,
which we annex to 1039, making 1039559, which, multiplied
by 2, produces 2.079118, the answer.
2. What number corresponds to the logarithm 4.031718?
Ans. 10757.673.
As the first figure in the decimal is 0, we had better use
Table II. at once.
The number next less in the table is the log. of 1075. As
the index is 4, we must have another figure for whole numbers.
Hence, from the given log., 4.031718
Subtract log. of 10750 (Table II.), which is 4.031408
.000310
This difference is less than any other logarithm in Table II.
Therefore, we can obtain no other factor, but we can correct
10750.
The next greater logarithm in the table is that of 10760,
which is .031812 ; the tablular diff. is .000404.
Therefore, .000404 : 000310 : : 10 : correction.
Or, 404 : 310 : : 10 : correction = 7.673.
LOGARITHMS. 255
3. The logarithm of the British gallon is 2.442909; a cubic
inch being the unit. How many cubic inches does it contain ?
Ans. 277.27+
(Art. 151^.) Logarithms are exponents; and as powers and
roots may be expressed by exponents, therefore, logarithms
may be used for finding powers and roots.
To extract the square root of any number, a, we divide its
exponent (one understood), by 2, and the result is a? . To
extract the third root, fourth root, &c., we divide the exponent
by 3, 4, &c., as the case may be.
The exponent of a number is the logarithm of that number.
Hence, to extract roots by logarithms, we have the following
rule :
RULE. Take the logarithm of the number and divide it by 2 for
the square root, by 3 for the third root, by 4 for the fourth root,
and so on.
The quotient will be the logarithm of the root sought, and the
number corresponding to this logarithm will be the root itself. *
(Art. 15U.) If x=a 3 Then log. x2> log. a.
By hypothesis, x=a.a.a.
By taking the logarithm of each member, we have
log. x= log. a + log. a -j- log. a = 3 log. a. ( 1 )
Therefore, in general terms,
If #=a n log. x=n log. ff.(2)
Now, let n be a fraction, then xcfo.
Cube both members ; then x 3 =a. That is, x.x.x=a.
By the property of logarithms we have
log. x + log. x + log. x = log. a.
Or, 3 log. x = log. a.
Or, log. x = i log. a. (3)
All these equations conform to the rule just given.
* Examples can be taken out of any Arithmetic.
256 ELEMENTS OF ALGEBRA.
USE AND APPLICATION OF LOGARITHMS.
(Art. 152.) The sciences of trigonometry, mensuration, and
astronomy alone, can develop the entire practical utility of lo-
garithms. The science of algebra can only point out their
nature, and the first principles on which they are founded. To
explain their utility, we must suppose a table of logarithms formed,
corresponding to all possible numbers, and by them we may re-
solve such equations as the following :
1. Given 2*= 10 to find the value of x.
If the two members of the equation are equal, the logarithms
of the two members will be equal, therefore take the logarithm
of each member ; but as a? is a logarithm already, we shall have
v log. 2=log. 10.
2. Given (729)* =3, to find the value of x.
Raise both members to the x power, and 3 X =729=9 3 ,
Or 3 Z =3 6 . Hence, #=6.
3. Given r +&*=c, and a x b v =d, to find the values of x
and y.
By addition, 2a x =c-\-d. Put c-\-d=2m ;
Then a z =m. Take the logarithm of each mem-
loff. m
her, and x log. =log. m, or x=^ - .
By subtracting the second equation from the first and
making c rf=2n, we shall find 2/ == f"^"~l*
3
4. Given (2 1 6) x =18, to find the value of x.
9 log. 6
Ans. x=- r -^~-
log. 12
9. Given ^=e, to find the value of x.
losf. m log. a /fix
Jtns. x=2 p.- m being equal to (rfe-f c]
LOGARITHMS. 257
6. Given 4 3 =16, to find the value of x. rftis. x=Q.
7 Given 6*= - to find the value of x.
18 log.24-{-log.l7--3 log.71
Ans. x ^ .
3 log. 6
8. Required the result of 23.46 multiplied by 7.218, and the
product divided by 11.17.
OPERATION.
23.46 log. 1.37033
7.218 log. 0.85842
Sum 2.22875
11.17 Subtr log. 1.04805
Result,. . . . 15.16 .log. 1.18070.
N". B. The log. of a vulgar fraction is found by subtracting
the log. of its denominator from the log. of its numerator.
For instance, T T is simply 9 divided by 11, and division in
logarithms is performed by subtraction.
Thus, from the log. of 9, .... 0.954343
Subtract log. of 11, .. . 1.041393
Log. of T T therefore is .... 1 .912850
The decimal part of a logarithm is never minus, but the
index is always minus when the number is less than unity.
Hence, the logarithm of a very small fraction has a large nega-
tive index, and the logarithm of is minus infinity. The loga-
rithm of 10 is 1, of 1 is 0, of .1 is I, of .01 it is 2, of .001
it is 3, and so on. As the decimal number goes down to
zero, the index of the logarithm goes up to minus infinity.
The previous example was nothing more than obtaining the
logarithm of a common fraction, where the numerator con-
sisted of two factors. That example might have been stated
thus :
Find the log. of the fraction, ( 23 - 46 ) ( 7 - 2 * 8 )
11.17
22 Ans. 1.18070.
258 ELEMENTS OF ALGEBRA.
CHAPTER V.
COMPOUND INTEREST.
(Art. 153.) Logarithms are of great utility in resolving some
questions in relation to compound interest and annuities ; but for
a lull understanding of the subject, the pupil must pass through
the following investigation :
Let p represent any principal, and r the interest of a unit of
this principal for one year. Then 1 -\-r would be the amouni
of $1, or J61. Put^=l-fr.
Now as two dollars will amount to twice as much as one dol
lar, three dollars to three times as much as one dollar, &c.
Therefore, 1 : Ji : : A : JP=the amount in 2 years,
And 1 : Ji : : JP : .# 3 =the amount in 3 years,
<fec. &c.
Therefore, /?" is the amount of one dollar or one unit of the
principal in n years, and p times this sum will be the amount for
p dollars. Let a represent this amount ; then we have this gen-
eral equation,
pA n =a.
In questions where n, the number of years, is an unknown
term, or very large, the aid of logarithms is very essential to a
quick and easy solution.
For example, what time is required for any sum of money
to double itself, at three per cent, compound interest?
Here a=2/?,and .#=(1.03), and the general equation becomes
/>(1.03) n =2/>
Or (1.03)"=2. Taking the logarithms
nk*. (1.03)=log. 2, or n=
years nearly.
2. A bottle of wine that originally cost 20 cents was put away
for two hundred years: what would it be worth at the end of
that time, allowing 5 percent, compound interest?
ANNUITIES. 259
This question makes the general equation staml thus :
(20 cts. being l of a dollar) i(1.05) 200 = a
Therefore (1.05) 200 =5
Taking the logarithms 200 log. (1.05)=log.5-{-log. a
Hence log. 0=200 log. (1.05) log. 5. JJns. $3458.10
3. A capital of $5000 stands at 4 per cent, compound interest;
what will it amount to in 40 years ? rfns. $24005.10.
4. In what time will $5 amount to $9, at 5 per cent, com-
pound interest ? Jlns. 12.04 years.
5. A capital of $1000 in 6 years, at compound interest,
amounted to $1800; what was the rate per cent?
Ans. log. (l+r)= log * L8 or 10 T \ nearly
6. A certain sum of money at compound interest, at 4 per
cent, for four years, amounted to $350. 95| ; what was the sum?
fins. $300.
7. How long must $3600 remain, at 5 per cent, compound in-
terest to amount to as much as $5000, at 4 per cent, for 12 years 1
Am. 16 years, nearly.
ANNUITIES.
(Art. 154.) An annuity is a sum of money payable peri
odically, for some specified time, or during the life of the re
ceiver. If the payments are not made, the annuity is said to h
in arrear, and the receiver is entitled to interest on the several
payments in arr3ar.
The worth of an annuity in arrear, is the sum of the several
payments, together with compound interest on every payment
after it became due.
On this definition we proceed to investigate a formula to be
applied to calculations respecting annuities.
Let p represent the annual principal or annuity to be
2GO ELEMENTS OF ALGEBRA.
paid, and \-\-r=*fl, the amount of annuity of principal for one
year, at the given rate r.
Let n represent the number of years, and put A' to represent
the entire amount of the annuity in arrear.
It is evident, that on the last payment due, no interest could
accrue, and therefore the sum will be p. The preceding pay-
ment will have one year's interest ; it will therefore be pJl ; the
payment preceding that will have two years' compound interest ;
and, of course, will be represented by pJP. (Art. 153.) Hence
the whole amount of A' will be
, &c., to
This is a geometrical series, and its sum (Art. 120.) is
This general equation contains four quantities, .#', p, r, and n ,
any three of them being given in any question, the others can be
found, except r.
EXAMPLES.
1. An annuity of $50 has remained unpaid for 6 years, at
compound interest on the sums due, at 6 per cent., what sum is
now due ?
By the general equation,
50[(t.06)'-l]
~:o6~
Faking the log. of both members, we have
log. A'= log. 50-flog. [(1.06) 6 1] log. .06.
The value of (1.Q6) 6 , as found by logarithms, is 1.41852, from
which subtract 1, as indicated, and take the log. of the decimal
number .41852, we then have
log. ^'=1. 69897 +( 1.62172) ( 2.778151)=2.54218,
From which we find, .#'=$348.56 Jim.
2. In what time will an annuity of $20 amount to $1000, at
4 per cent., compound interest ?
ANNUITIES. 261
The equation applied, we have
Dividing by 20, and multiplying by .04, we have
2=(1.04) 1 or (1.04)=3.
log. 3 .477121
3. What will an annuity of $50 amount to, if suffered to
remain unpaid for twenty years, at 3| per cent, compound in-
terest? Ans. $1413.98.
4. What is the present value of an annuity or rental of $50
a year, to continue 20 years, discounting at the rate of 3 per
cent., compound interest ?
N. B. By question 3d, we find that if the annuity be not paid
until the end of 20 years, the amount then due would be $1413.98.
If paid now, such a sum must be paid as, put out at compound
interest for the given rate and time, will amount to $1413.98.
Now if we had the amount of $1 at compound interest for 20
years, at 3j per cent., that sum would be to $1 as $1413.98 is
to the required sum, $710.62.
(Art. 1 55.) To be more general, let us represent the present
worth of an annuity by P. By (Art. 153.) the amount of one
dollar for any given rate and time, is Ji n ; A being \-\-r and n
the number of years. By (Art. 154.) the value of any annuity
p remaining unpaid for any given time, n years, at any rate of
pA n p
compound interest r, is -- or Jl'.
Now by the preceding explanation we may have this propor-
tion :
Jl* :l::J':P, or P=~- ...... (1)
Hence, to find the present worth of an annuity, we have this
RULE. Divide the amount of the annuity supposed unpaid
for the given number of years, by the amount of one dollar for
the same number of years.
2 ELEMKXTS OF ALCEBRA.
If in equation (1) we put the value of .#, we shall have
^nE L m Divide both members by .#*, and we have
(2)
r
This last equation will apply to the following problems :
5. The annual rent of a freehold estate is p pounds or dollars,
to continue forever. What is the present value of the estate,
money being worth 5 per cent., compound interest ?
Here, as n is infinite, the term, becomes 0, and equation
(2) becomes P=LJL^=2Qp that is, the present value of the
estate is worth 20 years' rent.
6. The rent of an estate is $3000 a year ; what sum could
purchase such an estate, money being worth 3 per cent., com-
pound interest ? Jim. $100000.
7. What is the present value of an annuity of $350, assigned
for 8 years, at 4 per cent. ? Jlns. $2356.46.
8. A debt due at this time, amounting to $1200, is to be dis-
charged in seven annual and equal payments ; what is the
amount of these payments, if interest be computed at 4 per cent. ?
Ans. $200, nearly.
9. The rent of a farm is $250 per year, with a perpetual lease.
How much ready money will purchase said farm, money being
worth 7 per cent, per annum 1 rfns. $3571 7
10. An annuity of $50 was suffered to remain unpaid for
20 years, and then amounted to $1413.98; what was the rate
per cent., at compound interest ?
N. B. This question is the converse of problem 3, and, ol
course, the answer must be 3i per cent. But the general equa-
tion gives us
GENERAL THEORY OF EQUATIONS. % 63
Or 28.2796 JH^ 2 ;
an equation from which it is practically impossible to obtain r,
except by successive approximations.
SECTION VII.
CHAPTER I.
GENERAL THEORY OF EQUATIONS.
(Art. 156.) In (Art. 101.) we have shown that a quadratic
equation, or an equation of the second degree, may be conceived
to have arisen from the product of two equations of the first de-
gree. Thus, if x=a, in one equation, and x=b in another
equation, we then have
x a=0,
and x &=0;
By multiplication, x 2 (a-{-b}x-{-ab=Q.
This product presents a quadratic equation, and its two roots
are a and b.
If one of the roots be negative, as x= , and x=b, the
resulting quadratic is
x z -\-(ab)xab=0.
If both roots be negative, then we shall have
Now let the pupil observe that the exponent of the highest
power of the unknown quantity is 2 ; and there are two roofs.
The, coefficient of the first power of the unknown quantity is
the algebraic sum of the two roots, with their signs changed ;
and the absolute, term, independent of the unknown quantity,
is the product of the roots (the sign conforming to the rules of
multiplication}.
264 ELEMENTS OF ALGEBRA.
When the coefficients and absolute term of a quadratic are not
largo, and not fractional, we may determine its roots by inspec-
tion, by a careful application of these principles
EXAMPLES.
Given #* 20#-f96=0, to find x.
The roots must be 12 and 8, for no other numbers will make
20, signs changed, and product 96.
Given y z 6y 55=0 to find y. Roots 11 and 5.
Given a? 6# 40=0 to find x. Roots 10 and 4.
Given s?+x 91=0 to find x. Roots 7 and 13.
Given y* 5y 6=0 to find y. Roots 6 and 1.
Given y z -\-\'2y 589=0 to find y.
Here it is not to be supposed that we can decide the values of
the roots by inspection; the absolute term is too large; but,
nevertheless, the equation has two roots.
Let the roots be represented by P and Q.
From the preceding investigation
P+$=- 12 (1)
And PQ= 589 (2)
By squaring eq. (1) P 2 -}-2PQ-{-Q 2 = 144
4 times eq. . . . (2) 4PQ =2356
By subtraction, p22PQ+Q 2 = 2500
By evolution, P #=50
But P-f-= .12
Hence P=19 or 31, and Q=3l or +19, the true
roots of the primitive equation ; and thus we have anothci
method of resolving quadratics.
(Art. 157.) In the same manner we can show that the product
of three simple equations produce a cubic equation, or an equa-
tion of the third degree. Conversely, then, an equation of the
third degree has three roots.
The three simple equations, #=, x=b, x=c,* may be put
* Of course, x cannot equal different quantities at one and the same time
and these equations must not be thus understood.
GENERAL THEORY OF EQUATIONS. 2 65
in the form of x a=Q, x &=0, and x c=0, and the pro-
duct of these three give
(x a}(x b] (x c)=0 ;
and by actual multiplication, we have
x 3 a
If one of the roots be negative, as x= c, or a?-j-c=0, the
product or resulting cubic will be
If two of them be negative, as x= b and #== e, the
resulting cubic will be
a^+C^+c a)x 2 -f(6c ab ac}x abc=0.
If all the roots be negative, the resulting cubic will be
Every cubic equation may be reduced to this form, and con-
ceived to be formed by such a combination of the unknown term
and its roots.
By inspecting the above equations, we may observe
1st. The first term is the third power of the unknown
quantity.
2d. The second term is the second power of the unknown
quantity, with a coefficient equal to the algebraic sum of the
roots, with the contrary sign.
3d. The third term is the first power of the unknown
quantity, with a coefficient equal to the sum of all the products
which can be made, by taking the roots two by two.
4th. The fourth term is the continued product of all the roots,
with the contrary sign.
It is easy, then, to form a cubic equation which shall have
any three given numbers for its roots.
Assuming x for the unknown quantity, find an equation which
shall have 1, 2 and 3 for its roots.
tins. a*(l+2+3)3M-(2+ 3+6)* 6=0 ;
Or
Find the equation which shall have 2, 3, and 4 for its roots.
23 Ans. x* 9? 14a?+24=0.
266 ELEMENTS OF ALGEBKA.
Find the equation which shall have 3, 4 y and +7 for its
roots. Am. x 3 0a; 2 37z 84=0,
Or x* 37# 84=0.
These four general cases of cubic equations may all be repre-
sented by the general form.
Thus: x s -\-px z +qx+r=0, ......... (1)
(Art. 158.) When the algebraic sum of three roots is equal U
zero, equation (1) takes the form of
x*+qx+r=0 ............. (2)
Equation (1) is a regular cubic, and is not susceptible of a
direct solution, by Cardan's rule, until it is transformed into
another wanting the second term, thus making it take the form
of equation (2). To make this transformation, conceive one of
the roots, or x, in equation (1), represented by u-{-v*
Then x*=u*+3u z v+3uv z -f-u 3
px*= pu 2 -\-2puv-\-pv 2
qx = qu -\-qv
r = r
By addition, and uniting the second member according to tho
powers of u, we shall have
for the transformed equation. But the object was to make such
a transformation that the resulting equation should be deprived
of its second power ; and to effect this, it is obvious that we
must make the coefficient of u 2 equal zero, or 3v-}-p=Q.
Therefore, v= ip.
Hence, we perceive that if x, in the general equation (1), be
put equal to u ~, there will result an equation in the form of
o
u*-\-qu-\-r=0, or the form of equation (2).
As x=u ^, and if , 6, and c represent the roots of equa-
o
tion (1), or the values of x, the roots of (2), or values of u will be
and
GENERAL THEORY OF EQUATIONS. 267
EXAMPLES.
1. Transform the equation ar 3 9x?-{-26x 30=0, into another
wanting the second term.
By the preceding investigation, we must assume
xu-}-3. Here p= 9 ; therefore, sp=3.
= 26w-f78
30 = 30
Sum, =w 3 u 6=0, the equation required.
2. Transform the equation a? 3 6r J +10a> 8=0, into another
not containing the square of the unknown quantity.
Put x=u+2. Result, u 3 2w 4=0.
3. Transform x s 3x z -}-6x 12=0, into another equation,
wanting the second power of the unknown quantity.
Put a?=w+l. Result, w 3 +3w 8=0.
(Art. 159. We have shown, in the last article, that any icgular
cubic equation containing all the powers of the unknown quan-
tity can be transformed into another equation deficient of the
second power ; and hence all cubic equations can be reduced to
the form of
We represent the coefficient of x by 3p, and the absolute term
by 2</, in place of single letters, to avoid fractions, in the course
of the following investigation.
Now, if we can find a direct solution to this general equation,
it will be a solution of cubic equations generally.
The" value of x must be some quantity ; and let that quantity,
whatever it is, be represented by two parts, v-\-y, or let
x=v-\-y. Then the equation becomes
By expanding and reducing, we have
Now as we have made an arbitrary division of x into two parts,
v and y we can so divide it, that
268 ELEMENTS OF ALGEBRA.
This hypothesis gives
v 3 +y*= 2q, (.#)
And V y=p, (J9)
Here we have two equations, (A) and (/?), containing two un-
known quantities, similarly involved, which admit of a solution
by quadratics. (Art. 108.) Hence we obtain v and y, and their
algebraic sum is x.
From equation (B],
This substituted in equation ($), gives
Or, V Q 2yu 3 =p 3 , a quadratic.
Hence v-
And y.
Or, as y-
Therefore x=(q+ jf+fi}* +(q
Or ,
These formulas are familiarly known, among mathematicians, aa
Cardan's rule.
(Art. 160.) When p is negative, in the general equation, and
its cube greater than <? 2 , the expression Jq* p s becomes imagi-'
nary ; but we must not conclude that the value of # is therefore
imaginary ; for, admitting the expression ,J(f -jo 3 imaginary, it
can be represented by aj 1, and the value of a?, in equation
(C), will be
(JENERAL THEORY OF EQUATIONS. 269
Now by actually expanding the roots of these binomials by the
binomial theorem, and adding their results together, the terms
containing J 1 will destroy each other, and their sum will be
a real quantity ; and, of course, the value of x will become real.
If in any particular case it becomes necessary to make the series
converge, change the terms of the binomial, and make J 1
stand first, and 1 second.
EXAMPLES.
Given x 9 6a?=5.6, to find the value of x.
Here, 3p= 6, and 2? 5.6, or p= 2, and #=2.8.
Then
*=(2.8+/7.84 8) +(2.877.848) , by equation (C)
Or tf=(2.8+.4,/ 1) +(2.8 .47 1)
Or "
Expand the binomials by the binomial theorem, (Art. 135.),
i d for the sake of brevity, represent 4 J 1 by b;
Then &*= , and fl='x-
270 ELEMENTS OF ALGEBRA
=24-0.0045350.000034=2.0045.
Therefore,
3 -=2.0045 or a?=(2.0045)(2.8)=2.8256, nearly
(Art. 161.) Every cubic equation of the form of
has three roots, and their algebraic sum is 0, because the equa-
tion is wanting its second term. (Art. 157.)
If the roots be represented by , b, and c, we shall have
If any two of these roots are equal, as b=c, then a= 2b (1),
and ab 2 =q (2). Putting the value of a taken from equation
(1), into equation (2), and we have 26 3 =g r .
Hence, in case of there being two equal roots, such roots must
each equal the. cube root of one half the quantity represented
byq.
EXAMPLES.
The equation re 3 48^=128 has two equal roots; what are
.he roots ?
Here, 26 3 ,=128, or b s = 64; therefore, b= 4.
Two of the roots are each equal to 4, and as the sum of the
three roots must be 0, therefore 4, 4, -\-8, must be the
three roots.
If the equation a: 3 27#=54 have two equal roots, what are
the roots ? Am. 3, 3, and +6.
Either of these roots can be taken to verify the equation ; and
if they do not verify it, the equation has not equal roots.
(Art. 162.) If a cubic equation in the form of
GENERAL THEORY OF EQUATIONS. 271
have two equal roots, each one of the equal roots will bo
equal to
The other root will be twice this quantity subtracted from p,
because the sum of the three roots equal p. (Art. 157.)
This expression is obtained from the consideration that the
tfirce roots represented by , 6, and c, must form the folloAving
equations: (Art. 157.)
a+b+c=p ......... (1)
ao-{-ac-\-bc=q ......... (2)
abcr ......... (3)
On the assumption that two of these roots are equal, tha* is,
a=b, equations (1) and (2) become
2a-\-c=p ........ (4)
And a 2 -{-2ac=q ........ (5)
Multiply equation (4) by 2, and we have
4a 2 -\-2ac=2ap ....... (7)
Subtract (5) 2 +2oc= q ....... (8)
And we have 3a 2 =2ap q.
This equation is a quadratic, in relation to the root or, and a
solution gives a=i(p.Jp 2 3q).
(Art. 163.) A cubic equation in the form of x s px=q can
be resolved as a quadratic, in all cases in which q can be resolved
into two factors, m and rc, of such a magnitude that m 2 -\-p=n.
For the values of p and </, in the general equation, put the
assumed values, mn=q, and p=n m z .
Then we have x 3 -{-nx m 2 x=mn.
Transpose m z x, and then multiply both members by x, and
= m z x z -{-mnx.
Add -- to both members, and extract square root ;
Then x*-\~-mx-\--. Drop -, and divide by a?, and x-=m.
272 ELEMENTS OF ALGEBRA.
Therefore, if such factors of q can be found, the equation is
already resolved, as x will be equal to the factor m.
EXAMPLES.
1. Given x?+6x=S&, to find the values of x
Here, wm=88=4X22, 4 2 -j-6=22. Hence, x=4
2. Given x 3 -\-3x=l4, to find one value of x.
2X7=14, 2 2 +3=7. Hence, x 2.
U. Given # 3 -|-6#=45, to find one value of x. *fl.ns. x=3.
4. Given x 3 13;r= 12, to find x. rfns. x=3.
5. Given 7/ 3 +48?/=104, to find y. Jlns. y=2.
In the above examples we have given only one answer, or one
root ; but we have more than once observed, that every equation
of the third degree must have three roots. Take, for example,
the 4th equation. We have found, as above, one of its roots to
be 3. Now we may conceive the equation to have been the
product of three factors, one of which was (x 3) ; therefore the
equation must be divisible by x 3 without a remainder, (other-
wise 3 cannot be a root) ; and if we divide the equation by x 3,
the quotient must be the product of the other two factors.
Thus, a: 3)* 3 I3x+12(x z -\-3x 4
2 I3x
4H-12
By putting x*-\-3x 4=0, and resolving the equation, we find
a?=l, or 4, and the three roots are 1, 3, 4. Their sum is
0, as it should be, as the equation is deficient of its second term.
In the general equation
If p and q are each equal to 0, at the same time, the equation
becomes a i3 -j-r=0, a binomial equation.
GENERAL THEORY OF EQUATIONS. 373
Every binomial equation has as many roots as there are
units in the exponent of the unknown quantity.
Thus a; 3 +8=0, and a 3 8=0, or o; 3 +l=0, and a; 3 1=0,
&c., are equations which apparently have but one root, but a
full solution will develop three.
Take, for example, a? =8
By evolution, x =2
4xS
4xS
Now by putting a; 2 +2^-}-4=0, and resolving the equation,
we find x= 1 + V ** an( ^ x ~ * >/"~~^ anc ^ tne tnree
roots of the equation X s 8 = 0, are 2, 1 + ^/ 3, and
1 J 3, two of them imaginary, but either one, cubed, will
give 8.
The three roots of the equation ic 3 +l=0, are
1 , 1 . 1 1
CHAPTER II.
GENERAL THEORY OF EQUATIONS CONTINUED.
(Art. 164.) In the last Chapter we confined our investigations
to equations of the second and third degrees ; and if they are
well understood by the pupil, there will be little difficulty, in
future, as many of the general properties belong to equations of
every degree.
All the higher equations may be conceived to have been formed
by the multiplication of the unknown quantity joined to each
of the roots of the equation with a contrary sign, as shown in
(Art. 157.).
274 ELEMENTS OF ALGEBRA.
Let a, b, c, d, e, &c., be roots of an equation, and a;
its unknown quantity, then the equation may be formed by the
product of (x a}(x b)(x c), &c., which product we may
represent by
Now it being admitted that equations ran be thus formed by
the multiplication of the unknown quantity joined to its roots,
conversely, when any of its roots can be found, such root, with
its contrary sign joined to the unknown term, will form a com-
plete divisor for the equation ; and by the division the equation
will be reduced one degree, and conversely.
If any quantity, connected to the unknown quantity by the
sign plus or minus, divide an equation without a remainder,
such a quantity may be regarded as one of the roots of the
equation.
The product of all the roots form the absolute term U.
(Art. 165.) Every equation having unity for the coefficient
of the first term, and all the other coefficients, whole numbers,
can have only whole numbers for its commensurable* roots.
This being one of the most important principles in the theory
of equations, its enunciation should be most clearly and distinctly
understood. Such equations may have other roots than whole
numbers ; but its roots cannot be among the definite and irre-
ducible fractions, such as f , J, U , &e. Its other roots must be
among the incommensurable quantities, such as J2, (3)% &c.,
i. e., surds, indeterminate decimals, or imaginary quantities.
To prove the proposition, let us suppose j- a commensurable
but irreducible fraction, to be a root of the equation
A, B, fec., being whole numbers.
Substituting this supposed value of x, we have
* Commensurable numbers are all those that measure or can be measured by
unity ; hence, all whole numbers and definite fractions are commensurable.
Surds, and imaginary quantities, are incommensurable.
GENERAL THEORY OF EQUATIONS. 275
Transpose all the terms but the first, and multiply by b m ~ } , and
we have
a m
y-
Now, as a and b are prime to each other, b cannot divide a,
or any number of times that a may be taken as a factor, for j
being irreducible, - X a is also irreducible, as the multiplier a
a?
will not be measured by the divisor b ; therefore y cannot be
expressed in whole numbers. Continuing the same mode of
a m
reasoning, y cannot express whole numbers, but every term in
the other member of the equation expresses whole numbers.
Hence, this supposition that the irreducible fraction = is a root
of the equation, leads to this absurdity, that a series of whole
numbers is equal to another quantity that must contain a fraction.
Therefore, we conclude that any equation corresponding to
these conditions cannot have a definite commensurable fraction
among its roots.
(Art. 166.) Any equation having fractional coefficients, can be
transformed to another in which the coefficients are all whole
numbers, and that of the first term unity.
For example, take the equation
m n p
Assume ar=--, and put this value of x in the equation,
mnp
And _
_..
mn*p p
Multiply every term by m*n*p*, and we have
276 ELEMENTS OF ALGEBRA.
When m, n, and p have common factors, we may put x equal
to y divided by the least common multiple of these quantities,
as in the following examples:
Transform the equation y*-\ -- .+ --- h-=0, into another
pm m p
which shall have no fractional coefficients, and that of the first
term unity.
To effect this, it is sufficient to put x=-^. With this value
pm
of x the equation becomes
__. = .
pV p*m 3 pm 2 p
Multiplying every term by j9 3 m 3 , we obtain
for the transformed equation required.
Transform the equation x*-\ - 1 'T"i"nA~^"Tn == ^9 into an-
other, having no fractional coefficients.
Result, y+20?/ 3 +18.24y4-7(24) 2 y-}-2(24) 3 =0.
(Art. 167.) Now as every commensurable root consists of
whole numbers, and as the coefficients are all whole numbers,
each term of itself consists of whole numbers, and the commen-
surable roots are all found among the whole number divisors of
the last term ; and if these divisors are few and obvious, those
answering to the roots of the equation may be found by trial. If
the factors are numerous, we must have some systematic method
of examining them, such as is pointed out by the following rea-
soning :
Take the equation * 4 +^ 3 +jar z +Ca?-j-Z)==().
Let a represent one of its commensurable roots. Transpose
all the terms but the last, and divide every term by a.
But, since a is a root of the equation, it divides D without a
GENERAL THEORY OF EQUATIONS. 277
remainder, the left hand member of this last equation is therefore
a whole number, to which transpose C, also a whole number,
and represent \-C by N.
Then N=a?rfa?a.
Divide each term by a, and transpose J3, and we have
-+= a 2 da.
a
The right hand member of this equation is an entire quantity,
(not fractional), therefore the other member is also an entire
quantity ; let it be represented by N 1 , and the equation again
divided by a.
Then = a A.
a
Transpose A ; reasoning the same as before, we can repre-
sent the first member by JV", and we then have
N"
Divide by a, and = 1. This must be the final result, in
case a is a root.
From these operations we draw the following rule for deciding
what divisors of the last term are roots of an equation.
RULE. Divide the last term by the several divisors, (each
designated by ,) and add to the quotient the coefficient of the
term involving x.
Divide this sum by the divisors (a), and add to the quotient
the coefficient of the term involving #*.
Divide this sum by the divisors (a), and add to the quotient
the coefficient of the term involving x?.
And thus continue until the first coefficient, .#, is transposed,
and the sum divided by a ; the last quotient will be minus one,
if a is in fact a root
278
ELEMENTS OF ALGEBRA.
EXAMPLES.
1. Required the commensurable roots (if any) of the equation
=0.
N
2 1
(M
7
1 1
CD
<N
_u "T3
O
1
7 i
1
7
CO
i
2 1
0*
o>
7
*
7
7 +7
i
0)
1
CO
^5 ift
i-H
,| , 7 ,_
1
7
c*
14
7 '
CO CO '-''-*
1 + 1 a
1
"c
^4
00 QO
05 05 .22
ij
7
tl I
bo
c*
CD
<N f-t
^UT 7 |
.s
CO
7
?
3
!i
1
o
a 2
*
1
05
7
1
|l
o
O^ "~^
CO
T
777
1 1
1 i
e* <M
r-l CD
^ ^
C |
1
1 7
i e
CD
" oT
a ^
il 1 Is
it
. 1 1 1 1 t
1 || 11 11 & |
_>: rJ r^ ^
o -5
O M
sl
CO
EQUAL ROOTS. 279
2. Required the commensurable roots of the equation
ar 5 * 6r4-lla: 6=0. dns. 1,2,3.
3. Required the commensurable roots of the equation
s_6;e 2 16^+21=0. Am. 3 and 1
Here the student might hesitate, as one regular term of the
equation is wanting, or rather the coefficient of x* is . hence,
the equation is # 4 dbOa; 3 6cr 2 16H-21=0.
Go through the form of adding 0.
4. Required the commensurable roots of the equation
a> 4 6a? 3 4-5^+2# 10=0. tins. 1, +5.
As the commensurable roots are only two, there must be two
incommensurable roots ; and they can be found by dividing the
given equation by a?-j-l, and that quotient by x 5, and resolv-
ing the last quotient as a quadratic.
EQUAL ROOTS.
(Art. 168.) In any equation, as
the roots may be represented by , &, c, d, e, and either one, put
in the place of x, will verify the equation.
Now, let y represent the difference between any two roots, as
a b ; then y=a &, and by transposition b-\-y=a. But as a
will verify the equation, it being a root, its equal, (b-\-y), sub-
stituted for x, will verify it also. That is,
By expanding the powers, and arranging the terms according
to the powers of y, we have
+106V
Cb 2 +2Cby
Db+Vy
E
=0.
We might have been more general, and have taken x m -{-Ax m 1 , &c., for
the equation ; but, in our opinion, we shall be better comprehended by taking
an equation definite in degree : the reasoning is readily understood as general
280 ELEMENTS OF ALGEBRA.
Now, as I is a root of the equation, the first column of this
transformation is identical with the proposed equation, on sub-
stituting the root b for x. Hence, the first column is equal to
xero ; therefore, let it be suppressed, and the remainder divided
byy.
We then have
=0.
2Cb
D
On the supposition that the two roots a and b are equal, y
becomes nothing, and this last equation becomes
As b is a root of the original equation, x may be written in
place of b ; then this last equation is
5zM-4^e 3 +3^+2Ca:+#=0. . . . . . (2)
This equation can be derived from the primitive equation by
the following
RULE. Multiply each coefficient by the exponent of x, and
diminish the exponent by unity.
Equation (2) being derived from equation (1), by the above
rule, may be called a derived polynomial.
(Art. 169.) We again remind the reader that b will verify the
primitive equation (1), it being a root, and it must also verify
equation (2) ; hence, b at the same time must verify the two
equations (1) and (2).
But if b will verify equation (1), that equation is divisible by
(x 6), (Art. 164.), and if it will verify equation (2), that equa-
tion also, is divisible by (x b), and (x 6) must be a common
measure of the two equations (1) and (2). That is, in case
the primitive equation has two roots equal to b.
(Art. 170.) To determine whether any equation contains equal
roots, take its derived polynomial by the rule in (Art. 168.), and
seek the greatest common divisor (Art. 27.), [which designate by
GENERA L THEORY OF EQUATIONS. ogj
(/)),] of the given equation and its first derived polynomial ;
and if the divisor D is of the first degree, or of the form of x h,
then the equation has two roots each equal to h.
If no common measure can be found, the equation contains no
equal roots. If D is of the second degree, with reference to x,
put Z)=0, and resolve the equation ; and if D is found to be in
the form of (x A) 2 ; then the given equation has three roots
equal to h.
If D be found of the form of (x h)(x h') t then the given
equation has two roots equal to h, and two equal to h'.
Let D be of any degree whatever ; put Z)=0, and, if possible,
completely resolve the equation ; and every simple root of D is
twice a root in the given equation ; every double root of D will
be three times a root in the given equation, and so on.
EXAMPLES.
1. Does the equation x 4 2x s 7x 2 -}-2Qx 12=0 contain
any equal roots, and if so, find them ?
Its derived polynomial is 4x 3 6x 2 14#-f20.
The common divisor, by (Art. 27.), is found to be x 2
therefore, the equation has two roots, equal to 2.
The equation can then be divided twice by x 2, or once by
(x 2) 2 , or by fl 2 4#-{-4. Performing the division, we find
the quotient to be x z -^-2x 3, and the original equation is now
separated into the two factors,
(xtlx+^tf+Zx 3) =0.
The equation can now be verified by putting each of these
factors equal to zero. From the first we have already x=2 r
and 2, and from the second we may find x=\ or 3; hence,
the entire solution of the equation gives 1, 2, 2, 3 for the four
roots.
2. The equation arH-2;r 4 lla? 8x*-{-2Qx+lG=Q has two
equal roots ; find them. Jlns. 2 and 2.
3. Does the equation a? 2# 4 +3r } 1x*+8x 3=0 contain
equal roots, and how many ?
Ans. It contains three equal roots, each equal to 1
24
282 ELEMENTS OF ALGEBRA.
4L Find the equal roots, if any, of the equation
x 3 ~\-tf 16#-f 20=0. tins* 2 and 2.
5. Find the equal roots of the equation
X 4_|_ 2 ^ 3 3x 2 4^+4=0.
Jlns. Two roots equal to 1, and two roots equal 2.
6. Find the equal roots of the equation
a? 53H-10a:--8==0.
JJns. It contains no equal roots.
(Art. 171.) Equations which have no commensurable roots, or
those factors of equations which remain after all the commensu-
rable and equal roots are taken away by division, can be resol-
ved only by some method of approximation, if they exceed the
third or fourth degree. It is possible to give a direct solution in
cases of cubics and in many cases of the fourth degree ; but, in
practice, approximate methods are less tedious and more conve-
nient.
We may transform any equation into another whose roots
shall be greater or less than the roots of the given equation by
a given quantity.
Suppose we have the equation
a*+Jla*+a*+Ca*+Dx+JE=0, ..... (1)
and require another equation, whose roots shall be less than those
of the above by a.
Put x=a-\-y, and, of course, the equation involving y will
have roots less than that involving x, by a, because y=x , or
?/ is less than x, by a.
In place of x t in the above equation, write its equal (a~\-y)
and we have
By expanding and arranging the terms according to the powers
of t/, we shall have, as in (Art. 168.),
EQUAL ROOTS.
E
(2)
After a little observation, these transformations may be made
very expeditiously, for the first perpendicular column may be
written out by merely changing x to a, in the original equation,
and then, each horizontal column run out by the law of the
binomial theorem.
Thus a 5 becomes 5a 4 , and this, again, 10a 3 , &c.
Now, the first column of the right hand member of this equa-
tion consists entirely of known quantities ; and the coefficients
of the different powers of y are known ; hence we have an
equation, involving the several powers of y, in form of equa-
tion (1),
Or, ^+^y+j&y+Cy+/)'y-H5;-==0; the equation
required; .#', B', &c., representing the known coefficients of
the different powers of y.
In commencing this subject, we took an equation definite in
degree, for the purpose of giving the pupil more definite ideas ;
but it is now proper to show the form of transforming an equation
of the most general character.
For this purpose, let us take the equation
1
Now let it be required to transform this equation into another
whose roots shall be less than the roots of this equation by a.
Put x=a-{-y, as before ; then
284
ELEMENTS OF ALGEBRA.
% .2
rd ^
CL> cC
+j 9
.f^f
rl ^
<;
'f3 TJ
eS ****
o . . g j>
CM 5^
1 ?
, . * 3 2
%> 1
2 ^53
s ^
I 2
u ?T
__. r^3 HH
55 ^N
1
*s " s
o s|
- : : : i! 1
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EQUAL ROOTS 285
If we should desire to make the third term (counting from the
highest power of y] of equation (2) to disappear, we must
Put 10rt 2 -Htf0+-#=0 ; and this involves the
solution of an equation of the second degree, to find the definite
value of a. To make the fourth term disappear would require
the solution of an equation of the third degree ; and so on.
If a is really a root of the primitive equation, then x=a, <y=0,
and each perpendicular column of the transformed equation is 0.
If we designate the first perpendicular column of the general
transformed equation by X, and the coefficients of the succeeding
columns by
X' X" X"'
The coefficients of the different powers of y, as X', X", X'", &c.
arc called derived polynomials, because each term of X' can be
derived from the corresponding term of X ; and each term of X"
can be derived from the corresponding terms of X', by the law of
the binomial theorem, as observed in the first part of this article.
But, to recapitulate :
X is derived from the given equation by simply changing x
to a.
X' is derived from X by multiplying each of the terms of X
by the exponent of a, in that term, and diminishing that expo-
nent by unity, and dividing by the exponent of y increased by 1.
X" is derived from X', in the same manner as X' is derived
from X ; and so on.
X' is called the first derived polynomial ; X" the second, &c.
To show the utility of this theorem, we propose to transform
the following equations :
1. Transform the equation
x 4 l2x 3 +l7x 2 9*4-7=0,
into another, which shall not contain the 3d power of the un-
known quantity.
286 ELEMENTS OF ALGEBRA.
By (Art. 172 ), put .r=7/+^ ..... or- ... x=3+y
Here =3 and m=4.
X =(3) 4 12(3) 3 +17(3) 2 9(3)-}-7 . or . . . X =110
X' =4(3) 3 36(3) 2 4-34(3) - 9 ...... or ... X' =123
~=6(3) 2 36(3)4-17 .......... or. . . = 37
z <&
Y"'
12 .............. m ---M
Therefore the transformed equation must be
2. Transform the equation
into another wanting its second term. Put x=2-{-y.
X =(2) 3 6(2) 2 +13(2) 12 ....... or. . .X =2
X' =3(2) 2 12(2)+13 .......... or. - X' =+1
6 ............... or...=
X'" X'
21T 1 ................... or ---2
Therefore the transformed equation must be
2/ 3 +
3. Transform the equation
into another whose roots shall be less by 2.
Put a?=24-y. Result, ?/ 4 4-4?/ 3 24y=0,
As this transformed equation has no term independent of y,
y=0 will verify the equation ; and x=2 will verify the original
equation, and, of course, is a root of that equation.
EQUAL ROOTS. 28 7
4. Transform the equation
into av.other whose roots shall be greater by 3.
Put x =3-}-y. Result, 2/ 4 +4?/ 3 -r-9i/ 2 42y=0.
5 Transform the equation
x 4 8z 3 -{-ar J +82;r 60=0,
into another wanting its second term.
Result, y 4 23*/ 2 +22i/-f-60=0.
(Art. 173.) We may transform an equation by division, as
well as by substitution, as the following investigation will show.
Take the equation
* 4 4-^ 3 4- Bx>+Cx+D=Q ........ (1)
If we put x=a-}-y, in the above equation, it will be trans-
formed (Art. D.) into
As x a-\-y, therefore T/ # a ; and put this value of y in
equation (2), we have
(*-)'+ J^-) 3 +^V-) 2 +X '(*-)+X=0. . .(3)
Now it is manifest that equation (3) is identical with equation
(1), for we formed equation (2) by transforming equation (l),and
from (2) to (3) we only reversed the operation.
Now we can divide equation (3), or in fact equation (1), by
(x a). and it is obvious that the first remainder will be X.
Divide the quotient, thus obtained, by the same divisor, (x a).
and the second remainder must be X'.
Divide the second quotient by (x a), and the third remaindei
nmst be -.
m
X'"
The next remainder mus', be , <fec., <fec., according to the
2.3
degrpe of the equation
238 ELEMENTS OF ALGEBRA.
Now if we reserve these remainders, it is manifest that they
may form the coefficients of the required transformed equation ;
taking the last remainder for the first coefficient ; and so on, in
reverse order.
For illustration, let us take the third example of the last article.
01 y=x 2.
16
2x* Sx
16H-32 4x 16
16#-{-32 4x+ 8
0=X 24=X
2)a:+2(l
a: 2
X'"
~
Hence the transformed equation is
0;
or, t/ 4 -f-4?/ 3 24i/=0, as before.
For a further illustration of this method, we will again operate
on the first example of the last article.
EQUAL ROOTS.
10a>-39
lOic 2 9x-
39a'-f- 7
_39.r-|-117
110 X. 1st Remainder.
;_3U- 3 9^ 1 O.r 39(0.- Gx 28
28^39
123 = X'. 2d Remainder.
2S(x 3 a> 3).r 3(1
3x28 X'"
__. 3d Remainder.
Hence 7/ 4 ()j/ 3 3 7/ 123^110 = 0, must be the trans-
formed equation.
We shall have a 4th remainder, if we operate on an equation
of the 4th degree; a 5th remainder with an equation of the f>tli
degree; and, in general, n number of remainders with an equa-
tion of the nth degree.
25
290 ELEMENTS OF ALGEBRA.
But to make this method sufficiently practical, the operator
must understand
SYNTHETIC DIVISION.
(Art. 174.) Multiplication and division are so intimately blended
that they must be explained in connection. For a particular
purpose ve wish to introduce a particular practical form of per-
forming certain divisions ; and to arrive at this end, we commence
with multiplication.
Algebraic quantities, containing regular powers, may be
multiplied together by using detached coefficients, and annexing
the proper literal powers afterwards.
EXAMPLES.
1 Multiply a 2 +2^+^ by a+x.
Take the coefficients. Thus
1+2+1
1 + 1
1+2+1
1+2+1
Product, ... 1+3+3+1
By annexing the powers, we have
J. Multiply 3t*+xy+y* by tfxy+if.
As the literal quantities are regular, we may take detached
coefficients, thus :
1 + 1 + 1
1 + 1 + 1
__!_ i i
1 + 1 + 1
Product,. 1+0+1+0+1
SYNTHETIC DIVISION. 291
Hero the second and fourth coefficients are ; therefore the
terms themselves will vanish ; and, annexing the powers, we shall
have for the full product
3. Multiply 3,r 2 2.r 1 by
321
4+2
1284
642
12282
Product, . . . 12^2^ Sx 2.
. Multiply x 4 aar'+flrV 3 #+a 4 by
1 1 + 1 1 + 1
11 + 1 1+ 1
+ 1 1 + 11 + 1
1+0+0+0+0+1
As all the coefFjcients are zero except the first and last, there-
fore the product must be
(Art. 175.) Now if we can multiply by means of detached
coefficients, in like cases we can divide by means of them.
Take the last example in multiplication, and reverse it, that is,
divide a^+a 5 by a-+a.
Here we must suppose all the inferior powers of x 5 and a 5
really exist in the dividend, but disappear in consequence of their
coefficients being zero ; we therefore write all the coefficients of
the regular powers thus :
292 ELEMENTS OF ALGEBRA.
Divisor. Dividend. Quotient.
Annexing the regular powers to the quotient, we have x* ax 3 -^-
a z x 2 3 vT+ a 4 , for the full quotient.
2. Divide a 5 5a 4 6+lO 3 6 2 10 2 6 3 +5a& 4 tf by a 8
2ab+b z .
12+1)15+1010+51(13+31
12+ 1
34-910
34-6 3
" 37+5
3_6+3
1+2-1
1+21
These coefficients are manifestly the coefficients of a cube,
therefore the powers are readily supplied, and are
N. B. If we change the signs of the coefficients in the divisor,
except the first, and then add the product of those changed terms,
we shall arrive at the same result.
Perform the last example over again, after changing the signs
of the second and third terms of the divisor. Thus,
SYNTHETIC DIVISION. 093
l-f2__l)l_5_}-10 10+5 1(1 3+ 31
l-j-2 1
Sum * 3-f- 910
_3_ 6+ 3
Sum . . . . ; 3 7+5
3-f 63
Sum ....... * J+2 J
12+1
Sum ....... ~~*~0+0
3. Divide or 3 Gr'+l 1# 6 by x2.
Change the sign of the second term of the divisor.
1+2)16+116(14+3
1+2
4+11
4 8
3+6
Let the reader observe, that when the first figure of the divisor
is 1, the first figure of the quotient will be the same as the first
figure of the dividend ; and the succeeding figures of the quotient
are the same as the first figures of the partial dividends.
Now this last operation can be contracted.
AY rite down the figures of the dividend with their proper signs,
and the second figure of the divisor, with its sign changed, on
the right. Thus
16+1 16(2.= Divisor
2 8+6
(144-3) o
The first figure, 1, is brought down for the first figure of
the quotient.
The divisor, 2, is put under 6 ; their sum is 4, which,
multiplied by 2, and the product 8 put under the next term,
294 ELEMENTS OF ALGEBRA.
the sum of +11 8 is 3, which multiplied by 2, gives 6, and the
sum of the last addition is 0, which shows that there is no
remainder.
The numbers in the lower line show the quotient, except the
last ; that shows the remainder, if any.
This last operation is called synthetic division.
4. Divide a,' 3 - r -2r ! -Sx 24 by rr 3.
COMMON METHOD.
a- 3 3.r 2
SYNTHETIC METHOD.
14-2 824(3
3+15-1-21
(14-54-7) 3
Now we are prepared to work the examples in (Art. E.) in a
more expeditious manner.
Transform again, the equation # 4 4x 3 8#4-32=0, to an
other, whose roots shall be less by 2.
This equation has no term containing a, 12 , therefore the coefli
cient of a? must be taken =0, if we use Synthetic Division.
FIRST OPERATION.
1 40 84-32 (2
24832
(12416), 0=X.
SECOND OPERATION.
12416(2
20 8
(l-f-0 4),-24=X'.
SYNTHETIC DIVISION. 295
THIRD OPERATION.
IrbO 4 (2
2+4
Y"
(1+2) 0=-.
FOURTH OPERATION.
1+2(2
2
(fence oui transformed equation is y 4 -f-4t/ 3 24*/=0, as
before.
To transform an equation of the fourth degree, we must have
four operations in division ; an equation of the nth degree n
opeiations, as before observed.
But these operations may be all blended in one. Thus
j __4 o 8 32 (2
248 32
0"=:X
-2
2
4
16
8
=X
2
2
2
4
+4
24
X"
=
2
2.3*
We omit the first column, except in the first line, as there aro
no operations with it.
The pupil should observe the structure of this operation. It
is an equation of the 4th degree, and there are four sums in nd-
dition, in the 2d column; three in the next; two in the next,
&.C., giving the whole a diagonal shape.
29 U ELEMENTS OF ALGEBRA.
Transform the equation x* l2x 3 -\-l7x 2 O.r+7 0, into an-
other whose root shall be 3 less.
OPERATION.
1 12 +17 9 _J- 7 (3
4. 3 27 30 117
_ y io 39 110=X
-f 3 18 84
6 28 123=X'
4.39
- 3 -37=^ ,
3
^51;
Hence the transformed equation is
10=0.
Transform tlie equation x 3 l'2x 28 = 0, into another whose
roots shall be 4 less.
12
28 (
4
+ 10
+ 10
4
4
12 = ]
4
32
IF
~~36 = :
V'
4
Hence the transformed equation must be ?/ 3 +12?/ s +30i/
12 0, on the supposition that we put y=x 4.
Transform the equation # 3 10a^+3a: 0946=0, into another
whose roots shall be less by 20.
Put #=20+?/.
SYNTHETIC DIVISION. 29?
10
3
6946
(20
20
200
4060
10
203
2886
20
30
600
803
20
=
50
The three remainders are the numbers just above the doulde
lines, which give the following transformed equation :
y+50?/ 2 +803?/ 2886=0.
Transform this equation into another, whose roots shall be
less by 3. Put y=3-\-z.
50
3
53
3
803
159
962
168
2886
+2886
(3
56
a
1130
59
Hence the second transformed equation is
This equation may be verified by making 2=0 ; which gives
y3 and #=20+3=23.
Thus we have found the exact root of the original equation by
successive transformations ; and on this principle we shall here-
after give a general rule to approximate to incommensuralle roots
of equations of any degree ; but before the pupil can be prepared
to comprehend and surmount every difficulty, he must pay more
attention to general theory, as developed in the following
Chapter
298 ELEMENTS OF ALGEBRA.
CHAPTER III.
GENERAL PROPERTIES OF EQUATIONS.
(Art. 170.) Jlny equation, having only negative roots, will
have all its signs positive.
If wo take a, b, c, &c., to represent the roots of an
equation, the equation itself will be the product of the factors ;
(z-r-a), (x+b), Or-fc), &c., =0 :
and it is obvious that all its signs must be positive.
From this, we decide at once, that the equation a? 4 -f-3a^-f-
6a?+6=0 ; or any other numeral equation, having all its signs
plus, can have no rational positive roots.
(Art. 177.) Surds, and imaginary roots, enter equations by
pairs.
Take any equation, as
and suppose (a-f-^/6) to be one of its roots, then ( Jb) must
be another.
In place of x, in the equation, write its equal, and we have
By expanding the powers of the binomial, we shall find some
terms rational and some surd. The terms in which the odd
powers of ,Jb are contained will be surd ; the other terms
rational ; and if we put R to represent the rational part of this
equation, and SJb to represent the surd part, then we must have
But these terms not having a common factor throughout, cannot
equal 0, unless we have separately /?=0, and #=0; and if this
be the case we may have
This last equation, then, is one of the results of
being a root of the equation.
GENERAL PROPERTIES OF EQUATIONS. 299
Now if we write (a Jb) in place of a?, in the original equ*
tion, and expand the binomials, using the same notation as
before, we shall find
But we have previously shown that this equation must be true ;
and any quantity, which, substituted for x, reduces an equation
to zero, is said to be a root of the equation ; therefore (a Jit)
is a root.
The same demonstration will apply to (-f~V a ) ( >/ a ) to
, J a, and to imaginary roots in the form of
(a+bjl),
(Art. 178.) If we change the signs of the alternate terms of
an equation, it will change the signs of all its roots.
At first, we will take an equation of an even degree.
If a is a root to the equation
x*+J33*+a*+Cx+D=0 ....... (1)
then will a be a root to the equation
x*x*+Bx*Cx+I)=Q ....... (2)
Write a for x, in equation (I), and we have
Now write a for re, in equation (2), and we have
a 4 rfa*+Ea 2 -}-Ca-{-I}=() ....... (4)
Equations (3) and (4) are identical ; therefore if a, put for .c in
equation (1), gives a true result, a put for x in equation (2),
gives a result equally true.
We will now take an equation of an odd degree.
If the equation x s -\-^x 2 -\-Bx-\-C=O t
have a for a root, then will the equation
have a for a root.
From the first
From the second a 3 4#a 2 /?a 6V.-0.
300 ELEMENTS OF ALGEBRA.
This second equation is identical with the first, if \ve change
all its signs, which does not essentially change an equation.
The equation x 4 -f a: 3 l9x 2 -{-l l;r-{-30=0, has 1, 2, 3,
and -5, for its roots ; then from the preceding investigation
\vc learn that the equation
must have 1 2, 3, and +5 for its roots.
(Art. 179.) If we introduce one positive root into an equa-
tion, it will produce at least one variation in the signs of its
term; if two positive roots, at least two variations.
The equation x 2 -\-x-\-l.=Q, having no variation of signs, can
have no positive roots. (Art. 176.) Now if we introduce the
root +2, or which is the same thing, multiply by the factor x 2,
x*+ a?+l
x 2
x-- x x
Zx 2 2x 2
Then a; 3 - a*- a: 2=0;
and here we find one variation of signs from -\-x* to 2 , and
one permanence of signs through the rest of the equation,
If we take this last equation and introduce another positive
root, say -|-5, or multiply it by x 5, we shall then have
11 1 2
5 -[-5 +5 +10
Here are two variations of signs, one from 4-a? 4 to O.r 3 , and
another from Gx 3 to -f-4;r 2 .
And thus we might continue to show that every positive root,
introduced into an equation, will produce at least one variation
of signs. But we must not conclude that the converse of this
proposition is true.
GENERAL PROPERTIES OF EQUATIONS. 3QJ
Every positive root will give one variation of signs ; but
every variation of signs does not necessarily show the existence
of a positive root.
For an equation may have
(a+bjV), ( V^f), c, d,
for roots ; then the equation will be expressed by the product of
the factors
(a* 2ax+a z +b*) (x+c) (x+d)=Q.
As one of these terms, ( 2ax), has the minus sign, it will
produce some minus terms in the product ; and there must neces-
sarily be variations of signs ; yet there is no positive root. At
the same time, the whole factor in which the minus term is found,
must be plus, whatever value be given to x, as it is evidently
equal to (x a) 2 -\-b*, the sum of two squares.
The equation
^ 2x 3 x 2 -\-2 x +10=0,
has two variations of signs, and two permanences, but the roots
are all imaginary, viz.,
T and -.1-=T.
If it were not for imaginary roots, the number of variations
among the signs of an equation would indicate the number of
plus roots : and this number, taken from the degree of the equa-
tion, would leave the number of negative roots; or the number
of permanences of signs would at once show the number of
negative roots.
To determine a priori the number of real roots contained in
any equation, has long bafHed the investigations of mathemati-
cians ; and the difficulty was not entirely overcome until 1829,
when M. Sturm sent a complete solution to the French Academy.
The investigation is known as Sturm's Theorem, and will be pre-
sented in the following Chapter.
LIMITS TO ROOTS.
(Art. 180.) All positive roots of an equation are comprised
between zero and infinity ; and all negative roots between zera
302 ELEMENTS OF ALGEBRA.
and minus infinity ; but it is important to be able at once to
assign much narrower limits.
We have seen, (Art. 179.), that every equation, having a posi-
tive root, must have at least one variation among its signs, and
at least one minus sign.
If the highest power is minus, change all the signs in the
equation.
Now we propose to show that the greatest positive root must
be less than the greatest negative coefficient plus one.
Take the equation
=Q.
It is evident, that as the first term must be positive for all de-
grees, x must be greater and greater, as more of the other terms
are minus : then x must be greatest of all when all the othei
terms are minus, and each equal to the greatest coefficient, (D
being considered the coefficient of a?).
Now as A, B, &c., are supposed equal, and all minus, we shall
have
For the first trial take x=J2, and transpose the minus quantity,
and we have
Divide by ^ 4 , and we have
1=1+ xh/F*~;#~ 3 *
Now we perceive that the second member of the equation is
greater than the first, and the expression is not, in fact, an equa-
tion. x=rf proves x not to be large enough.
For a second tiial put x=
Then
Dividing by (^-fl)' 1 , we have
A Jl A A
-* 8 - 4 '
We retain the sign of equality for convenience, though the
GENERAL PROPERTIES OF EQUATIONS. 303
members are not equal. The second member consists of terms
in geometrical progression, and their sum, (Art. 120), is
1 T~aTT~\\' Hence tne fi rst member is greater than the second,
which shows that (.#+1) substituted for x, is too great. But Jl
was too small, therefore the real value of x, in the case under
consideration, must be more than Jl and less than (.#+1).
77m/ is, the greatest positive root of an equation, in the
most extreme case, must be less than the greatest negative
coefficient plus one.
In common cases the limit is much less.
From this, we at once decide that the greatest positive root of
the equation x 5 S-r'+To: 3 8x z 9a? 12=0, is less than 13.
Now change the second, and every alternate sign, and we
have the equation
ar 5 +3.r 4 +7r 3 +8;r 2 9^+12=0.
The greatest positive root, in this equation, is less than 10 ; but,
by (Art. 178.), the greatest positive root of this equation is the
greatest negative root of the preceding equation; therefore 10 is
the greatest limit of the negative roots of the first equation ; and
all its roots must be comprised between +13 and 10 ; but as
this equation does not present an extreme case, the coefficients
after the first are not all minus, nor equal to each other ; there-
fore the real limits of its roots must be much within +13 and
-10. In fact, the greatest positive root is between 3 and 4, and
the greatest negative root less than 1 .
If it were desirable to find the limits of the least root, put
0,=-, and transform the equation accordingly. Then find, as
y
just directed, the greatest limit of y, in its equation ; which will,
of course, correspond to the least value of x in its equation.
(Art. 181.) If we substitute any number less than the least
root, for the unknown quantity, in any equation of an even
degree, THE RESULT WILL BE POSITIVE. *ftnd if the degree of the
equation be odd, THE RESULT WILL BE NEGATIVE.
Let , b, c, &c., be roots of an equation, and x the unknown
3J4 ELEMENTS OF ALGEBRA.
quantity. Ayso, conceive a to be the least root, /; the next
greater, and so on. Then the equation will be represented by
(x a)(x b)(x c)(a? rf), &c., =0.
Now in the place of x substitute any number h less than a, and
the above factors will become
(ha)(hb)(hc}(hd), &c.
Each factor essentially negative, and the product of an even
number of negative factors, is positive ; and the product of an
odd number is negative ; therefore our proposition is proved.
Scholium. If we conceive h to increase continuously, until it
becomes equal to a, the first factor will be zero ; and the product
of them all, whether odd or even, will be zero, and the equa-
tion will be zero, as it should be when A becomes a root.
If h increases and becomes greater than a, without being
iqual to I), the result of substituting it for x will be NEGATIVE,
in an equation of an even degree, and positive in an equation
of an odd degree.
For in that case the first factor will be positive, and all the
other factors negative ; and, of course, the signs of their product
will be alternately minus and plus, according as an even or odd
number of them are taken.
If h is conceived to increase until it is equal to b, then the
second factor is zero, and its substitution for x will verify the
equation. If h becomes greater than b, and not equal to c, then
the first two factors will be positive ; the rest negative ; and the
result of substituting h for x will give a positive or negative
result, according as the degree of the equation is even or odd.
If we conceive h to become greater than the greatest root,
then all the factors will be positive, and, of course, their product
positive.
For example, let us form an equation with the four roots 5,
2, 6, 8, and then the equation will be
(^+5)(x 2) (a? 6)(ar 8)=0,
Or. ... # 4 11^ 4z 2 +284z 480=0.
(The greater a negative number is, the less it is considered.)
GENERAL PROPERTIES OF EQUATIONS. 305
Now if we substitute 6 for x, in the equation, the result must
be positive. Let 6 increase to 5, and tho result will be 0.
Let it still increase, and the result will be negative, until it has
increased to +2, at which point the result will again be 0.
If we substitute a number greater than 2, and less than 6, for
#, in the equation, the result will again be positive. A number
between 6 and 8, put for x, will render the equation negative ;
and a number more than 8 will render the equation positive ;
and if the number is still conceived to increase, there will be no
more change of signs, because we have passed all the roots.
If in any equation we substitute numbers for the unknown
quantity, which differ from each other by a less number than
the difference between any two roots, and commence with a
number less than the least root, and continue to a number greater
than the greatest root, we shall have as many changes of signs
in the results of the substitution as the equation has real roots.
If one real root lies between two numbers substituted for the
unknown quantity, in any equation, the results will necessarily
show a change of signs.
If one, or three, or any odd number of roots, lie between the
two numbers substituted, the results will show a change of signs
If an even number of roots lie between the two numbers sub-
stituted, the results will show no change of signs.
In the last equation, if we substitute 6 for a?, the result
will be plus.
If we substitute -J-3, the result will also be plus, and give no
indication of the two root? 5 and -f-2, which lie between.
(Art. 182.) If an equation contains imaginary roots, the
factors pertaining to such roots will be either in the form of
(a-f-) or in the form of [(a? a) 2 -{-& 2 ], both positive, whatever
numbers may be substituted for x, either positive or negative ;
hence, if no other than imaginary roots enter the equation, all
substitutions for x will give positive results, and of course, no
changes of sign. It is only when the substitutions for x pass
real roots that we shall find a change of signs.
26
306 ELEMENTS OF ALGEBRA.
CHAPTER IV.
GENERAL PROPERTIES OF EQUATIONS CONTINUED.
(Art. 183.) If we take any equation which has all its roots
real and unequal, and make an equation of its first derived poly-
nomial, the least root of this derived equation will be greater than
the least root of the primitive equation, and less than the next
greater.
If the primitive equation have equal roots, the same root will
verify the derived equation.*
We sha)l form our equations from known positive roots.
Let <i, b, c, d, &c., represent roots ; and suppose a less than
&, b less than c, <fec., and x the unknown quantity. An equation
of the second degree is
x 2 (
Its first derived polynomial is
If we make an equation of this, that is, put it equal to 0, we
shall have
Now if b is greater than , the value of x is more than a,
and less than 6, and proves our proposition for all equations of
the second degree. If we suppose a=6, then x=a, in both
equations.
An equation of the third degree is
^(a+b+c^+^b+ac+bclxabc^ ..... (i)
Its first derived polynomial is
This equation, being of the second degree, has two roots, and
only two.
To ensure perspicuity and avoid too abstruse generality, we ope-ate on
equations definite in degree ; the result will be equally satisfactory to the
learner, and occupy, comparatively, but little space.
GENERAL PROPERTIES OF EQUATIONS. 30?
Now if we can find a quantity which, put for x, will verify
equation (2), that quantity must be one of its roots. If we try
two quantities, and find a change of signs in the results, we are
sure a root lies between such quantities. (Art. 181.) Therefore
we will try a, or write a in place of x. As b and c are each
greater than a, we will suppose that
With these substitutions, equation (2) becomes
Reduced, gives ...... -j-/j/i'=0.
Therefore a cannot be a root; if it were we should have 0=0.
If we now make a substitution of b for a*, or rather (a-\-li)
for a*, and reduce the equation, we shall find
It is apparent that this quantity is essentially minus, as h'
is greater than h. Hence, as substituting a for a?, in the equa-
tion, gives a small plus quantity, and b for x gives a small
minus quantity, therefore one value of a?, to verify equation (2),
must lie between a and b.
This proves the proposition for equations of the third degree :
and in this manner we may prove it for any degree; but the labor
of substituting for a high equation would be very tedious.
If we suppose =6, and put c=a-\-h', and then substitute a
in place of a?, we shall find equations (1) and (2) will be verified.
Therefore in the case of equal roots, the equation and its
first derived polynomial will have a common measure, as before
shown in (Art. 168).
If all the roots of an equation are equal, the equation itself
i:.5ay be expressed in the form of
(x a) m =0.
Its first derived polynomial, put into an equation, will be
m ( x a) m ~'=0.
It is apparent that the primitive equation has m roots equal to
a; and the derived equation, (ra 1), roots also equal to a.
308 ELEMENTS OF ALGEBRA.
Lastly, take a general equation, as
Its first derived polynomial, taken for an equation, will be
mx m - l +(m l)rfx m ~* ..... 7?=0.
We may suppose this general equation composed of the
factors
(x a)(x b) (x c), &c., =0,
and also suppose b greater, but insensibly greater, than a ; c
insensibly greater than b, &c. Then the equation will be nearly
(x ) OT =0 ;
and its derived polynomial,
m(x )" l ~ 1 =0,
cannot have a root less than (a), the least root of the primitive
equation ; but its root cannot equal a, unless the primitive equa-
tion have equal roots ; therefore it must be greater.
By the same mode of reasoning we can show that the greatest
root of an equation is greater than the greatest root of its derived
equation ; hence the roots of the derived equation are interme-
diate, in value, to the roots of the primitive equation, or contained
within narrower limits.
(Art. 184.) If we. take any equation, not having equal roots,
and consider its first derived polynomial also an equation,
and then substitute any quantity less than the least root of
either equation, for the unknown quantity, the result of such
mbstilution will necessarily give opposite signs.
Let a, b, c, &c., represent the roots of a primitive equation,
and a', b', &c., roots of its first derived polynomial ; x the un-
known quantity. Then the equation will be
(x a)(x b)(x c), &c., to m factors =0 ;
the derived equation will be
(x a')(x b')(x c'), &c., to (m 1) factors =0.
Now if we substitute h for x, and suppose h less than either
root, then every factor, in both equations, will be negative.
The product of an even number of negative factors is positive
and the product of an odd number is negative ; and if the factor?,
GENERAL THEORY OF EQUATIONS 309
in the primitive equation are even, those in the der ved equation
must be odd.
Hence any quantity less than any root of either equation, will
necessarily give to these functions opposite signs.
(Art. 185.) Now if we conceive h to increase until it becomes
equal to a, the least root, the factor (x ) will be 0, and reduce
the whole equation to 0. Let h still increase and become
greater than a, and not equal to a', (which is necessarily greater
than a, (Art. 184.), and the factor (x a) will become plus, while
till the other factors, in both equations, will be minus, and, of
course, leave the same number of minus factors in both functions,
which must give them the same sign. Consequently, in passing
the least root of the primitive equation a variation is changed into
a permanence.
Sturm's Theorem.
(Art. 186.) If we take any equation not having equal roots, and its
first derived polynomial, and operate with these functions as though
their common measure was desired, reserving the several remain-
ders with their signs changed, and make equations of these func-
tions, namely, the primitive equation, its first derived polynomial
and the several remainders with their signs changed, and then
substitute any assumed quantity, h, for x, in the several functions,
noting the variation of signs in the result ; afterwards substitute
another quantity, h', for x, and again note the variation of signs ;
the difference in the number of variation of signs, resulting
from the two substitutions, will give the number of real roots
between the limits h and h'.
If * QQ and +00 are taken for h and h', we shall have the
whole number of real roots ; which number, subtracted from the
degree of the equation, will give the number of imaginary roots.
DEMONSTRATION.
Let X represent an equation, and X' its first derived polyno-
mial.
In operating as for common measure, denote the several quo-
* Symbols of infinity.
310 ELEMENTS OF ALGEBRA.
tients by Q, Q' Q", &c., and the several remainders, with their
sign$ changed, by R, R', R", &c.
In these operations, be careful not to strike out or introduce
minus factors, as they change the signs of the terms ; then a
re-change of signs in the remainder would be erroneous.
From the manner of deriving these functions, we must have
the following equations
X R
R R
IF* -IF
<fcc. &c.
Clearing these equations of fractions, we have
X =X' Q R
X'=RQ' R'
R =nR'Q" R"
R'=R"Q"' R'"
As the equation X=0 must have no equal roots, the functions
X and X' can have no common measure (Art. 168.), and we shall
arrive at a final remainder, independent of the unknown quantity,
and not zero.
Proposition 1. No two consecutive functions, in equations (A),
can become zero at the same time.
For, if possible, let such a value of h be substituted for x, as
to render both X' and R zero at the same time ; then the second
equation of (A) will give R'=0. Tracing the equations, we must
finally have the last lemainder R TO =0; but this is inadmis-
sible ; therefore the proposition is proved.
Prop. 2. If lien one of the functions becomes zero, by giving
GENERAL PROPERTIES OF EQUATIONS. 311
a particular value to x, the adjacent functions between which it
is placed must have opposite signs.
Suppose R' in the third equation, (A), to become 0, then the
equation still existing, we must have R= R".
The truth of Sturm's Theorem rests on the facts demonstrated
in Arts. 184, 185 and in the two foregoing propositions.
If we put the functions X, X', R, R', &c., each equal to ;
that is, make equations of them, and afterwards substitute any
quantity for #, in these functions, less than any root, the firs
and second functions, X and X', will have opposite signs, (Art.
184.) ; and the last function will have a sign independent of x,
and, of course, invariable for all changes in that quantity. The
other functions may have either plus or minuSj and the signs
have a certain number of variations.
Now all changes in the number of these variations must
come through the variations of the signs in the primitive func-
tion X. A change of sign in any other function will produce
no change in the number of variations in the series.
For, conceive the following equations to exist:
(B)
Now take x=h, yet h really less than any root of the equa-
tions, (B), and we may have the following series of signs :
X = 1
X' = 4-
R =
R' =
R"=
R'"= 4-
Or we mav have any other order of signs, restricted only to the
fact that the signs of the two first functions must be opposite, and
the last invariable, or unaffected by all future substitutions.
are three variations of signs.
312 ELEMENTS OF ALGEBRA.
Now conceive h to increase. No change of signs can take
place in any of the equations, unless h becomes equal and passes
a root of that equation ; and as there are no equal roots, no two
of these functions can become at the same time (Prop. 1.) ;
hence a change of sign of one function does not permit a change
in another ; therefore by the increase of A, one of the functions,
(C), will become 0, and a further increase of h will change its
sign.
In the series of signs as here represented, X' cannot be the
first to change signs, for that would leave the adjacent functions,
X and R, of the same sign, contrary to Prop. 2 ; nor can the
function R' be the first to change sign, for the same reason.
Hence X or R or R" must be first to change sign.
If we suppose X to change sign, the other signs remaining the
same, the number of variations of signs is reduced by unity.
If R or R" change sign, the number of variations cannot be
changed ; a permanence may be made or reduced, and all cases
that can happen with three consecutive functions may be ex-
pressed by the following combinations of signs ;
+ )
Or + j
either of which gives one variation and one permanence.
Now as no increase or decrease in the number of variations of
signs can be produced by any of the functions changing signs,
except the first, and as that changes as many times as it has real
roots, therefore the changes in the number of variations of signs
show the number of real roots comprised between h and h'.
If h and h' are taken at once at the widest limits of possibility,
from infinity to -1- infinity, the number of variations of signs
will indicate the number of real roots; and this number,
taken from the degree of the equation, will give the number of
the imaginary roots.
(Art. 187.) The foregoing is a full theoretical demonstration
of the theorem ; but the subject itself being a little abstruse,
some minds may require the following practical elucidation.
GENERAL PROPERTIES OF EQUATIONS. 3J3
Form an equation with the four assumed roots, 1, 3, 4, 6.
The equation will be
or X = a? 4 14# 3 +67# 2 126aH-72=0 Roots 1, 3, 4, 6.
X = 4x s 42r4-J34# 126 =0. . Jfoofa 2, 3.3, 5 nearly.
R =13y?9lx +153 ......... Moots 2.8, 4.1 nearly.
R'=72# 252 ............. Hoot 3.5
R"= +
\
Let the pupil observe these functions, and their roots, and see
that they correspond with theory. The least root of X is less
than the least root of X'. (Art. 183.) The roots of any func-
tion are intermediate between the roots of the adjacent functions.
This corresponds with (Prop. 2.) ; for if three consecutive func-
tions have the same sign as , -, , or -f-, +, -K the middle
one cannot change first and correspond to (Prop. 2.) ; but signs
change only by the increasing quantity passing a root, and it must
pass a root of one of the extreme functions first ; therefore the roots
of X' must be intermediate in value between the roots of X and
R ; and the roots of R intermediate in value between the roots
of X' and R' ; and so on. But the roots of X' are within nar
rower limits than the roots of X (Art. 183.) ; therefore the roots
of all the functions are within the limits of the roots of X.
We will now trace all the changes of signs in passing all the
roots of all the functions.
We will first suppose x or h = ; which is less than any
root ; then as we increase h above any root, we must change the
sign of that function, and that sign only
We represent these changes thus :
27
314 ELEMENTS OF ALGEBRA.
X X' R R' R"
* When xQ + + -{-.... 4 variations
#=1.1 * 4- 4-. . . .3
" #=2.1 4-* 4. 4- .... 3 "
" #=2.9 4- * _)_.... 3
" #=3.1 4-* + + .... 2
#=3.4 4" * 4" 2 "
" #=3.7 4- -p _{-.... 2
" #=4.1 * + 4. . . . . i
" #=4.11 4-* 4- 4-....1 "
#=5.i 4-* 4- 4- -i- .... i
#=6.1 4-* 4- 4- 4- 4 o
We commenced with 4 variations of signs, and end with
variations, after we have passed all the roots ; therefore the real
roots, in the primitive equation, must be 4 0=4.
By this it can be clearly seen, by inspection, that the changes
of sign in all the functions, except the first, produce no change
in the number of variations.
In making use of this theorem we do not go through the inter-
mediate steps, unless we wish to learn the locality of the roots as
well as their number. We may discover their number by sub-
stituting a number for x less than any root, and then one greater ;
tne difference of the variations of signs will be equal to the num-
ber of real roots.
If we take oo and 4-QD, the sign of any whole function
will be the same as that of its first term.
* In making this table, we did not really substitute the numbers assumed for
x, as we previously determined the roots ; and as passing any root changes the
sign in that function, we write a star against that sign which has just
changed.
APPLICATION OF STURM'S THEOREM. 315
CHAPTER V.
APPLICATION OF STURM'S THEOREM.
(Art. 188.) In preparing the functions, remember that we are
at liberty to suppress positive numeral factors.
EXAMPLES.
! How many real roots has the equation x 3 -}-9x=Q ?
Here X =
X'=
R= ar+1
R'=
Now for x substitute co or 100000, and we see at once
^ X X' R R'
-j- ~J- 2 variations.
Again, for x put -f-co or +100000, and the resulting signs
must be , ,
f- + 1 variation.
Hence the above equation has but one real root ; and, of course,
two imaginary roots.
To find a near locality of this root, suppose x=Q, and the
signs will be ,
4- =t 2 variations.
x=l -|- 4~ rb .1 variation.
Hence the real root is between and 1
Now as we have found x, in the equation x 3 -}-Qx 6=0, to
be less than 1, X s may be disregarded, and 9x 6=0, will give
us the first approximate value of a?; that is, a?=.6, nearly.
2. How many real roots has the equation a; 4 3x z 4=0 ?
X= a; 4 Zx 2 4
X'=4a: 3 Qx
R=+25
If XCQ -f- +2 variations.
#=-1-03 -|- -}- + variation.
Hence there must be 2 real roots, and 2 imaginary roots.
316 ELEMENTS OF ALGEBRA.
3. How many real roots has the equation x" 4# 3 621=0 1
(See Art. 103.)
X=z 6 4^621
X'=x 5 23?
R =+625
When x= co + +2 variations.
When #=-fco + + +0 variation.
Hence there are two real roots and four imaginary roots.
4. How many real roots has the equation x 9 15#-|-21=0?
Arts. 3.
5. How many real roots are contained in the equation
6. How many real roots are contained in the equation
=0? dm. 2.
7. Find the number and situation of the roots of the equation
=0.
X=
X'=3^ 2 +22a? -102
R=122o? 393
R=+
Putting x= co H- +3 variations.
ar=-|-co -t- + + + variation.
Hence all the roots are real.
To obtain the locality of these roots there are several principles
to guide us, there is (Art. 180), but the real limits are much
narrower than that article would indicate, unless all the coeffi-
cients after the first are minus, and equal to the greatest.
A practised eye will decide nearly the value of a positive root
by inspection ; but by (Art. 183.) we learn that the root of R,
or 122# 393=0, must give a value to x intermediate between
the roots of the primitive equation.
From this we should conclude at once that there must be a
root between 3 and 4.
NEWTON'S METHOD OF APPROXIMATION. 317
Substituting 3 for #, in the above functions, we have
+ 4-2 variations.
#=4 -h + 4- -{- variation.
Hence there are two roots between 3 and 4.
As the sum of the roots must be 11, and the two positive
roots are more than 6, there must be a root near 17.
As there are two roots between 3 and 4, we will transform
the equation, (Art. 175), into another, whose roots shall be less
by 3; or put a?=3-J-y. Then we shall have
X=
X'=
R I22y 27
R'=+
The value of y, in this transformed equation, must be near
the value of y in the equation 122y=27, (Art. 183.) ; that is, y
is between .2 and .3
2/=.2 gives 4~ + 2 variations.
7/=.3 gives 4-4~4-4-0 variation.
Hence there are two values of y between .2 and .3 ; and, of
course, two values of x between 3.2 and 3.3.
We may now transform this last equation into another whose
roots shall be .2 less, and further approximate to the true values
of x r in the original equation.
Having thus explained the foregoing principles, and, in our
view, been sufficiently elaborate in theory, we shall now apply it
to the solution of equations, commencing with
NEWTON'S METHOD OF APPROXIMATION.
(Art. 189.) We have seenf m (Art. 175.), that if we have any
equation involving x, and put x=a-{-y, and with this value
transform the equation into another involving y, the equation
will be
If a is the real value of x. then i/=0, and X=0.
318
ELEMENTS OF ALGEBRA.
If a is a very near value to a?, and consequently y very
small, the terms containing y 2 , i/ 3 , and all the higher powers of
T/, become very small, and may be neglected in finding the op-
proximate value of y.
Neglecting these terms, we have
X-f-X'2/=0,
Or ..... y=~ ......... (1)
In the equation x=a-{-y, if a is less than x, y must be posi-
tive ; and if y is positive in the last equation, X and X' must
have opposite signs, corresponding to (Art. 184.).
Following formula (1), we have an approximate value of y ;
and, of course, of x. The value of x, thus corrected, again call
, and find a correction as before ; and thus approximate to any
required degree of exactness.
EXAMPLES.
1. Given 3a? 5 +4a? 3 5x 140=0, to find one of the approxi
mate values of x.
By trial we find that x must be a little more than 2.
Therefore, put x=2+y.
X = 3(2) 5 -{- 4(2) 3 5(2) 140. . . or . . . X = 22
X'=15(2) 4 +12(2) 2 5 ........ or. . . X'= 283.
X 22
Whence y= _= =0.07 nearly.
For a second operation, we have
X= 3(2.07) 5 -{- 4(2.07) 3 5(2.07) 140...By log. X = -0.854
X'=15(2.07) 4 +12(2.07) 2 5 ....... By log. X'= 321.82
Q &A
Hence the second value, or 2/=xHT^-r=0.00265-}-
o21o.2
And ............... #=2.07265+
2. Given x*-}-2x 2 23#=70, to find an approximate value
of x. Am. 5.1 3454-.
HORNER'S METHOD OF APPROXIMATION. 319
3. Given x 4 '3x z -\-75x=lQQQQ, to find an approximate
value of x. Ans. 9.886+
4. Given 3jJ* 35# 3 llx 2 14rr-f30=0, to find an approxi-
mate value of x. dns. 11.998-}-.
5. Given 5x 3 3x* 2x=1560, to find an approximate value
of x rfns. 7.00867+.
CHAPTER VI.
HORNER'S METHOD OF APPROXIMATION.
(Art. 190.) In the year 1819, Mr. W. G. Homer, of Bath,
England, published to the world the most elegant and concise
method of approximating to roots of any then known.
The parallel between Newton's and Homer's method, is
this ; both methods commence by finding, by trial, a near value
to a root.
In using Homer's method, care must be taken that the number,
found by trial, be less than the real root. Following Newton's
method we need not be particular in this respect.
In both methods we transform the original equation involving
#, into another involving y, by putting xr-{-y, as in (Art. 175),
r being a rough approximate value of #, found by trial.
The transformed equation enables us to find an approximate
value of y, (Art. 189.).
Newton's method puts, this approximate value of y to r, and
uses their algebraic sum as r was used in the first place ; again
and again transforming the same equation, after each successive
correction of r.
Homer's method transforms the transformed equation into
another whose roots are less by the approximate value of y ; and
again transforms that equation into another whose roots are less,
and so on, as far as desired.
By continuing similar notation through the several transforma-
tions we may have
320 ELEMENTS OF ALGEBRA.
x=r+y
y=s-\-z
z = t+z'
z'=u+z"
&c. &c.
Hence x=r-\-s-\-t, &c. ; r, s, t, &c., being successive figures
of the root. Thus if a root be 325, r=300, s=20, and /=5.
On the principle of successive transformations is founded the
following
RULE for approximating to the. true value of a real root
of an equation.
1st. find by Sturm 1 s Theorem, or otherwise, the value of
the first one or two figures of the root, which designate by r.
2d. Transform the equation (Art. 175), into another whose
roots shall be less by r.
3d. With the absolute term of this transformed equation for a
dividend, and the coefficient of y for a divisor, find the next
figure of the root.
4th. Transform the last equation into another, whose roots
shall be less, by the value of the last figure determined; and
so proceed until the whole root is determined, or sufficiently
approximated to, if incommensurable.
NOTE 1. In any transformed equation, X is a general symbol
to represent the absolute term, and X' represents the coefficient
of the first power of the unknown quantity. If X and X' be-
come of the same sign, the last root figure is not the true one,
and must be diminished.
NOTE 2. To find negative roots, change the sign of every alter-
nate term, (Art. 178.) : find the positive roots of that equation,
and change their signs.
(Art. 191.) We shall apply this principle, at first, to the solu-
tion of equations of the second degree ; and for such equations
as have large coefficients and incommensurable roots, it will fur-
nish by far the best practical rule.
HORNER'S METHOD OF APPROXIMATION. 321
EXAMPLES.
1. Find an approximate root of the equation
x*-\-x 60=0.
We readily perceive that x must be more than 7, and less
than 8, therefore r=7.
Now transform this equation into another whose roots shall be
less by 7.
Operate as in (Art. 175.), synthetic division
i i .60 (7
7 56
8 4
7
15
Trans., eq. y* -f- 15y 4=0
Here we find that y cannot be far from T 4 ^, or between .2 and
.3 ; therefore transform the last equation into another whose
roots shall be .2 less ; thus,
s
1 15 4 (.2
0.2 3.04
15.2 0.96
2
15.4
The second transformed equation, therefore, is
z 2 -f- 15.4* 0.96=0
.96
To obtain an approximate value of z, we have ^ or 0.06.
15.4
In being thus formal, we spread the work over too large a
space, and must inevitably become tedious. To avoid these diffi-
culties, we must make a few practical modifications.
1st. We will consider the absolute term as constituting the
second member of the equation ; and, in place of taking the
algebraic sum of it, and the number placed under it, we will take
their difference.
322 ELEMENTS OF ALGEBRA.
2d. We will not write out the transformed equations ; that is,
not attach the letters to the coefficients ; we can then unite the
whole in one operation.
3d. Consider the root a quotient ; the absolute term a dividend,
and, corresponding with these terms, we must have divisors.
In the example under consideration, 8 is the first divisor ; 15
.s thejirst trial divisor ; 15.2 is the second divisor, and 15.4 is
the second trial divisor ; 15.46 is the third divisor, &c.
Let us now generalize the operation. The equation may be
represented by
Transform this into another whose roots shall be less by r ; that
equation into another whose roots shall be less by s, &c., &c.
SYNTHETIC DIVISION.
1 a n ( r-j-s
r ar-f-r 8
1st divisor, .... a-\-r n'
r (a-\-2r-}-s}s
1st trial divisor, . . a-f-2r n"
0-t-s &c.
2d divisor, . . . .a-\-2r-}-s
s
2d trial divisor, . .a+2r+2
&c.
In the above we have represented the difference between n
and (ar-{-' 2 ) by n', &c. As n', n", n'", &c,, with their corre-
sponding trial divisors, will give 5, t, u, &c., the following for-
mula will represent the complete divisors for the solution of all
equations in the form of
IIORJNER'S METHOD OF APPROXIMATION 3^3
1st divisor, .... a+ r
add .... r-\- s
2d divisor,
add
3d divisor, .... a+2r+2*-f- t
add .... t-\-u
4th divisor, ....
&c. &c. &c.
Equations which have expressed coefficients of the highest
power, as
the formulas will be :
1st divisor, .... a-h cr
add .... cr-f- cs
2d divisor, .... a-f 2cr+ cs
add .... cs-\-ct
3d divisor, .... a-f 2cr-j-2cs-t-c*
&c. &c.
To obtain trial divisors we would add cr only, in place of
(cr+cs), &c.
We will now resume our equation for a more concise solution.
1
7
n r
60 ( 7.
56
stu
262
1st divisor, . .
add . .
8
7.2
4
304
2d divisor, . .
add . .
15.2
26
96
9276
3d divisor, . .
add . .
15.46
62
324
31044
4th divisor, . . 15.522 1356
324 ELEMENTS OF ALGEBRA.
We can now divide as in simple division, and annex the quo-
tient figures to the root, thus :
15522)1356 (08734
124176
11424
10865
559
465
94
Hence #=7.2620873-f-
2. Find #, from the equation x z 700#=59829.
On trial, we find x cannot exceed 800 ; therefore, r 700.
n rst
-h r -700+700= 59829(777
r+ s 700-H 70=770 00)000
=770 770)59829=n'
77 5390
=847 847)5929=n"
5929
Hence, #=777.
3 Find # from the equation x 2 1283#= 16848.
By trial, we find that # must be more than 1000, and
than 2000; therefore, r=1000.
n rstu
4- r = 283 16848(1296
H- 5= 1200 283
a-r-2r4- ~s= 917 917)2998
s+t 290 1834
= 1207 1207)11644
96 10863
1303 1303)
Hence,
HORNER'S METHOD OF APPROXIMATION. 305
4. Given x 2 5;r=8366, to find x.
By trial, we find x must be more than 90, and less than 100
Therefore, a-f- r = 85 ) 8366 ( Q4=x
r+s . . 94
+2H-s=179
765
716
716
5. Find a?, from the equation x 2 375#+ 1904=0.
Here the first figure of the root is 5.
5
375
1st divisor, 370.
5.1
2d divisor, 364.9
.14
1904(5.1480052207
1850
5400
3649
3d divisor, 364.76
48
4th divisor, 364.712
8
5th divisor, 364.7039
175100
145904
2919600
2917696
190400
1823519
80491
72941
7550
7294
256
255
1
6. Given ^+7^1194=0, to find x. Ans. 31.2311099.
7. Given x 3 21^=214591760730, to find x. rfns. 463251.
It might be difficult for the pupil to decide the value of r, as
applied to the last example, without a word of explanation : x
must be more than the square root of the absolute term, that is,
326 ELEMENTS OF ALGEBRA.
more than 400000 ; then try 500000, which will be found too
great.
(Art. 192.) When the coefficient of the highest power is not
unity, we may (if we prefer it to using the last formulas for di-
visors), transform the equation into another, (Art. 166.), which
shall have unity for the coefficient of the first term, and all the
other coefficients whole numbers.
8. Given 7x* 3#=375.
Put #=- and we shall have y 2 3^=262 5.
One root of this equation is found to be 52.7567068+, one-
seventh of which is 7.536672+ ; the approximate root of the
original equation.
9. Given 7# 2 83#+ 187=0, to find one value of a?.
Jlns. 3.024492664
1O. Given x* T 3 T a?=8, to find one value of x.
Am. 2.96807600231
11. Given 4# 2 +J#=^, to find one value of x.
Jlns. Jlns. .14660+
12. Given \x*-\-\x T 7 T , to find one value of x.
Jlns. .6042334631
13. Given 115 3x* 7#=0, to find one value of x.
Jlns. 5.13368606
(Art. 193.) We now apply the same principle of transforma-
tion to the solution of equations of the third degree.
EXAMPLES.
1. Find one root of the equation
We find, by trial, that one root must be between 3 and 4.
HORNER'S METHOD OF APPROXIMATION.
327
1st Transformation.
1 1
3
3
~5
3
~S
70
*76
15 J
91 =X'
r
300 (3.
228
72=X
.... _- 72 _A
to
o.
H
1
I 8
0.7
91 '
6.09-|
s
72 (0.7
67.963
8.7
7
*97.09 ,
6.58J
4.037=X
9.4
7
To.7
103.67=X'
1 10.1
.03
103.67
.3039-}
t
4.037 (0.03
3.119217
10.13
3
*103.9739
.3048J
0.917783=X
10.16
3
HU9"
104.2787=X'
1 10.19
.008
104.2787
.081584
u
0.917783(0.008
0.834882272
10.198
8
10.206
8
10.214
*104.360284
81648
104.441932
.082900728
328 ELEMENTS OF ALGEBRA.
The terms here marked X are trial divisors ; we have pro
fixed stars to the numbers that we may call complete divisors.
We rest here with the equation
(z") 3 +10.214(z // ) 2 4-104.4419z" 0.0829=0.
The value of z" is so small that we may neglect all its powers,
except the first, and obtain several figures by division, thus :
104 ) 829 ( 797
728
74
r s tu
Hence, one approximate value of # is .... 3.738797+.
(Art. 193.) We may make the same remarks here as in (Art.
191.), and, as in that article, generalize the operation.
Let x s -\-J2z? -\-Bx-N represent any equation of the third
degree, and transform it into another whose roots shall be r less ;
thus,
1 A B = N ( r
r
N'
The transformed equation is
If we put (
and
we shall have . . . y*-\-*ft'y z -{-B'y=N t , an equation similar
to the primitive equation.
If we transform this equation into another whose roots shall be
less by s, we shall have
Or, . . z*-\-"z*-\-B"z=N" ; an equation also similar to
HORNER'S METHOD OF APPROXIMATION. 329
the first equation. And thus we may go on forming equation
after equation, similar to the first, whose roots are less and less.
The quantities N', N", &c., are the same as X in our previous
notation, and the quantities B', B", &c., are the same as the
general symbol, X' ; but we have adopted this last method of
notation to preserve similarity.
Observe, that as (r-{-JT)r-{-JB is the first complete divisor, N
is the number considered as a dividend and r the quotient ; and
therefore... r^-^, nearly.
The next equation gives us
' '
( ' "
JV N" very nearly.
And the next, /=
N'" N"
&c. = &c. &c.
The denominators of these fractions are considered complete
divisors, and the quantities, B', B", B'", are considered trial
divisors. The further we advance in the operation the nearer
will the trial and true divisors agree.
Before the operation is considered as commenced, we must
find the first figure of the root (r) by trial. Then the operator
can experience no serious difficulty, provided he has in his mind
a clear and distinct method of forming the divisors ; and these
may be found by the following
RULE. 1st. Write the number represented by B, and directly
under it, write the value of r(r+A) ; the algebraic sum of these
two numbers is the first complete divisor.
2d. Directly under the first divisor write the value of r 2 , and
the sum of the last three numbers is B 1 , or the first trial divisor.
3d. Find by trial, as in simple division, how often B' is con-
tained in N, calling the first figure s, (making some allowance
for the augmentation o/'B'), and s will be a portion of the
root under trial.
28
330 ELEMENTS OF ALGEBRA.
4th. Take the value of the expression (3r+s-[-A) and add
it to the first trial divisor ; the sum is the second divisor (if
we have really the true value of s).
IN GENERAL TERMS;
Under any complete divisor, write the square of the last figure of
the root ; add together the three last columns, and their sum is
the next trial divisor. With this trial divisor decide the next
figure of the root.
Take the algebraic sum, of three times the root previously
found, the present figure under trial, and the coefficient Jl, and
multiply this sum by the figure under trial, and this product,
added to the last trial divisor, gives the next complete divisor.
EXAMPLES.
1. Given a? 3 +2a^+3a?= 13089030, to find one value of x.
By trial, we soon find that x must be more than 200 and less
than 300 ; therefore we have
r=200, .#=2, .5=3.
By the rule,
N rs
B ....... 3 40403 ) 13089030 ( 235
40400 - 80806
-j
1st divisor ....... 40403 f 139763 ) 500843
r 2 ........ 40000J 419289
1st trial divisor . . B' =120803 163108 ) 815540=^
(3H-s-M)s 18960^ 815540
2d divisor ....... 139763 f
s 2 ........ 900J
2d trial divisor JB"=I 59623
3485
3d divisor ....... 163108
Hence, ........ . ............... z=235,
HORNER'S METHOD OF APPROXIMATION. 331
3. Given ^-(-173^=14760638046, to find one value of x
Here .#=0, I?=173, and we find, by trial, that x must be
more than 2000, and less than 3000 ; therefore r=2000.
B ........ 173
4000000
1st divisor ......... 4000173 f
r 2 ......... 4000000J
B' ........ 12000173
2560000
2d divisor ......... 14560173 f
s 2 ......... leooooJ
B" ....... 17280173
. 362500>
3d divisor ......... 17642673 f
2500 J
B'" ....... 18007673
.... 22059
4th divisor ........ 18029732
4000173 ) 14760638046 ( 2453=z
8000346
14560173 ) 67602920 =N'
58240692
17642673 ) 93622284 =N"
88213365
18029732 ) 54089196=^'"
54089196
3. Given a? 3 +2a^ 23#=70, to find an approximate value
of x. rfns. #=5.134578-K
4. Given x* 17a^-i-42iC=185, to find an approximate value
of x. Jns. r== 15.02407.
332 ELEMENTS OF ALGEBRA.
(Art. 194.) When the coefficient of the highest power is not
unity, we may transform the equation into another, (Art. 166.),
in which the coefficient of the first term is unity, and all the
other coefficients whole numbers ; but it is more direct and con-
cise to modify the rule to suit the case.
If the coefficient of the first power is c, the first divisor will
be (cr-f^)r-f-^, in place of (r+A)r-\-B.
In place of (3r-{-s-\-JI)s, to correct the first trial divisor, we
must have (3cr-\-cs-\-tf)s ; and, in genera], in place of using 3
times the root already found, we must use 3c times the root ; and,
in place of the square of any figure, as r 2 , s 2 , &c., we must use
cr 2 , cs 2 , &c.
EXAMPLES.
5. Find one root of the equation, 3:e 3 -{-2;r 2 -}-4;r=75,
By trial, we find that x must be more than 2, and less than 3;
therefore , a A
r=2, c=3, #=2, /j4.
N r stu
B ........ 4. 75 ( 2.577
16. >
1st divisor 20.
cr 2 12. J
40
35=7V'
29375
B 1 ....... 48. 5625=^"
. .10.75 5038579
2d divisor ...... 58.75 f . 586421 =N'"
cs
75J .517301099
B" ....... 70.25 69119901
3d divisor ..... 71.9797 [ Continue, by simple division, thus
ct * ....... _ Jfy 739 ) 6911 ( 935
B>" ....... 73.7241 6651
176057 ~
4th divisor ..... 73.900157 221
Hence, ................. z =2. 577935 -f.
* R is a symbol to represent the entire root, as far as determined.
HORNER'S METHOD OF APPROXIMATION. 333
6. Find one root of the equation, 5# 3 Qx?-\-3x 85.
rfns. x= 2.16399-.
7. Find one root of the equation, \2x*-\-x* 5a?=330.
dns. #=3.0364754-
8 Find one root of the equation, 5# 3 -f 9# 2 7#=2200.
Ans. #=7.107353G-f-.
9. Find one root of the equation, 5# 3 3a? 2 2.r=1560.
Ans. #=7.0086719-1- .
(Art. 195.) This principle of resolving cubic equations ma)'
be applied to the extraction of the cube root of numbers, and
indeed gives one of the best practical rules now known.
For instance, we may require the cube root of 100. This
gives rise to the equation
in which .#=0, and 5=0, and the value of x is the root
sought.
As Ji and B are each equal to zero, the rule under (Art. 193.)
may be thus modified.
1st. Keeping the symbols as in (Art. 193.), and finding r by
trial, r 2 will be the first divisor, and 3r 2 is B', or the first TRIAL
divisor.
2d. By means of the dividend (so called), and the first trial
divisor, we decide s the next figure of the root.
3d. Then (3r-{-s)s ; that is, three times the portion of the
root already found, with the figure under trial annexed, and
the sum multiplied by the figure under trial, will give a sum,
which, if written two places to the right, under the last trial
divisor, and added, will give the next complete divisor.
4th. After we have made use of any complete divisor, write
the square of the last quotient figure under it ; the sum of the
three preceding columns is the next trial divisor ; which use,
and render complete, as above directed, and so continue as far as
necessary.*
* In case of approximate roots after three or four divisors are found, we
may find two or three more figures of the root, with accuracy, by simple division.
334 ELEMENTS OF ALGEBRA.
We may now resolve the equation
1st divisor
(3r-f-s> .
2d divisor
. 16
. .48
. . 756
. . 5556
r stu
100 ( 4.6415889-t-
64
36
33336
B" 6348
(3R+t)t . 5536]
3d divisor .... 640336 f
? 16 J
B"' 645888
. . 13921
4th divisor. .64602721
64616643
646166
2. Extract the cube root of
1st divisor * * 64
B' 192
(3r-f-s)s 1729
2d divisor
B" .
2664000
2561344
102656
64602721
38053279
32308321
5744958
5169331
575627
516933
58694
58154
673373097125.
N rstu
673373097125 (8765
512
3d divisor
B'"
20929 \
49 j
161373
146503 rNO TE.-To deter-
22707
15696]
mine s, we have
14870097 1 i92)1613(
13718376 | Some allowance
2286396 \
36J
11 51721 1251 crease of 192.
1151721125
2302128
131425
230344225
HORNER'S METHOD OF APPROXIMATION. 335
3. Extract the cube root of 1352605460594688.
tins. 110592.
4. Extract the cube root of 5382674. tins. 175.25322796.
5 Extract the cube root of 15926.972504.
Ans. 25.16002549.
6. Extract the cube root of 91632508641.
Ans. 4508.33859058.
7. Extract the cube root of 483249. Am. 78.4736142
(Art. 196.) The method of transforming an equation into an-
other, whose roots shall be less by a given quantity, will resolve
equations of any degree ; and for all equations of higher degrees
than the third, we had better use the original operation, as in
(Art. 192.), and attempt no other modification than conceiving the
absolute term to constitute the second member of the equation ;
and the difference of the numbers taken in the last column in
place of their algebraic sum.
The following operation will sufficiently explain :
1. Find one value of x from the equation
r
1
3
75
= 10000 (9
9
81
702
6993
9
78
777
3007=^V
9
162
2160
18
240
2937
9
243
27
483
9
36
(Continued on the next page.)
336
ELEMENTS OF ALGEBRA.
1 36
483.
2937.
9
=3007. ( 0.8
to
c-
t -^
.8
29.44
409.952
2677.5616
36.8
512.44
3346.952
329.4384=^"
ansformation.
8
3776
8
38.4
30.08
542.52
30.72
573.24
434.016
coefficients to their nearest
3780.968
Take the
8
unit.
39.2
t
1 39
573
3781
= 329.4384 ( 0.08
+
3
46
306.16
39
576
3827
23.2784=^'"
+
3
46
39
579
3873
u
1 39
579
3873
= 23.2784(0.00600-1-
3
23.256
3876
224
3
3879
Hence, a-=9.88600+.
N. B. We went through the first and second transformations in full. Had
we been exact, in the third, we should have added .08 to 39.2, and multiplied
their sum, (39.28), by .08, giving 3.1424 ; we reserve 3. only to add to the
next column. By a similar operation we obtain 46. to add to the next column.
EXAMPLES.
1. Given x & x 3 x 4 -J-500=0, to find one value of a;. Ans. 4.46041671
2. Given a; 4 5.r 3 -}-9;r=2.8, to find one value of x. Ans. .32971055072
3. Given 20#-{-ll;r 2 -j-9z 3 ^=4, to find one value ofx.
Ans. .17968402502
4. Required the 5th root of 5000; or, in other terms, find one root of the
equation ^=5000. Ans. 5.49280-}-
5. Given 2?=
to find one value of x.
Ans. 2.120003355
THEORY OF EQUATIONS-SUM OF COEFFICIENTS. 337
In following out general principles, we have, thus far, omit-
ted some artifices which apply in particular cases, or depend
on particular circumstances, which we will now set forth, as
they will enlarge the views of all who pay attention to them.
The first circumstance to be observed is the sum of the
numerical coefficients.
(Art. 197.) Transpose every term to the first member of the
equation, and if the sum of the coefficients is zero, we are sure
that unity is one of the roots of the equation.
For example, we are sure that the equation
a;4+7a; 3 133 2 -79*+84==0,
has unity for one of its roots, because l-J-7 13 79+84=0.
Assume x=l, and substitute 1 for x in the equation, and the
equation will be verified. But assume x, a little greater or a
little less than 1, and substitute, and the equation cannot be
verified ; therefore the following principle is established :
In any equation having numerical coefficients, if the sum of the
coefficients is zero, -\-l is one of the roots of the equation.
We can now divide the given equation by (x 1) and thus
depress it one degree.
For another example we will recall the 4th, on page 266,
which is
Here 1 13-(-12=0, whence x=\. The answer on page
266 is 3 ; that is another root. The third root must, therefore,
be 4 ; the sum of all the roots must be 0, because the second
term of the equation is zero. (Art. 157, second observation.)
We perceive at once, that unity is one root in each of the
following equations :
2. x 3 Qx^ + Ux 6=0.
, 3. x *Q x 3J r \4 x 28 x }5Q.
4. X 3_ i3^2_j_3 9a ._ 27=0.
(Art. 198.) When the absolute term of an equation is
numerically too large (whether it be plus or minus), to make
29
338 ELEMENTS OF ALGEBRA.
the sum of the coefficients zero, we can diminish it, as well as
other coefficients, by the following transformation.
For instance : In the equation
the sum of the coefficient is obviously not equal to zero ; 70 is
too large, but we can diminish it by placing x=2y, or x=ny,
and substituting this value of x in the equation, we have
H 3y3 iOw 2
Dividing by n 3 we obtain
n n
Now, we can assume n equal to 2, 3, 4, or any other number
whatever.
Let w=2, and the preceding equation becomes
y'-5y s +"H-=o. (i)
let=-2,
Let 7i=5, and y 3 -^+y +=0. (3)
Zo ~o
The sum of the coefficients in each of the equations (2) and
(3), is zero. Therefore, y equals 1 in each of them.
But x=ny, therefore z 2, and #=5,
which are two of the three roots in the original equation.
But the sum of the three roots must be 10, by the genera]
theory of equations ; therefore,, the third root must be 7,
Because, 2+5-f 7= 1 0.
If we take the first equation in (Art. 197), and depress it
one degree, by dividing by (x 1), we shall obtain the equation
z*--8x 2 oxU=Q. (a)
Here it is obvious that the coefficients require depressing to
make their sum zero.
Therefore, place x=nP, and
p 3 i 8 j p 2 _l p __84 =a ^
w n 2 n 3 v '
THEORY OF EQUATIONS SUM OF COEFFICIENTS. 339
Making n=3, this equation becomes
2_5 p _28 =0
99
Here ?+?i - =0. Whence P=l, but =3,
99 9 9
and z=nP=3, another of the roots of the original equation.
We may now depress equation (a) one degree, by dividing
by (x 3), and then find the other two roots.
Or, we may assume n 4, and substitute this value for n
in (b), then that equation becomes
Here, because H~ff r 5 6-h?6=- ^=1, and x= 4,
which is a third root of the original equation. x= 7 is the
fourth root.
Thus we can find all the roots in the equation contained in
the last article.
If it were our object to transform an equation so as to
increase the coefficients in place of diminishing them, as
p
in this article, we should place #= , but this has already
n
been shown in (Art. 166).
This method of solution may be considered a sequence to
Newton's method of dividers as shown in (Art. 147).
(Art. 199.) An equation which has one for a root, can bo
changed into another, which will have minus one for a root, by
changing its second, and every alternate sign. (Art. 178.)
Consequently, then, an equation which has minus one for
one of its roots, can be changed into another, which will have
plus one for one of its roots, by changing the sign of its second,
and every alternate term.
Therefore, if changing the sign of the second, and every alter-
nate term in any equation ivill make the sum of the coefficients
ZERO, one root of that equation must be 1.
For example, the sum of the coefficients of the equation
4.c 3 20;r 2 -{-ll.r+35=0, is not zero;
340 ELEMENTS OF ALGEBRA.
but change the signs as above directed, and we have the
equation,
4z 8 -f20# 2 -)-l 1^35=0,
and the sum of the coefficients is now zero, and consequently
unity is one root of this equation, and therefore 1 is a root
of the first mentioned equation.
Hence, in seeking to solve any equation of this kind, if the
sum of the coefficients is not zero, we may change the sign of
the second, and each alternate term, and if the sum is then
zero, 1 may be taken as one root of the equation ; and then
we may depress the equation one degree by dividing by (x-\-\).
(Art. 200.) Another circumstance to be observed in rela-
tion to coefficients, is that of their recurrence, producing
what is called
RECURRING EQUATIONS.
In a recurring equation the coefficients of those terms which are
equally distant from the extreme terms, are numerically equal.
Thus x 5 3x*+7x 3 +7x 2 3a?-fl=0, (1)
are all recurring equations of an odd degree.
If we substitute 1 for x, in each of the equations (1) and
(2), they will be verified; and -f- 1 put for x will verify (3)
and (4). Therefore, any equation of an odd degree has either
1 or _j_ i for one of its roots, and the equation itself is redu-
cible to an even degree, by dividing by (x-\-\ ) or by ( x ] ).
Minus one is a roof, when the like coefficients have the same sign;
and plus one is a root when the like coefficients have opposite signs.
(Art. 201.) Recurring equations, of an even degree, in ivhich
the like coefficients have opposite signs, and ivhoss middle term is
ivanting, are divisible by (x 2 1), and therefore +1, and 1,
are both roots of every such equation.
THEORY OF EQUATIONS SUM OF COEFFICIENTS. 341
The following is an equation of this description :
which can be put in the following form :
(x*l)+8x(x*l)-\-llx 2 (x 2 l)=Q.
and this is obviously divisible by (x 2 1). Whence we may
place x 2 1=0, and x=l, or 1.
And thus the principle enunciated in this article is estab-
lished. Dividing the equation by (x 2 1), will reduce it two
degrees lower, and leave it an equation of an even degree.
(Art. 202.) Every recurring equation of an even degree, above
the second, can be reduced to an equation of half that degree, by the
following artifice.
We take the following equation for an example :
#6_7 a .5_|_8 2; 4 9z 3 -far 2 72+1=0. (1)
Divide each term by the square root of the highest power of
the unknown quantity. In this case divide by x 3 , and we have
. (2)
Which can be put in the following form :
Now assume, /Vfi'Wz. Then, fx 2 -\-^\=z 2 2.
And Sx 3 -\-- )=z 3 3z. Whence (3) becomes
v as 3 /
( 2 3_3 2 )7(s 2 2)4-8^ 9=0, (4)
an equation of the third degree.
Thus (1), an equation of the 6th degree is reduced to (4), an
equation of half that degree ; and this process will reduce any
other recurring equation of an even degree to another of half
that degree.
(Art. 203.) A recurring equation of the 4th degree can be
reduced to a quadratic by the following rule.
Divide by the coefficient of x 4 , and transpose the term containing
-x 1 . Add to each member +2#-. Use the plus sign when the signs
342 ELEMENTS OF ALGEBRA.
of the second and fourth term are alike, and use the minus sign
when tJtey are unlike. Also add to each member the square of
half the term containing x. Extract the square root of each mem-
ber, and the result is an equation of the second degree.
EXAMPLES.
a a
Add
.x, a quadratic.
We are indebted to Professor Stevens, of Greenmount Col-
lege, Indiana, for this article. It is original with him.
BINOMIAL EQUATIONS.
(Art. 204.) Binomial equations have but two terms, such as
a; 1=0, z 2 1=0, z 3 1=0, &c.
Or, such as #a=0, #"a n =0, &c., &c.
Every binominal equation is also a recurring equation, for
the coefficient certainly recurs.
A binomial equation has apparently but one root, but a full
investigation will discover as many roots as there are units
in the exponent of the unknown quantity, as is explained by an
example in (Art. 163). We here give one more practical
illustration.
The equation x 5 1=0, is a binomial equation of an odd
degree, and it is also a recurring equation.
It is also divisible by (x 1), having apparently but one
root, -f-1. The division produces
THEORY OF EQUATIONS SUM OF COEFFICIENTS. 343
O. (1)
a recurring equation of the 4th degree.
Dividing again by x 2 , we obtain
l=0. (2)
Assuming fx+-\=P. Then z 2 +-L=P 2 2,
\ x / x
And (2) becomes P 2 +P 1=0, a quadratic.
Again: Operating under (Art. 203), we obtain
Add *
x
Square root,
Or x 2 4-1(1^:^/5)2:= 1, a quadratic.
Whence the five roots in (x 5 1)=0, or the five-fifths roots
of unity are
x=\, or x=
or s=i(,y5 1 V
or x=
or a?= 1(^5+1 +V 10+2 V 6 )-
EXAMPLES.
1. Find the roots of the equation
Ans. Three of its roots, are each equal 1, and f. _=h- 75 j
2.
3.
See (Art. 201). Jws. a?=l, or
4. a; 4_|, r 3_j_22; 2 --fa;+l=0.
r=2, or , or db^/
344 ELEMENTS OF ALGEBRA.
5. 5x<+8x*+9x*+8x+5=Q.
(Art. 203) reduces this to the following quadratic :
6. 4* 24* 5 -f-57z 4 73z 3 -f57# 2 24*+4=0.
Roots, 2 ,^l^,
7. 4# 4 4-3* 3 8z 2 3^+4=0.
Ans. #=rhl, or ~
o
8.
;=, or
Some one, or more of the foregoing principles will apply to
the solution of the following examples.
9. a? 4 2s 3 7a a 824-16=0.
Ans. ar=l, or 4, or ^f
10. x 4 +2a: 3 3^ 2 _4a:+4=0.
^4ws. ar=l, 1, or 2, 2.
11. x* 2x 3 25a: 2 4-26a:+120=0.
Ans. #=3 or 5, or 2, or 4.
12. x*
3.
. #=0, or 1, or
13.
. .r=0, or 4, or db^/ 8.
14. o; 3 +5a: 2 -f3^ 9=0.
^Iw5. =!, or 3, or 3.
15. a^Gz 2 7# 60=0.
^4tts. ar=3, 4, or 5.
16. 2 3 +8.r'-fl7;r-f-10=0.
Ans. x= 1, or 2, or 5.
17. x 3 29# 3 -|-198# 360=0.
Ans. x=3, 6, or 20.
THEORY OF EQUATIONS SUM OF COEFFICIENTS. 345
27=0.
Ans. #=-, -J, or 27.
37=0.
Ans. x=%, -J, or 37.
20. 4x*-{-3x 3 -\-ttx 2 3.r-[-4=0.
Ans. x=aj+a 2 and2a=(dr N / 247).
21. *5 13z 4 +67z 3 171* 2 +216z 108=0.
(See Art. 198.) Ans. 3, 3, 3, or 2, 2.
REMARK. This treatise has hitherto omitted one process
of elimination, which Bourdon, and some others, have intro-
duced, and which, some teachers regard as very scientific and
interesting. We cannot quite agree with this class of teachers,
for whatever is most simple, and most practical, we regard as
most scientific, and this method is neither very simple, nor
practical. Still, in a work like this, it is worthy of a passing
investigation like the following.
GENERAL METHOD OF ELIMINATION AMONG EQUATIONS
ABOVE THE FIRST DEGREE.
(Art. 205.) Suppose we have two equations, each contain-
ing x and y, like the following :
x+y-6=0. (1)) Ans x=4, or 2.
#3+2/3 72=0. (2)j ' y=2, or 4.
To make the illustration clear, we give the values of # and
y. Now by the theory of equations (Art. 156), if x=4, equa-
tion (1) is divisible by (# 4), without a remainder.
That is, whatever the remainder appears to be, it is in value
zero.
Dividing (x-\-y 6) by (x 4) the quotient is 1, and the
apparent remainder is (y 2), but this must be zero ; therefore
y=2, corresponding to x=4.
But x=4, corresponds to both the given equations ; there-
fore equation (2) is divisible by (x 4), and (x 4) is- a
common measure to the two equations (1) and (2).
316 ELEMENTS OF ALGEBRA.
Now, therefore, if we take equations (1) and (2), and ope-
rate with them for common measure, as taught in (Art. 27),
the last remainder must be zero ; and in fact we shall show,
that each and every remainder, throughout the whole operation, must
be zero. And if we take care that only one letter shall appear
in the last remainder that is, operate until one letter only
appears in the remainder that remainder can be put equal to
zero, and the desired elimination is effected.
OPERATION".
(A)
y 3 72 (B) remains.
Observe that the divisor is in value zero, and it being mul-
tiplied by x 2 , the product (A) must be zero in value ; and that
product, subtracted from the dividend (zero), the remainder
(B) must be zero.
Thus we might establish the fact, that each remainder (and
in fact each product also), throughout the operation, must be
zero.
Continuing the operation we obtain
x*y xy 2 -\-6x 2 +6y* 72. 2d remainder.
Or, xy(x-\-y)-\-Q(x 2 -\-y 2 ) 72 =2d remainder.
But we learn by equation (1) that (x-{-y)=6, therefore
we can divide this second remainder by 6, and we have
Qx
6x 2xy-\-y 2 12 3d remainder.
Or, (6 2y)z-f-y 2 12 3d remainder.
(6 2y )#-]-( 6 2y)(y 6)
y 2 12 (6 2y)(y 6). 4th remainder.
This last remainder does not contain x, therefore that letter
is completely eliminated, and the remainder put equal to zero
will give the values of y.-
THEORY OF EQUATIONS SUM OF COEFFICIENTS. 347
N. B. When this method is clearly analyzed, it will be
found to be nothing more, than elimination by addition and
subtraction.
Here we have two equations, (1) and (2), each equal to
zero, and we multiply one of them, and subtract the product
from the other. The remainder is an equation equal to zero,
and if its terms have a common divisor, we can divide the
equation by it. Hence we have no new principle, nothing to
gain by keeping these forms in view. The common method
gives us much more freedom of action.
We give one more example.
Given, ( f *+ f *)(*- y )-1220=0> d and
And, x 2 -\-y 2 -^-x y 132=0[
Assume, % 2 -}-y 2 =P, and x y=Q
Then, P+Q 132=0,
And, PQ 1220=0 (Q,
2 -f 1320 1220=0, a quadratic.
We extract the following from Prof. Perkin's Algebra.
Indeed, we have observed it in several other works, and it is
given as an appropriate example to illustrate this method of
elimination.
We give it to show the utter inutility of this mode of ope-
ration, in a practical point of view.
Given, a.*+ X y+y*l=0, (1)
And, .^+y 3 =0, (2)
to find a single equation in terms of y.
FIRST OPERATION.
1) x*+y*(x y
x 2 y y 2 x -f- yy
y=first remainder.
348 ELEMENTS OF ALGEBRA.
SECOND OPERATION.
/ 2 ' !(* (2y 3 2y)
_(2t/
second remainder.
Whence, 4y 6 6y 4 -|-3y 2 1=0, is the equation sought.
As the sum of these coefficients equal zero, therefore, y=l,
and if it were not for this lucky circumstance the values of x
and y, in (1) and (2) would still be far away.
We hare observed, that all remainders in these operations,
must be zero ; therefore, our first remainder must be zero, and
xy 2y 3 .
Or, x=(l 2y 2 )y. This value of x put in (2) and we
have (l-f2y 2 ) 3 2/ 3 -f?/ 3 =0. Or, (1 22/ 2 ) 3 +l=0.
Expanding and dividing by 2, produces
4y6_6 2/ 4_|_3 y 2_ 1 _o. The same as before.
But if the object is to solve equations (1) and (2), and find
the values of x and y, the common method is incomparably
the best.
s a -H*y+y a = i. (1)
Equation (2) is divisible by (#-f-y), and the other factor is
x 2 xy-\-y 2 . Therefore we have
x+y=o. (2)
Or, xtxy+yV^Q, (3)
Subtracting (3) from (1) and we dbtain
2ary=l, or, xy=^.
This last added to (1), and subtracted from (3) will produce
And,
A.lso, x= 1, andy=l.
APPENDIX.
DESIGNED TO SHOTf THE UTILITY OF ALGEBRA IN PHILOSOPHICAL
INVESTIGATIONS.
INEQUALITY.
Expressions of inequality sometimes occur in the mathematics, but we
can always make equations of them by adding a symbol to the smaller
quantity. Then we can reduce the expressions the same as equations.
without danger of confusion.
EXAMPLES.
1. Find the limit of the value of x in the inequation
Let h represent the difference between these two expressions,
en,
Whence,
then, ..... lx 8?=?f+5+A, is a perfect equation
o o
or
That is, x is greater than the number 2.
2. The double of a number diminished by 5 is greater than 25, and triple of
the number diminished by 1 is less than the double increased by 13. What
numbers will satisfy the conditions?
Let ar= the number, and h and k the difference of the expressions.
Then, ...... 2;c 5=25-^. (1)
3x 7-t-& 2*4-13. (2)
From .... (1) *=15-H/i. From (2) *=20 k.
That is, x must be more than 15 and less than 20, consequently, 16, 17,
18 and 19 will each answer the conditions.
3. The sum of two numbers is 32, and if the greater be divided by the less,
the quotient will be less than 5, but greater than 2. What are the numbers?
Let x= the greater, y= the less, then ar-f-y=z32. (1)
And .... f=5-nft. (2) f=3+Jfc. (3)
y y
From (2) x=5yhy. This put in (1) gives 6yhy=32.
on
Therefore, y=_. From (3) x=2y+ky; this, put in (1),
32
gives JF=f
From these last equations we perceive that 32 divided by less than 6 will
give the limit of y in one direction, and divided by more than 3 will give
its limit in the opposite direction; that is, y must be more than 5 and less
than 10 ; consequently x must be more than 22 and less than 27.
349
350 - APPENDIX.
' DIFFERENTIAL METHOD OF SERIES.
A series purely of the arithmetical order has been fully treated, as also
those of the geometrical order; but a series in general exist, when each of
the terms is derived from one or more of the preceding terms, according to
some definite law. (Art. 141 )
Therefore, a variety of series may exist, neither arithmetical nor geo-
metrical, and these can be investigated by means of finite differences; and
the word differential at the head of these remarks, is not identical with the
same word applied to the differential calculus. For example,
1, 5, 15, 35, 70, 126,
and so on, is a regular series, and if we wished to find its 10th, 12th, or
nth terms, it would be too unscientific and tedious to lind it through the
succession of terms ; we must, therefore, investigate for general principles.
The series is 1, 5, 15, 35, 70, 126, 210,
1st order of diff. 4, 10, 20, 35, 56, 84,
2d 6, 10, 15, 21, 28,
3d " " " 4, 5, 6, 7,
4th " " " 1, 1, 1,
5th " " 0, 0,
Here it will be observed that the first term of the series is 1. The first
term of the first order of differences is 4. The first term of the 2d order
of differences is 6. The first term of the 3d order of differences is 4. The
first term of the 4th order is 1 j and lastly, the first term and all the terms
of the 5th order are 0.
To obtain general principles, however, we must take general notation
as follows : (The symbol a 4 is read a sub. 4, and so on for other like
symbols.)
Let ..... a,, a 2 , a 3 , <* 4 , a^ a a , be a series.
ij, 6 2 , 6 3 , & 4 , 6 a , 1st differences
c i C 2 c s' C 4' 2< * differences.
d lt d 2 , cZ 3 , 3d differences.
&c. &c.
By this notation we perceive that d^ would represent the 4th term of
the third order of differences, d being the third letter of the alphabet.
From the foregoing notation, we readily derive the following equations :
and so on, and so on, and so on.
By transposition (writing the letters in alphabetical order), we have
and so on, and so on, and so on*
In the 2d equation of the first column we will substitute the values of
an I 6^, then we shall have fl 3 =ff 1 4~'W>i~h c i t (1)
DIFFERENTIAL SERIES. 351
By the same process we find
^i-H'i-M- (2)
In the 3d equation of the first column we find
Substituting the values of a 3 , 6 3 from equations (1) and (2) gives us
a 4 =a 1 -f36 1 -|-3c 1 -t-rf 1 . (3)
By the same process,
6 4= 6 ,-i- 3c rt-3<VK ( 4 )
Also by referring back we find
a 5= a 4-H4-
Substituting the values of a 4 and b 4 from (3) and (4), we find
(5)
By inspecting equations (1), (3) and (5), we perceive that the coefficients
in theSd members correspond to binomial coefficients, and if we should con-
tinue the process to any degree, they would still correspond, and the letters
represent the first terms of the several orders of differences.
Therefore, in general,
And . . 6 n+l =H-nc-f-n . ^~d-\- n . "f . ?=-V}- &c.
We drop the sub. ones in the second members.
The series terminates with that order of difference which becomes con
etant ; for the succeeding order becomes zero.
EXAMPLES.
1. Required the 12th term of the series.
1, 5, 15, 35, 70, 126,
4, 10, 20, 35, 56,
6, 10, 15, 21,
4, 5, 6..
1, 1.
Here a=l, 6=4, c==6, rf=4, =1, /=0, n=ll.
Whence, a 12 =l-{-ll . 4+11 . . 6+11 . {? . L . 4-J-ll . 5 |.f
Or, ... a l2 =l-f-44-j-330-f-660-t-330=1365, Ans
2. Required the 15th term of the series
1, 4, 10, 20, 35, &c. Ans. 670
3 Required the 20th term of the series
6, 10, 15, 21, Ac. Ans. 253.
4. Required the nth term of the series
1, 3, 6, 10, Ac. AntJ
5. Required the nth term of the series
2, 6, 12, 20, 30, <fe. Ans. n"-\-n.
N. B. To solve the 4th and 5th examples, write (n 1) for n in the
formula.
352 APPENDIX
6. Find the 20th term of the series 1, 8, 27, 64, 125, Ac. Ans. 8000.
7. Required the 50th term of the series
1, 3, 6, 10, 15, Ac. Ans. 1275.
Compare examples 4 and 7.
Our investigation thus far has been limited to finding a particular term
of a series.
We now propose to find the sum of any proposed number of consecutive
terms.
Let . . . Oj, a^, a 3 , a 4 , a., . . . a n , <fec., represent a series.
With the terms of this series form a new series commencing with 0,
for its first term ; adding the succeeding term of the proposed series to the
preceding term of the new series, thus,
0, (0+flj), (0+ ar f-a 2 ), (0+ flr f-a 2 -{-a 3 ), and so on.
Here it is obvious that the (n-{-l) th term of this new series is the sum
of n terms of the proposed series. Therefore, we will find the (/i-|-l/ lh
term of the new series by the formula, and that will be n terms of the
proposed series as required.
EXAMPLES.
1. Find the sum of 10 terms of the series 3, 5, 7, 9, 11, Ac.
The new series to be formed is
0, (O-j-3), (0-f3-l-5), (0-f3-}-5-|-7), (0-j-3-{-5-f 7+9), Ac
That is, .... 0, 3, 8, 15, 24, Ac.
We must now find the llth term of this series, which will be the sum
of 10 terms of the proposed series as required. Ans. 120.
2. Find the sum of 20 terms : also the sum of n terms of the series
1, 3, 6, 10, 15, Ac.
Ans. 20 terms =1540, n
3. Find n terms of the series 2, 6, 12, 20, 30, Ac.
Ans.
4. Find the sum of n terms of the series of cubes
P, 23, 33, 43, & c . Ans.
5. Find the sum of 10 terms, also the sum of n terms of the series
1, 4, 9, 16, 25, 36, Ac.
N. B. The new series is 0, 1, 5, 14, 30, 55, 91, &c
Ans, Sum of 10 terms =385. Sum of n terms =L 2re2 + 3ra + 1 ) ra .
6
The series of numbers in example 2 are called triangular numbers, be-
cause they may be represented by points forming equilateral triangles, thus:
t *
36 10
DIFF1 RENTIAL SERIES. 353
In like ma.iner the square numbers in example 5 can be represented by
points, thus:
In forts and armories shot and shells are piled on triangular, square, or
oblong bases.
When the base is a triangle, and the pile complete, the number of balls
in the pile is expressed by the answer to example 2, in which re represents
the number of layers in the pile.
When the base is a square, and the pile complete, the number of shot in
the pile is expressed by the answer to example 5, in which n represents the
number of layers in the pile.
When the pile is an oblong and complete, the upper layer is a single row.
The next larger has two rows and one more ball in each row. The third
layer consists of three rows, and each row one more ball than the preceding,
Now if the top layer consists of ..... m-j-1 balls,
The 2d must consist of ... 2(m+2) or 2ro+4 "
The 3d must consist of ... 3(m+3) or 3m+9 "
The 4th must consist of . . . 4(m+4) or 4m-4-l6 "
The 5th must consist of ... 5(m+5) or 5m+25 " <fcc.
The number of balls in the whole pile must then consist of the sum of
the series (m+1), (2w+4), (3m-J-9), <fec., to n terms n representing the
number of layers in any pile.
We therefore require the sum of this series to n terms.
The new series to be formed is
0, (0+m+l), (0+3771+5), (0+6/71+14), <fcc.
1st diff. (m+1), (2771+4), (3m+9), (4771+16), &c.
2d diff. (>H-3)> (+5), (wi+7).
3d diff. 2 2
4th diff.
Now by the formula, the (n+l) th term of this series is
a n+1 =0+n(m+l)+n . ^.(w+^+n . n ~ l . n -^X2.
The right hand member of this equation is the general formula for an
oblong p'le of shot or shells, in which (wi+1) represents the number in the
top row, and n the number of layers.
We have previously found that rciw+I) (M+^) re p resen t s a triangular
pyramidical pile and, n ( n "f" w +_) represents a square pyramidical pile
6
EXAMPLES.
1 . How many shot are in a triangular pyramidical pile consisting of 1 8
layers? Ans. 1140.
2. How many shot or shells in a square pile consisting of 15 layers ?
Ans. 1240.
3. The top row of an oblong pile of shot or shells consists of 31 balls, and
the number of layers is 30. How many balls are there in the whole pile ?
30 An
354 APPENDIX.
4. There is an oblong pile of balls consisting of 20 layers, and 644*
balls ; how many balls are in its base ? Ans. 740.
5. A square pile of shells consists of 12 layers, the upper layer has 8
shells on a side ; how many shells are in the pile ? Ans. 2330.
6. The number of balls in a triangular pile is to the number in a square
pile (having the same number of layers) as 5 to 9 j required the number
in each pile. Ans. 455 and 819.
Our object is now to express the several orders of differences by the terms
of the series,
a i a , a 3 4 fl 5 6 &C '
1st order, (a aj, (a 3 2 ), (a 3 ), (a, a 4 ) &c.
2d order, (a 3 2^+^), (a 2 3 +a 2 ), (a 2a 4 +a 3 ).
3d order, ( 4 3a 3 +3a 2 a,), (a^^a.^aa )
4th order, (a. 4a^- r -6a 3 ia-j-Cj).
Here it can be observed that the coefficients correspond to those of the powers
of a binomial.
Also observe that to compose thejirst term of the 2d order of differences,
we must use the first three terms of the series. To compose the first term
of the 3d order, we must use the first four terms of the series. To com-
pose the first term of the 4th order of differences, we must use the first
five terms of the series : and in general, to compose the first term of the
n th order of differences, we must use the first (n-f-1) terms of the series.
Observing these facts, the first term of the 5th order of differences of the
preceding general series, must be expressed thus :
o 6 5a.+lOa 4 10a 3 +5 2 aj.
If the first term of any order vanishes, or becomes very small, the
expression for it may be put equal to zero, and any term of the series com-
prised in it, can be found, provided the other terms are given.
For example, suppose the 4th order of differences in a series becomes 0,
then
Now suppose the 3d term of the series lost or unknown, the others
being given, it is found thus :
__
The series 1, 8, a, G4, 125, has lost its third term, its 4th order of dif-
ference vanishes ; what is that lost term ?
Ans.
6 6
One of the most important applications of this calculus of finite differ-
ences is, that of inserting between the terms of a series some new term
or terms subject to the same law. This is called
INTERPOLATION.
We have already seen that the general equation is
n 1. , _ n 1 n 2
T
This series will apply to any term beyond a. If we required the 3d
INTERPOLATION.
355
term beyond a, we must put n=3. If the 4th beyond a, we put n=4,
and so on.
If part of a term beyond a, we put w= to that fractional part.
EXAMPLE .
Given the logarithms of 102, 103, 104, and 105, to find the logarithm
of 103.6, which is the 1.6th term beyond a.
NO.
LOGARI'MS.
1ST DIFF.
2o DIFF.
3D DIFF.
102
103
104
105
2. 0086002
2. 0128372
2. 0170333
2. 0211893
.0042370
.0041961
.0041560
.0000409
.0000401
.0000008
a=2.0086002, 6= .0042370, c .0000409, d .0000008, n=1.6
a 2. 0086002
nb +0. 0067792
1
c 0. 0000196
Ld ....... +0. 00000005
log. 103.6 2 01535985
In most cases the 3d difference may be omitted : in this case it only in-
fluenced the 8th decimal place. If we were to require the logarithm of
102.3, then n=0.3. If 104.7, then n=2.7, and so on.
Interpolation is much used in astronomy, as the following example will show:
At noon (Greenwich time), May 10th, 1846, the moon's declination was
3 27' 17.5" north; at midnight following it was 5 46' 38" north; on
the llth, at noon, 7 38' 57" north, and the following midnight 10 2'
31.7" north. What was the declination on the 10th of May at 3h P. M?
Here a=3 27' 17.5", 6=2 19' 20.5", c= T 1.5", d 1' 43.2", and
the interval of 12 hours must be considered as the unit of time : 3 hours
is, therefore, J of this unit. Hence, n=0.25, and the required result must
be the sum of the following series :
The 1st term a =3 27' 17.5".
2d termn6 34' 50.1"
3d term 39.5".
4th term 5.6".'
Sum~42' 41.5"T Ans.
To show tne utility and application of this formula, we give the following
EXAMPLE S.
Given the logarithmic sine of 5, 5 12', 5 24', 5 36', 5 48' and 6, ft
find the sine of any or all intermediate arcs.
a.
Sine 5 0'=8. 9402360
Sine 50 12'=8. 9572843
Sine524'=8 9736280
Sine5036'=8 9893737
Sine548'=9 0045634
Sine GO 0'=9 0192346
+6.
1st diff.
.0169883
.0163437
.0157457
.0151897
.0146712
2d diff.
6446
5980
5560
5185
-M.
3d diff.
466
420
375
4th diff
46
45
356
APPENDIX.
In applying the formula to this example, 12' must be taken as the unit
of space between the terms. Then if we require the sine of 5 3', we
must put n=y 3 .2=2, and the formula gives
Sine53'=8.9402960- r -2(.Ol69883)-H . tz?
4G6=8.9402960+.0042471+604.3-f 25.6=8.9446061.
Sine54'=8.9402960+i(.0169883)-H . tz?
456=8.9402960+.0056628+716+29=8.9460333.
For the sine 5 13', we put n= T 1 2 and
fin. 50 13'=8.9572843+ T 1 3 (.0163437)+ T 1 2 . j
jti(420)=8.9572843+.0013619+S29+13=8.9586704.
O
To find the log. sine of 5 14' 13", we could put n=?I?=!??. '
12 720
The work of finding sines, cosines, and tangents, &c., is already done;
and, therefore, in respect to this application, the formula for interpolation
has comparatively lost its importance ; but with the practical astronomer,
this formula is of the greatest importance, and must forever be in use.
We give the following
EXAMPLE.
For the English Nautical Almanac the moon's declination is required
for every hour of Greenwich time; but to compute it directly from the lunar
tables and trigonometry, would present an apalling amount of labor, and to
surmount this difficulty, the declination is computed from the lunar tables
for each noon and midnight of Greenwich time ; and the declination in-
serted for the intermediate hours by interpolation.
Opening the English Nautical Almanac for 1854, at page 91, I find the
moon's declination as follows :
3) dec. N. +6
May 23, mean moon, 6 13' 14.4"
May 23, midnight, 8 55' 6.4" 2 41' 52.0"
May 24, noon, 11 30' 26.2" 2 35' 19.8"
May 24, midnight, 13 57' 44.4" 2 27' 16.2"
May 25, noon, 16 15' 33.6" 2 17' 49.2"
6' 32.2"
8' 3.6"
9' 27.0"
1' 31.4"
1' 23.4'
From the above data we require the moon's declination at the commence-
ment of every hour between noon and noon of May 23d and 24th.
For the first hour, or 1, afternoon, we must put n=J .
The D's dec. at 1 P. M. =6 13' 14.4''+^ (2 41' 52.0")+T2 . itzl X
14.9" 2.2"r=6 26' 56.4" North.
SPECIFIC GRAVITY.
357
The following are the results taken from the Nautical Almanac which
serve as answers to twenty-two examples in the application oi this formu-la
Hours.
f)Dec.
Hours.
Dec.
1
6 C
Ofi'
565"
13
9
8' 19.3"
2
6
40
365
14
9
21 29.4
3
6
143
15
9
34 36.6
4
7
7
498
16
9
47 41.0
5
7
91
23.0
17
10
42.5
6
7
'M
53.9
18
10
13 40 9
7
7
48
22.3
19
10
26 36.4
8 . . .
8
1
48.3
20
10
39 28.7
9 .
8
15
11.7
21
10
52 17.9
10
8
32.6
22
11
5 3.9
11
8
-11
50.8
23
11
17 46.7
12
8
rr
64
24
.. 11
30 26.2
SPECIFIC GEAVITY.
Gravity is weight. Specific gravity is the specified weight of one body,
compared with the specified weight of another body (of the same bulk), taken
as a standard.
Pure water, at the common temperature of 60 Fahrenheit, is the standard
for solids and liquids ; common air is the standard for gases.
Water will buoy up its own weight. If a body is lighter than water, it
will float ; if heavier than water, it will sink in water.
If a body weighs 16 pounds, in air, and when suspended in water weighs
only 14 pounds, it is clear that its bulk of water weighs 2 pounds ; and the
body is 8 times heavier than water ; therefore the specific gravity of this body
is 8, water being 1.
If the specific gravity of a body is n, it means that it is n times heavier
than its bulk of water. Therefore
If we divide the weight of any body by its specific gravity, the quotient
will be, the weight of its bulk of water.
On this fact alone we may resolve all questions pertaining to specific gravity
EXAMPLES.
1. Two bodies, whose weights were A and B, and specific gravities a and b,
were put together in such proportions as to make the specific gravity of the
compound mass c. What proportions of A and B were taken?
A quantity of water, equal in bulk to A, must weigh
A quantity "
A+B
c
A quantity of water, equal in bulk to (A-^-B), will weigh
A B A4-B
Therefore, - ) = ! ; Or, bcA-\-acB=abA-\-abB ;
Or, b(o a)A=a(b c}B.
3-5?; AITENDIX.
Hence the quantities of each must be reciprocal to these coefficients; or if we
a /b c\
take one, or unity of B, we must take j ( - J units of A.
U \,C"~~flyX
2. Hiero, king ef Sicily, sent gold to his jeweler to make him a crown ; he
afterwards suspected that the jeweler had retained a portion of the gold, and
substituted the same weight of silver, and he employed Archiwedes to ascer-
tain the fact, who, after due reflection, hit upon the expedient of specific gravity
He found, by accurately weighing the bodies both in and out of water, that
the specific gravity of gold was 19, of silver 10.5, and of the crown 16.5.
Fromthesedata he found what portion of the king's gold was purloined. Re-
peat the process.
The preceding problem is the abstract of this, in which A may represent the
weight of the gold in the crown, B the weight of the silver, and (./i-J-fi) the
weight of the crown ; a=19, 6=10$, c=16.
Then if we take J5=l, one pound, one ounce, or any unity of weight, of
silver, the comparative weight of the gold will be expressed by -j( - J.
That is, for every ounce of silver in the crown, there were 4 ^ 2 j ounces of
gold. If clearer to the pupil, he may resolve this problem as an original one,
without substituting from the abstract problem.
3. I wish to obtain the specific gravity of a piece of wood that weighs 10
pounds ; and as it will float on water, I attach 21 pounds of copper to it, of a
specific gravity of 9. The whole mass, 31 pounds, when weighed in water,
weighs only 4 pounds ; hence 27 pounds of the 31 were buoyed up by the
water ; or we may say, the same bulk of water weighed 27 pounds. Require-1
the specific gravity of the wood.
Let s represent the specific gravity of the wood.
10
Then = the weight of the same bulk of water.
21
And = the weight of water of the same bulk as the copper.
y
10 7 30
Hence, ........ -{--=27, Or, s==^=.405, nearly ,
4. Granite rock has a specific gravity of 3. A piece that weighs 30 ounces,
being weighed in a fluid, was found to weigh only 21.5 ounces. What waa
the specific gravity of that fluid ?
The weight of the fluid, of the same bulk as the piece of granite, was evi-
dently 8.5 ounces. Let a represent its specific gravity.
8.5 80
Then = the weight of the same bulk of water; also - Q =10= tlio
3 o
weight of the same bulk of water.
8.5
Hence =10, Or, s=.85 Ans. t which indicates impure alcohol
MAXIMA AND MINIMA.
5. The specific gravity of pure alcohol is .797 ; a quantity is offered of the
specific gravity of .85, what Proportion of water does it contain?
Let A= the pure alcohol, and W = the water.
Then
Ans. The resolution of this equation shows 1 portion of water to 2.255-j-
portions of alcohol.
6. There is a block of marble, in the walls of Balbeck, 63 feet long, 12 wide,
and 12 high. What is the weight of it in tons, the specific gravity of marble
being 2.7 and a cubic foot of water 62^ pounds, Ans. 683 y\ tons.
7. The specific gravity of dry oak is 0.925 ; what, then, is the weight of a
dry oak log, 20 feet in length, 3 feet broad, and 2 feet deep ? Ans. 867 if Ibs.
We may now change the subject, to make a little examination into maxima
and minima. For this purpose, let us examine Problem 2 (Art. 114).
1. Divide 20 into two such parts that their product shall be 140. It may
be impossible to fulfil this requisition, therefore we will change it as follows :
Divide 20 into two such parts, that their product will be the greatest possible.
Let x-\-y one part, and x y= the other part.
Then 2.r=20, and #=10, and the product, x 2 y*, is evidently the
greatest possible when y=Q. Hence the two parts are equal, and the greatest
product is 100, or the square of one half the given number.
2. Given the base, m, and the perpendicular, n, of a plane triangle, to find
the greatest possible rectangle that can be inscribed in the triangle.*
Let ABC be the triangle, BC=m, A
AF=n, AD=x, and AE=x', De be a
very small distance, so that x' is but insen-
sibly greater than x.
As Z), comparatively, is not far from the
vertex, it is visible, that the rectangle
a'b'c'd' is greater than the rectangle abed.
If we conceive the upper side of the
rectangle to pass through D 1 , in place of
D, and we represent AD' by x, and At
by or', it is visible that the rectangle
e'fg'h' is less than the rectangle efg h.
If we subtract the rectangle abed from the rectangle a'b'c'd', we shall have
a positive remainder.
If we subtract the rectangle efg h from the rectangle t'f'g'h', we shall have
a negative remainder.
* We do not introduce this problem to show its solution ; it bt.ongs to the calculus,
and, in its place, is extremely simple ; we introduce it to show a p-inciple of reasoning
extensively used in the higher mathematics, and, perchance our illustration may uj]
a pupil in his progress in the calculus.
A A
360 APPENDIX.
The rectangle abed cannot be the greatest possible, so long as we can have
a positive remainder by subtracting it from the next consecutive rectangle
immediately below.
After we pass the point on the line AF where the greatest possible rectangle
comes in, the next consecutive rectangle immediately below, will become less ;
and by subtracting the upper from it, the difference will be negative.
Hence, when abed becomes the greatest possible rectangle, the difference
between it and its next consecutive rectangle can be neither plus nor minus
but must be zero.
Therefore, it is manifest, that if we obtain two algebraical expressions for tho
two rectangles abed and a'b'c'd', and put their difference equal to 0, a resolu
lion of the equation will point out the position and magnitude of the maximum
rectangle required.
Put the line ab=y, and a'b'=y'. As AD=x and AE=x l , DF=n x
Wid EF=n x f . The rectangle abcd=y(n x), and a'b'dd'=y'(n a:/).
From the consideration just given, the maximum must give
y'(n -a;') y(n #)=0.
mx
By proportional triangles, we have x : y : : n : m or. y =-
n
By a like proportion, we have y=x'.
Put these values of y and yi in the above equation, and, dividing by ,
we have x'(n x'}=x(n x} ;
Or, x 2 x'* =n(xx').
By division, x -{- x' =n
As x' is but insensibly greater than x, 2x=n ; which shows that AD is one
half AF, and the greatest rectangle must have just half the altitude of the triangle.
3. Required the greatest possible cylinder that can be cut from a right cone.
Conceive the triangle (of Prob. 2.) to show the vertical plane cut through
the vertex of the cone, and ab=y the diameter of the required cylinder. Then,
the end of the cylinder is .7854?/ 2 , and its solidity is .7854# 2 (n *). The next
consecutive cylinder is .7854i/'2(?i x"). Hence y' 2 (n a:')=;y 2 (n ;r).
By similar triangles x : y : : n : m, Or, y 2 = ^ and y' 2 = -z' 9 .
Hence, *' 2 (n a:')=r 2 (n *), 3r, x 3 *'3=n(z 2 z' 2 ) ;
Divide both members by (z z'), and a -f zz4-z' 2 ==n(z-|-z').
As x=x' infinitely near, 3z 2 =2wz, or, x=|n; which shows tnat the
altitude of the maximum cylinder is 5 the altitude of the cone.
In this way all problems pertaining to maxima and minima can be resolved ; bui
the notation and language of the calculus, in all its bearings, is preferable to this. We
had but a single object in view that of showing the power of algebra.
(r ,
/
XC 49563
961661
THE UNIVERSITY OF CALIFORNIA LIBRARY