Charles Hutton.

A course of mathematics for the use of academies, as well as private tuition .. online

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1. Given 2x — 5 + 16 = 21 ; to find x. Ans, x^5.

2. Given 6x — 16 = x + 6; to find x. Ans. «=4J.
6. Given 8— 3x+ 12=30- 5x+4; to find x. An8.x=s7.

4. Given x + ^x + Jx=:13 ; to find x. Ans. x:sl2.

5. Given 3x+ix+2=5x — 4; to find x. Ans. x=s4.

6. Given 4ax+ Ja — 2=ax — hx ; to find x.

^"•^ = 95+85-

7. Given |x — i* + i* = i ; to find x. Ans. x = ^.

8. Given v/(4+x)=4— ^'^ ; to find x. Ans. x == 2^.

X*

9. Given 4a + x = -r ; to deter, x. Ans. x = — 2a.

4a + aj

10. Given v'(4a*+ x*)= */(4ft* + x*) ; to find x.

6*— 4a*
Ans. x = v^— 2^i —

4a

11. Given ^x + ^(2a-f x) = ,^ , \ ; to find x.

^ . v^(2a+x)

Ans* X :s |a.

12. Given ~- + —^ = 26 ; to find x.

• l+2x 1 — ^2x

Ans. xs= ^ V — T" •
18. Given a+x= v'Caa+x v^(46*+x*)) ; to find x.

Ans. x=s — — a.

a



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aMPiiX savATioivB. 287



OV RKDUCmO DOITBIiB, TRIPLE, &C. BQUATIONS, COHTAIRIKO
TWO, THREE, OR MORE UNKNOWir QUANTITIES.

FR^BLEX I.

7b es^mttfiate Uoo Unknown Quantities; Or, 1o redmce ike
Ueo Simple EqutOione containing ihemj to a Single one*

RULE I.

Fom the value of one of the unknown letters, in terms of
the other quantities, in each of the equations, by the methods
already explained. Then put those two yalues equal to
each other for a new equation, with only one unknown quan-
tity in it, whose value is to be found as before.

Note. It is evident that we must first begin to find the
values of that letter which is easiest to be found in the two
proposed equations.

^ XXAXPLE8.

I. Given jJigJ^IJj ; tofindajftndy.

17— -Sir
In the l8t equat. trans. 3y and div. by 2, gives x =^ — 5—? ;

144-2y
In the 2d trans. 2y and div. by 6, gives x =s — — i ;

« . t^ 1 . . 14+2y 17-8y

Putting these two values equal, gives — r - ^ = — « 5

Then mult, by 2 and 5, gives 28 + 4j^ = 85 — 15y ;

Transposing 28 and Iby, gives 19y :s 57 ;

And dividing by 19, gives y = 3.

And hence as =4.

Or, effect the same by finding two values of y, thus

17— 2«
In the 1st equat. tr. 2x and div. by 8, gives y = — g— - ;

59—14

In the 2d tr. 2y and 14, and div. by 2, gires y = _ — •

^ . ^ , , . 6ap-14 17-4iv

rutting these two values equal, gives — 5- — « - ^ — |



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SS8 AI.OBBBA.

Mult b3r2andby 3,giye8 15« — 42«84 — 4x;
Tnmsp. 42 and 4ff, gives 19x = 76 ;
Dividing by 19, gived x = 4.
Hence y = S, as before.

2. Given J J* Jl2j«6| ' tofindscandy.

Ans. x&a + bf and y « ^€1 — H*
B. GKveii3x + y = 22,and3y + ar=:l8; tofindxandy.

Ans. X sa 6, and y « 4.

4. Given |}^ + |5 = Jj| ; tofindxandy.

Ans. X = 6) and y = 3.

«L«» 2x,3y 22 ,3x,2y 67 ^ .

5. Given-5. + f =yand- + J^=-; to find «

and y. Ans. x =s3t and y ss 4.

6. Givenx+2y = «, andx' — 4y^s=^; tofindxandy.

Ana. X i= — ;r — , and y =s — - — .

7. Given x — 2y = if, and xzyiiaih; to find x and y.

ad . hd

^*^s:::26'"^^y^5=56-

XVLS n.

Fnm the value of one of the unknown' letters, in only one
of the equations, as in the former rule ; and substitute this
value instead of that unknown quantity in the other equation,
and there will arise a new equation, with only one unknown
quantity, whose value is to be found as before.

Note. It is evident that it is best to begin first with that
letter whose value is easiest found in the given equations.

BXAXTLXS.

1. Given i &^ ZI 2!! = 14 C » to find x and y.
Tliis will admit of four ways of solution ; thus ; First,
in the 1st eq. trans. 3y and div. by 2, gives x « T^ •

Thii val. subs, for x in the 2d, gives — ^^^ — 2y» 14;
JifiiU. by 3, thM becomes 85 ~ 16y - 4y a 9S;



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tftXPLE BaVATlONS. 9|^

Transp. 15y and 4y and 28, gires 57 = IQj^ $
And dividing by 19, gives 8 = 3^.

2dly, in the 2d trans. 2y and div. by 5, gives « = 11 + ^ .

This subst. for x in the Ist, gives — ^^ + 8j^ = 17 ;

Mult, by 6, gives 28 + .4y + 15y = 85 ;
Transpo. 28, gives ]9y « 57 ;
And ^viding by 10, gives y = 3.

Then «« li~?? = 4, as before.



Sdly, in Ihe 1st trans. 2x and div. by 3, gives y as



17 — 2j?



.^ , 34 4^

This subst. for y in the 2d, gives 5x ~- » 14 ;

3

Multiplying by 3, gives 15jc — 34 + 4x » 42 ;
Transposing 34, gives 19x s= 76 ;
And dividing by 19, gives a; =s 4.

ir 17 — 2* ^ . ^

Hence y = — = 8, as before.

o

4thly, in the 2d tr. 2y and 14and div. by2,givesy= ^^— •

Tliis substituted in the Ist, gives 2x + ^^^""^^ -le 17 ;

Multiplying by 2, gives 19a: — 42 = 34 ;
Transposing 42, gives 19x = 76 ;
And dividing by 19, gives x =» 4 ;

•• 5x — 14

Hence y = — - — =: 3, as before.

2. Given 2r + Sy = 29, and 3x — 2y ^ 11 : to find x
and y. ^ Ans. a; == 7, and y = 6.

3. Given ^^jlj^^ll' tofindxandy.

Ans. X s: 8, andy =s 8.



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k



240 ALGBBHA.

4. Given | ^'^'^l/^ | ; to find x and y.

Ans. X = 6, and y ss 4.

5. 6iven| + 8y s= 21, and | + 3x = 29 ; to find x
and y. Ans. x = 9, and y=^S.

6. Given 10 — | =^ | + 4, and ?L=lif + | _ 2 =

3tf *— X

-2^ — — 1 ; to find X and y. Ana. x = 8, and y = 6.

7. Given x : y : : 4 : 3, and x* — y' = 37 ; to find x
and y. Ans. x = 4, and y = 3.



BULE m.

Lit the given equations be so multiplied, or divided, diz;c.
and by such numbers or quantities, as will make the terms
which contain oile of the unknown quantities the same in
both equations ; if they are not th^ same when first pro-
posed.

Then by adding or subtracting the equations, according
as the signs may require, there will result a new equation,
with only one unknown quantity, as before. That is, add
the two equations when the signs are unlike, but subtract
them when the signs are alike, to cancel that common
term.

iVbCe. To make two unequal terms become equal, as
abovoi multiply each term by the co-efficient of the other.



KXAMPLBS.

Given i 2x + ^ = 16 ( * ^ ''^^ * *°^ ^*

Here we may either make the two first terms, containing x,
equal, or the two 2d terms, containinff y, equal. To make
tb^ two first terms equal, we must multiply the 1st equation
by 2, and the 2d by 5 ; but to make the two 2d terms equaJ,
we must multiply the Ist equation by 5, and the 2d by 3 ;
as follows.



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8IMPLB BaVATIORS. 241

1. By making the two first tenns equal :

Multr the 1st equ. by 2,.gives lOx — Qy =48

Aad mult, the 2d by 5, gives lOx + 25^ =s 80

Subtr. the upper from the under, gives Sly = 62

And dividing by 31, gives y = 2.

Hence, from the 1st given equ. x = — r-^ = 3.

2. By making the two 2d terms equal :

Mult, the 1st equat. by 5, gives 25r — 15y == 45 ;
And mult, the 2d by 3, gives 6x + 15y == 48 ;
Adding these two, gives Six = 93 ;

And dividing by 31, gives x =s 3.

Bence, from the Ist equ. y = -— — = 2.



mSCELLANEOVS EXAMPLES.

1. Given ^^ + 6y = 21, and ?~^ + 6x = 23 ; to find
X and y. Ans. « = 4, and y = 3.

2. Given2f^ + 10 = 13,and??^J^ + 5==12; to find

4 »

X and y. Ans. x = 5, and y = 3.

3. Given?^y+*=10,and?f:=:?? + | = 14;tofind

5 4 So

X and y. Ans. a; = 8, and y = 4.

4. Given 3a;-i-4y = 38, and 4x — 3y ^ 9 ; ao find x and y.

Ans. X ^ 6, and y = 5.

PEOBLESI III.

To exterminate three or more Unknoum QMontUie^ ; Or^ to
reduce the simple E^uationSf containing them^ to a Single
one.

EULB.

This may be done by any of the three methods in the last
problem: viz.

1. ApTim the manner of the first rule in the last problem,
find the value of one of the unknown letters in each of the
given equations ; next put two of these values equal to each
other, and then one of these and a third value equal, and so
on for all the values of it ; which gives a new set of equations.

Vol. I. 32



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243 ALGB9EA.

with which the same process is to be repeated, and so on till
there is only one equation, to be reduced by the rules for a
single equation.

2. Or, as in the 2d rule of the same problem, find the value
of one of the unknown quantities in one of the equations only ;
then substitute this value instead of it in the other equations ;
which gives a new set of equations to be resolved as before,
by repeating the operation.

3. Or, as in the 3d rule, redutk the equations, by multi-
plying or dividing them, so as to make some of the terms to
agree : then, by adding or subtracting them, as the sigas
may require, one of the letters may be exterminated, dec. as
before.

EXAMPLES.



1. Given



< « 4- 2y +32 == 16 > ; to find ar, y, and sr.
^« + 3y+42 = 20



1. By the 1st method :
Transp. the terms containing y and z in each equa. gives
a: = 9 — y — s,
« = 16 — 2y — 32,
« = 21 — 3y — 4»;
Then putting the 1st and* 2d values equal, and the 2d and 3d
values equal, give

9— y— 2=16— 2jf — 32,
16 — 2y— 32=^21 — 3y — 42;
In the 1st trans. 9, z, and 2y, gives y = 7 — 2z;
In the 2d trans. 16, 32, and 3y, gives y = 5 - z;
Putting these two equal, gives 6 — 2 = 7 — 22.
Trans. 5 and 22, gives 2 = 2.
Hence y = 6 — 2 = 3> and a: = 9-— y— 2=4.
2dly. By the 2d method :

From the 1st equa. a? = 9 — y — 2 ;
This value of x substit. in the 2d and 3d, gives
9 + y + 22 = 16,
9 + 2y + 32 = 21 ;
In the 1st trans. 9 and 22, gives y = 7 — 22 ;
This substit. in the last, gives 23 — 2 = 21 ;
Trans. 2 and 21, gives 2 = 2.
Hence again y = 7 — 22 = 3, and a? = 9 — y — 2 = 4.



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SIMPLE EaUATIONS. 243

8dly. By the 3d method : subtracting the Ist equ. from
the 2d, aod the 2d from the 3d, gives
y + 22; = 7,
y+ « = 5;
Subtr, the latter from the former, gives z = 2.
Hence y = 5 — » =^ 3, and x = 9 — y — a;=5=4.
ix+ y+ z=^l8)

2. Given < x + 3y + 2z = 38 > ; to find x, y, and x.

<« + 4y + i«.= ioi

Ans. « = 4, y = 6, « = 8.
= 27^

3. Given -J a? + Jy + Ja; = 20 > ; to find ar, y, and z.




Ans. X ^ 1, y = 12, z = 60.
4. Given x — y = 2, x — « = 3, and y + « = 9 ; to
find X, y, and 2. Ans. x =? 7, y <=: 5, z s 4.



(2x+3y + 4;K = 34)
i^3x+4y + 6z=:46)

^2x+6y + 8!» = fi8)

ix(x + y + a^)= 45)
L c y (« + y + «) = 70 ) ; to find x, y, and «.

<«(* + y + «) = 105)



Given -J 3x + 4y + 6z = 46 J ; to find x, y, and «•
Given <



A OOLUtCnON OF QUESTIONS PRODUCING SIMPLE
EQUATIONS.

Quest* 1. To find two numbers, such, that their sum
shall be 10, and their difference 6.

Let X denote the greater number, and y the less* .
Then, by the 1st condition x + y = 10,
And by the 2d - - - x — y « «,
Transp. y in each, gives x = 10 — y,

and X =5 6 + y ; .
Put these two values equal, gives 6 + y n 10 — y;
Transpos. 6 and — y, gives - 2y = 4 ;
Dividing by 2, gives - - y = 2.
And hence • - - • x = 64-y = 8*



* In Aete solntions, as many unknown letters are always used as
there ore unknown numbers to be found, purposely for exercise in the
modes of reducing the equations : avoiding the short wsvs of notation,
whicht though they may cive neater solutions, afford less exercise Im
pfaetlsing the several ndesTn reducing equations,



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244 ALGSBBA.

Quest. 2. Divide lOOZ among a, b, c, so that a may have
201 more than b, and b lOZ more than c.

Let X = a's share, y = b's, and % = c's.
Then» + y + % = 100,

X =r y + 20,

y = « + 10.
In the Ist suhstit. y + 20 for x, gives 2y + « + 20 = 100 ;
In this substituting z + 10 Tor y, gives dz + 40 ss 100 ;
By transposing 40, gives . . 3z = 60 ;
And dividing by 3, gives • - z =s 20.
Hencey » z + 10 aSO, and x s= y + ^ == ^•

Quest. 8. A prize of 500Z is to be divided between two
persons, so as their shares may be in proportion as 7 to S ;
required the share of each.

Put X and y for the two shares ; then by the question,

7 : 8 : : X : y, or mult, the extremes,
and the means, 7y « 8x,

and X + y = 500 ;
Transposing y, give? x =s 500 — y ;
This substituted in the 1st, gives ly &= 4000 -^ 8y ;
By transposing 8y, it is 15y = 4000 ;
By dividing by 15, it gives y = 260| ;
And hence x == 500 — y^ 283^.

Quest. 4. What fraction is that, to the numerator of
which if 1 be added, the value will be \ ; but if 1 be added
to the denominator, its value wilLbe \ ?

X

Let — denote the fraction.

y

X -+- 1 a:

Then by the quest. = 4, and — r-; = *•

y y + 1*

The Ist mult, by 2 and y, gives 2x + ^ = y ;
The 2d mult, by 3 and y + 1, is 3x = y + 1 ;
The upper taken from the under leaves x — 2 = 1;
By transpos. 2, it gives x == 3.
And hence y = 2x -{- 2 = 8 ; and the fraction is |.

s

Quest. 5. A labourer engaged to serve for 30 days on
these conditions : that for every day he worked, he ;was to
receive 20ji, but for every day he played, or was absent, he
was to forfeit 10(2. Now at the end of the time he had to
receive just 20 shillings, or 240 pence. It is required to



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]



SIMPLE EQUATIONS* 245

find how many days he worked, and how many he was
idle?

Let X be the days worked, and y the days idled.
Then 20x is the pence earned, and lOy the forfeits ;
Hence, by the question - x + y =^ SO,

and 20x - lOjf = 240 ;
The 1st mult, by 10, gives lOx + 1(^ = 300 ;
These two added, give - 30x = 540 ;
This div. by 30, gives - x = 18, the days worked ;
Hence - y = 30 — x == 12, the days idle J. •

Quest. 6. Out of a cask of wine which had leaked away
I, 30 gallons were drawn ; and then, being guaged, it appear,
ed to be half full ; how much did it hold ?

Let it be supposed to have held x gallons.

Then it would have leaked ^x gallons,

Conseq. there had been taken away ^x + 90 gallons.

Hence jx = Jx + 30 by the question.

Then mult, by 4, gives 2x =» x + 120 ;

And transposing x, gives x = 120 the gidlons it held.

QxTSST. 7. To divide 20 into two such parts, that 3 times
the one part added to 5 times the other may make 76.

Let X and y denote the two parts.
Then, by the question - - x + y = 20,

and 3x + 5y = 76.
Mult, the 1st by 3, gives - 3x + 3j^ = 60 ;
Subtr. the latter from the former, gives 2y = 16 ;
And dividing by 2, gives - - y = 8.
Hence, from the Ist, - x = 20 — y = 12.

Quest. 8. A market woman bought in a certain number
of eggs at 2 a penny, and as many more at 3 a penny, and
sold them all out again at the rate of 5 for two-pence, and
by so doing, contrary to expectation, found she lost 3d ; what
number of eggs had she 7

Let X = number of eggs of each sort.
Then will ^x = cost of the first sort,
And }x = cost of the second sort ;
But 5:2 : : 2x (the whole number of eggs) : |x;
Hence |x = price of both sorts, at 5 for 2 pence ; ^
Then by the question |x + Jx — jx = 3 ;
Muh. by 2, gives - x + J« — |x = 6 ;
And mult, by 3, gives 5x — y x = 18 ;
Also mult, by 5, gives x ss 00, the number of eggs of
each sort.



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246 ALGEBRA.

Quest. 9. Two persons, a and b, engage at play. Bo-
fore they begin, a has 80 guineas, and b has 60. After a
certain number of games won and lost between them, A rises
with three times as many guineas as b. Query, how many
guineas did a win of b ?

Let X denote the number of guineas a won.
Then a rises with 80 + x,
And B rises with 60 — x;
Theref. by the quest. 80 + x = 180 — 3x ;
Transp. 80 and 3x, gives 4x = 100 ;
And dividing by 4, gives x = 25, the guineas won.



QUESTIOr^S FOR PRACTICE.

1. To determine two numbers such, that their difference
may be 4, and the difference of their squares 64.

Ans. 6 and 10.

2. To find two numbers with these conditions, viz. that
half the first with a third part of the second may make 9,
and that a 4th part of the first with a 5th part of the second
may make 5. Ans. 8 and 15*

3. To divide the number 20 into two such parts, that a
3d of the one part added to a 5th of the other, may make 6.

Ans. 15 and 5.

4. To find three numbers such, that the sum of The Ist
and 2d shall be 7, the sum of the 1st and 3d 8, and the sum
of the 2d and 3d 9. Ans. 3, 4, 5.

5. A father, dying, bequeathed his fortune, which was
2800Z, to his son and daughter, in this manner ; that for eve-
ry half crown the son might have, the daughter was to have
a shilling. What then were their two shares ?

Ans. The son 2000Z and the daughter 800L

6. Three persons, a, b, c, make a joint contribution, which
' ^' whole amounts to 400Z ; of which sum b contributes

IS much as a and 20Z more ; and c as much as a and
her. What sum did esnch contribute ?

Ans. A 60Z, b 140Z, and c 200Z.



L person paid a bill of lOOZ with half guineas and
I, using in all 202 pieces ; how many pieces were there
h sort ? Ans. 180 half guineas, and 22 crowns.



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8I91PLE EQUATIONS. S47

8. Says a to b, if you give rae 10 guineas of your money,
I shall then have twice as much as you will have left : but
says B to A, give me 10 of your guineas, and then I shall
have 8 times as many as you. How many had each ?

Ans. A 22, B 26.

9. A person goes to a tavern with a certain quantity of
money in his pocket, where he spends 2 shillings ; he then
borrows as much money as he had led, and going to another
tavern, he there spends 2 shillings also ; then borrowing
again as much money as was leA, he went to a third tavern,
where likewise he spent 2 shillings ; and thus repeating the
same at a fourth tavern, he then had nothing remaining.
What sum had he at first ? Ans. 3^. 9d.

10. A man with his wife and child dine together at an
inn. The landlord charged 1 shilling for the child ; and lor
the woman he charged as much as for the child and j- as
much as for the man ; and for the man he charged as much
as for the woman and child together. How much was that
for each ? Ans. The woman 20d and the man '32(1.

11. A cask, which held 60 gallons, was filled with a
mixture of brandy, wine, and cyder, in this manner, viz.
the cyder was 6 gallons more than the brandy, and the
wine was aB much as the cyder and ^ of the brandy. Ho\^
much was there of each ?

Ans. Brandy 15, cyder 21, wine 24.

12. A general, disposing his army into a square form,
finds that he has 284 men more than a perfect square ; but
increasing the side by 1 man, he then wants 25 men to be a
complete square. How many men had he under his com-
mand ? Ans. 24000.

13. What numl)er is that, to which if 3, 5, and 8, be
severally added, the three sums shall be in geometrical pro-
gression? Ans. 1.

14. The stock of three traders amounted to 7601 : the
shares of the first and second exceeded that of the third
by 240 : and the sum of the 2d and 3d exceeded the first
by 360. What was the share of each ?

Ans. The 1st 200, the 2d 300, the 3d 260.

15. What two numbers are those, which, being in the
ratio of 3 to 4, their product is equal to 12 times their sum 7

Ans. 21 and 28.



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246 AL6EBB1..

16. A certain company at a tavern, when they canrie to
settle their reckoning, found that had there been 4 more inr
company, they might have paid a shilling each less than
they did ; but that if there had been 3 fewer in company,
they must have paid a shilling each more than they did.
What then was the number of persons in company, what
each paid, and what was the whole reckoning ?

Ans. 24 persons, each paid 7^, and the whole
reckoning 6 guineas.

17. A jockey has two horses : and also two saddles, the
one valued at 16/. the other at 3/. Now when he sets the
better saddle on the 1st horse, and the worse on the 2d, it
makes the first horse worth double the 2d ; but when he
places the better saddle on the 2d horse, and the worse on
the first, it makes the 2d horse worth three times the 1st.
What then were the values of the two horses ?

Ans. The 1st 6/, and the 2d 91.

16. What two numbers arc as 2 to 3, to each of which if
6 be added, the sums will be as 4 to 5 ? Ans. 6 and 9.

19. What are tliose two numbers, of which the greater is
to the less as their sum is to 20, and as their difference is to
10 ? Ans. 15 and 45.

20. What two numbers are those, whose difference, sum,
and product, are to each other, as the three numbers 2,
3, 5 ? . ' Ans. 2 and 10.

21. To find three numbers in arithmetical progression, of
which tho first is to the third as 5 to 9, and the sum of b\\
three is 63. Ans. 15, 21, 27.

22. 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. Ans. 16 and 8.

.23. A gentleman being asked the age of his two sonS)
answered, that if to the sum of their ages 16 be added, the
result will be double the age of the elder ; but if be taken
from the difference of their affes, the remainder will l^e equal
to the age of the younger. What then were their ages 1

Ans. 30 and 12.

24. To find four numbers such, that the sum of the Ist,
2d, and 3d shall be 13 ; tho sum of the 1st, 2d, and 4th,
15 ; the sura of the 1st, 3d, and 4th, 18 ; and lastly, the sum
of the 2d, 3d, and 4th, 20. Ans. 2, 4, 7, 9.



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2& To divide 48 into 4 such parts, that, the first increased
bj 3, tike second diminished by 8, the third multiplied by 3,
and the 4th divided by 3, may be all equal to each other.

Ans. 6, 12| 3, 27.



QUADRATIC EQUATIONS.

Qd[ADBatic Equations are either simple or compound.

A simple quadratic equation, is that which involves the
square only of the unknown quantity. As oo^ = 6. The
solution of such quadratics has been already given in simple
equations.

A compound quadratic equation, is that which contains
the. square of the unknown quantity in one term, and the
first power in another term. Aaoj^ + bx^c.

All compound quadratic equations, after being properly
reduced, fall under the three following forms, to which they
must always be reduced by preparing them for solution.

1. «»+ax = 5

3. «« — a« = — 5

The general method of solving quadratic equations, is by
what is called completing the square, which is as follows :

1. RsnucE the proposed equation to a proper simple form,
.as usual, such as the forms above ; namely, by transposing '
«){ the terms which contain the unknown quantity to one
tide of the equation, and the known tefros to the or her ;
placing the square term first, and the single power second ;
difiding the equation by the co-cfiicieDt of the square or
first term, if it has one, and changing the signs of airthe
terms, when that term happens to be negative, as that
term must alwitys be made positive before the solution.
Then the proper solution is by completing the square as
follows, viz.

2. Complete the unknown side to a square, in this man-
ner, viz. Take half the co-efficient of the aecbnd term, and
square it ; which square add to both sides of the equation,
then thatiside which contains the unknown quantity will be
a complete square.

VOL.L 33



Digitized by VjOOQ IC



250 AteBBBA.

3. Then extract the square root on h<itb si^efi of the
equation'*'| and the value of the unknown quantity will be
determined, making the root of the known side either 4* or
w, which will give two roots of the equation, or two values of
the unknown quantity.



* As the square root of any quantity may be either + or — , there-
fore all quadratic equations admit of two solutions. Thas, the square
root of + n» is either + n or — «; for + n X + « and — « X — «
are each equal to -+- n^. But the square root of — n«, or V — «* is
imaginary or impossible, as neither -f ^ nor — ft, when squared, gives
— n«.

So, in the first form, x^-\-ax = b, where a: -|- J|o is foand = V(^ -



the root may be either 4- V(b + 4a2), or — V(5 + {a^Y since either
of them being multipliea by itself produces &-{- i^i'. And this ambi-
guity is eipressed by writing the uncertain or double sign zt, before
V(6 4- Jfl«) ; thus x = ± V(6 + ja') — Ja.

In this form, where a: = Hh \/(h + Ja^) — Ja, the first value of x, vis.
X = -f- V(6 -}- ifl') — i* " always affirmative ; for since Jfl^-f ft is
greater than ^a', the greater square must necessarily have the greater
root ; therefore v(b -J- ^a^), will always be greater than Via^ or its
equal j|a ; and consequenUy -f V{b + i^) ~ i^ will always be afirm-
ative.

The second value, viz. x r-z — ^(h -{■ |fl«) — ^a will always be nega-
tive, because it is composed of two negative terms. Therefore when
:fi + ax sz: b, vre shall have x = -\- \/{h + ia^) — ^ for the affiroiativ«
value of X, and x = — V(p + ia-) — ^n for the negative value of x.

In the second form, where x = ± V(6 -|^ jji«) -f ^a the first valae«
viz. X = + V(6 + jff^) + i« is always affirmative, since it is composed



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