Charles Davies.

First lessons in algebra, embracing the elements of the science online

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many dollars per yard as the piece contained yards. Now,
he gets four times as much for one piece as for the other :
how many yards in each piece ?

Let x= the number in the larger piece ;
y=z the number in the shorter piece.

Then, by the conditions of the question,

x+y—12.

xXx—x 2 z=z what he got for the larger piece ;

yXy=y 2 = what he got for the shorter.

And x 2 z=z\y 2 , by the 2nd condition.

x =^y, by extracting the square root

Substituting this value of x in the first equation, we have

%y+y = 12;

and consequently, y= 8,

and x=z 4.

Ans. 8 and 4.

3. What two numbers are those whose product is 30, and
quotient 3^ ? Ans. 10 and 3.

4. The product of two numbers is a, and their quotient
b : what are the numbers !

Ans. -\/ ab and \ / — .
V b

5. The sum of the squares of two numbers is 117, and
the difference of their squares 45 : what are the numbers ?

Ans. 9 and 6.



172 FIRST LESSONS IN ALGEBRA.

6. The sum of the squares of two numbers is a, and the
difference of their squares is b : what are the numbers ?

Ans. x= -\/a-\-b, y= -yja — b.

7. What two numbers are those which are to each other
as 3 to 4, and the sum of whose squares is 225 ?

Ans. 9 and 12.

8. What two numbers are those which are to each other

as m to n, and the sum of whose squares is equal to a 2 1

ma na

Ans. , — — , ■==..
ym 2 -\-n 2 ym 2 -\-n 2

9. What two numbers are those which are to each other
as 1 to 2, and the difference of whose squares is 75 ?

Ans. 5 and 10.

10. What two numbers are those which are to each other
as m to n, and the difference of whose squares is equal
to b 2 !

mb nb



Ans.



ym 2 — n 2 yn



'm* — n*

11. A certain sum of money is placed at interest for six
months, at 8 per cent, per annum. Now, if the amount be
multiplied by the number expressing the interest, the pro-
duct will be 562500 : what is the amount at interest 1

Ans. $3750.

12. A person distributes a sum of money between a num-
ber of women and boys. The number of women is to the
number of boys as 3 to 4. Now, the boys receive one-
half as many dollars as there are persons, and the women
twice as many dollars as there are boys, and together they
receive 138 dollars : how many women were there, and
how many boys ?

(36 women.
I 48 boys.



EQUATIONS OF THE SECOND DEGREE. 173

Of complete Equations.

1 22. We have already seen (Art. 117), that a complete
equation of the second degree, after it has been reduced,
contains three terms, viz : the square of the unknown
quantity in the first term ; the first power of the unknown
quantity in the second term ; and a known quantity, in a
third term.

Thus, if we have the equation

r ox 2 — 2tf 2 +8 = 9#-|-32,
we have, by transposing and reducing,

3* 2 — 9#=24,
and by dividing by 3,

x 2 —3xz=S.
which contains but three terms.
2. If we have the equation

a 2 x 2 + Sabx -f- x 2 = ex + d,
by collecting the coefficients of x 2 and x, we have

(a 2 + l)x 2 +(3ab — c)x = d ;
and dividing by the coefficient of x 2 , we have

3ab — c d

a?H 9 — t-«=-



a 2 +l « 2 +l



Quest. — 122. How many terms does a complete equation of the
second degree contain 1 Of what is the first term composed 1 ? The
second 1 The third 1

15*



174 FIRST LESSONS IN ALGEBRA.

If we represent the coefficient of x by 2p, and the known
term by q, we have

a 2 -\- 2px=zq,

an equation containing but three terms ; and we see, from
the above examples, that every complete equation of the
second degree may be reduced to this form.

123. We wish now to show that there are four forms
under which this equation will be expressed, each depend-
ing on the signs of 2p and q.

1st. Let us for the sake of illustration, make
2p—+4, and q=-\-5:
we shall then have # 2 +4#=5.
2nd. Let us now suppose

2pz=— 4, and q = -{-5 :
we shall then have a? 2 -— 4a? =5.
3rd. If we make

2q= +4, and ^=—5,
we have a? 2 +4#= — 5.

4th. If we make

2p—— 4, and q= — 5,
we have # 2 — 5a?=— 5.



Quest. — 123. Under how many forms may every equation of the
second degree be expressed 1 On what will these forms depend 1 "What
are the signs of the coefficient of x and the known term, in the first
form 1 What in the second 1 What in the third 1 What in the fourth !
Repeat the four forms.



EQUATIONS OF THE SECOND DEGREE. 175

We therefore conclude that every complete equation of
the second degree may be reduced to one of these forms :

x 2 -f 2px = + q, 1st form.

x 2 — 2px as + q, 2nd form .

x 2 +2px = —q, 3rd form.

x 2 — 2px =—q, 4th form.

124. Remark. — If, in reducing an equation to either of
these forms, the second power of the unknown quantity
should have a negative sign, it must be rendered positive
by changing the sign of every term of the equation.

125. We are next to show the manner in which the
^alue of the unknown quantity may be found. We have
>*een (Art. 38), that

(x-\-p) 2 —x 2 +2px+p 2 ;

and comparing this square with the first and third forms, we
see that the first member in each contains two terms of the
square of a binomial, viz : the square of the first term plus
twice the product of the 2nd term by the first. If, then, we
take half the coefficient of x, viz : p, and square it, and add
to both members, the equations take the form

x 2 -\-2px-{-p 2 =q +p\

x 2 +2px +p 2 =-~q 2 +p 2 )

in which the first members are perfect squares. This is



Quest. — 124. If in reducing an equation to either of these forms the
coefficient of z 2 is negative, what do you do !— - 125. What is the square
of a binomial equal to 1 What does the first member in each form con-
tain * How do you render the first member a perfect square I What is
this called 1



176 FIRST LESSONS IN ALGEBRA.

called completing the square. Then, by extracting the
square root of both members of the equation, we have

and x-\-p= ± -y/^q+p 2 ,

which gives, by transposing p,

oc=—pzk<\/q+p 2 ,
x = — p ± -yj — q~\-p 2 .

126. If we compare the second and fourth forms with
the square

(x — p) 2 = x 2 — 2px +p 2 ,

we also see that half the coefficient of x being squared and
added to both members, will make the first members perfect
squares. Having made the additions, we have

x 2 — 2px -\-p 2 =s q -\-p 2 ,

x 2 — 2px -\-p 2 =—q -fp 2 .

Then, by extracting the square root of both members, we
have

X— pz=±<y/q+p*,

and x—p= zb -\J — q+p 2 ;

and by transposing — p, we find

x=p±i/^-~p 2 ',
and x —p ± -}/— q-\-p 2 .



Quest. — 126. In the second form, how do you make the first mem-
ber a perfect square 1



EQUATIONS OF THE SECOND DEGREE. 177

127, Hence, for the resolution of every equation of the
econd degree, we have the following

RULE.

I. Reduce the equation to one of the known forms.

II. Take half the coefficient of the second term, square it t
*nd add the result to both members of the equation.

III. Then extract the square root of both members of the
equation ; after which, transpose the known term to the second
member.

Remark. — The square root of the first member is always
equal to the square root of the first term, plus or minus half
the coefficient of #.

EXAMPLES IN THE FIRST FORM.

I . What are the values of x in the equation

2x 2 +8x=64 ?
If we first divide by the coefficient 2, we obtain
x 2 +4* =32.
Then, completing the square,

* 2 +4*-f4=32+4=36.

Extracting the root,

x+2 = ±i/W= + 6 or —6.
Hence, x=z— 2 + 6 = + 4 ;

or, x= — 2— 6 = — 8.



Quest. — 127. Give the general rule for resolving an equation of the
second degree. What is the first step 1 What the second 1 What the
third] What is the square root of the first member always equal to t



178 FIRST LESSONS IN ALGEBRA.

That is, in this form the smaller root is positive, and the
larger negative.

Verification.

If we take the positive value, viz : #=+4,

the equation x 2 + 4x = 32

gives 4 2 +4 x 4 = 32 :

and if we take the negative value of x, viz : x= — 8,

the equation # 2 +4a?=32

gives (_8) 2 +4(— 8)=64— 32 = 32.

From which we see that either of the values of x, viz ;
«r=+4 or x= — 8, will satisfy the equation.

2. What are the values of x in the equation

3# 2 +I2*— 19=— x 2 — 12#+89 ?

By transposing the terms, we have

3a? 2 +tf 2 +12a: + 12tf = 89+19 ;

and by reducing,

4a 2 +24* =108;

and dividing by the coefficient of a? 2 ,

x 2 +6x=27.

Now, by completing the square,

z 2 +6a: 2 +9 = 36;
extracting the square root,

a+3 = ±V36 = + 6 or —6:
hence, #=+6— 3 = + 3;

or, «= — 6— 3= — 9.



EQUATIONS OF THE SECOND DEGREE. 179

Verification.
If we take the plus root, the equation

x 2 +6xz=27
gives (3) 2 + 6(3)=27;

and for the negative root,

x 2 +6x=27
gives (— 9) 2 +6(-9)=81 -54=27.

4. What are the values of x in the equation



X'



>2



x 2 — 10ir-i-15=: 34*+ 155.

5

By clearing the fractions, we have

5x 2 —50x+75 = x 2 — 170*+775 :

by transposing and reducing, we obtain

4a 2 + 120a?=700;

then, dividing by the coefficient of x 2 , we have

x 2 +30xz=zl75 ;

and by completing the square,

a: 2 + 30^+225=400;

and by extracting the square root,

x+l5 = dz V400 = +20 or —20.
Hence, # = + 5 or —35.

Verification.
For the plus value of x, the equation
* 2 +30;r=175
gives (5) 2 +30x5=25+150 = 175.



180 FIRST LESSONS IN ALGEBRA.

And for the negative value of x, we have

(— 35) 2 +30(-35):=1225-1050=175.

5. What are the values of x in the equation
5 2 l j_ 3 g 2 2 . 273 t

Clearing the fractions, we have

10a 2 -6tf+9=:96 - 8tf-12;r 2 +273 ;

transposing and reducing,

22x 2 +2x=360 ;
dividing both members by 22,

. 2 360

X 2 -f-—X=-



22 22

/ 1 \ 2
Add ( — ) to both members, and the equation becomes

■ 2 , / 1 \ 2 360 / 1 \ 2

x+ 22 x+ k2) =*w + m ;

whence, by extracting the square root,

, 1 /360 J / IT 2

^+22 = ± V'22- + V22) >

therefore,



1 , /360 / 1 \2



1\ 2



and a= _J__ v /g , +Q



EQUATIONS OP THE SECOND DEGREE. 181

It remains to perform the numerical operations. In the
first place, T^r+f™) must De reduced to a single num-
ber, having (22 ) 2 for its denominator.

N — ( 1 Y_ 36 °X 22 + 1 _7921 >

° W ' 22 + \22/~~ (22) 2 ~*(22) 2;

extracting the square root of 7921, we find it to be 89;
therefore,



: V 22 + V22/ ~~



89



Consequently, the plus value of x is



a? ~'~22 + 22~22"~ '
and the negative value is

__ r l 89____45

°°~ 22 22"~~~TT ;

that is, one of the two values of x which will satisfy the
proposed equation is a positive whole number, and the other
a negative fraction.

6. What are the values of x in the equation

3x 2 +2x-9=76.

-4

7. What are the values of x in the equation



Ans. { *-'



2* 2 +8a?+7=~ - ~ + 197.
4 o



Ans. \*= 8



16



182 FIRST LESSONS IN ALGEBRA.

8. What are the values of x in the equation

a? x x 2

___ +1 5 = __8* + 95*.



Ans. $



x=9

-64fr

9. What are the values of x in the equation

x 2 5x x

T"~T~ = ~2

Ans. 5 x = 2

( ««-?&

10. What are the values of x in the equation



-8=— -7(r+6i



x 2 x x 2 #,13
T + T = 1T~10~'~20'



Ans. J

}x= —



x=l

2|.



EXAMPLES IN THE SECOND FORM.



1. What are the values of x in the equation
#2—8*4-10=19.
By transposing,

ff 2 -8a?=19-10=9,
then by completing the square

# 2 - 8a:4- 16 = 94- 16=25,
and by extracting the root

a?—4=± V25 = + ^ or —5.
Hence,

a?=4-f-5=9 or x=z4— 5 = — 1.

That is, in this form, the largest root is positive and the
smaller negative.



EQUATIONS OF THE SECOND DEGREE. 183

Verification.

If we take the positive value of x, the equation
x 2 — 8x—9 gives (9) 2 — 8x9 = 81 — 72 = 9;
and if we take the negative value, the equation

x 2 — 8*=9 gives (~l)2_8( — 1) — 1 + 8=9 ;

from which we see that both values alike satisfy the equa-
tion.

2. What are the values of x in the equation

By clearing the fractions, we have

6x 2 +4x— 180 = 3z 2 +12tf— 177
and by transposing and reducing

3x 2 — 8a; = 3,
and dividing by the co-efficient of x 2 , we obtain
2 8 i

X 2 —X=zl.

o

Then, by completing the square, we have

X 3 ;r ~ h 9 __+ "9 9'
and by extracting the square root,

x ~~lf - \/ 9"" + T or ""3'



Hence,



4 , 5 4 5 1

#=—+ - -= + 3, or a=— —= — —.

3 3 ' 3 3 3



184 FIRST LESSONS IN ALGEBRA,

Verification.
For the positive value of x, the equation

2 8 1
x 2 — —x=l

8
gives 3 2 ~x3=9-8=l:

and for the negative value, the equation

x 2 — —x=l
o



1 \* 8
gives



/ L\ 2 _JL 1 _ 1 , 8 _i

I 3J 3 X ~ 3~T + T" 1 '
3. What are the values of x in the equation



L73— « ?

Clearing the fractions, and dividing by the coefficient of
a? 2 , we have

2 2 11

Completing the square, we have

\ 2 1 1 49

or x-\ — 14 - 4 = — :

3 ^9 4 ^9 36*

then, by extracting the square root, we have



1 /%? 7 7

x—

hence,



*-T =± V36= + -6- ° r ~6'



1,7 9 17 5



EQUATIONS OF THE SECOND DEGREE. 185

Verification.
If we take the positive value of x, the equation

x 2 -—x=l±

gives (li^^xii^i-l:^:

and for the negative value, the equation
x 2 — 3"^=li
/ 5\ 2 2 5 25,10 45 ,,

4. What are the values of x in the equation

4a 2 —2x 2 +2ax=18ab — ISb 2 I

By transposing, changing the signs, and dividing by 2, it
becomes

x 2 — ax=2a 2 — 9ab+9b 2 j

whence, completing the square,

a 2 9a?

x 2 —ax+—=z — 9ab+9b 2 ;

4 4

extracting the square root,



9a 2
Now, the square root of -— 9ab-\-9b 2 , is evidently

——3b. Therefore,

« . /3a ^,\ C a?s= 2a—35,

2 V 2 / ( aw— a+3b.

16*



186 FIRST LESSONS IN ALGEBRA.

What will be the numerical values of x, if we suppose
a=6 and 5=1 ?



5. What are the values of x in the equation

1 4

~x— 4 — x 2 -\-2x — -
3 5



—x— 4— x 2 -\-2x — — x 2 = 45 — 3x 2 + 4# ?



. ( spsb 7,12 ) to within

(x=z -5,73) 0,01.

6. What are the values of x in the equation

8a 2 — 14*+ 10=2*+34 ?

. ( x= 3.

.Arcs. <

<a?= — 1.

7. What are the values of x in the equation

x 2

- 30+a=2tf— 22 ?

4

Ans - llZ-t

8. What are the values of x in the equation

2

**=-l.

9. What are the values of x in the equation

2ax—x 2 =—2ab — b 2 1

Ans. m 2a+b '

t X——-D.

10. What are the values of x in the equation

m 2 x 2 .



Ans. <



a 2 + b 2 —2bx+x 2 =-

nr

: — —\bn-\- V a 2 m 2 -\-b 2 m 2 — a 2 n 2 ).

n 2 —m 2 \ /



c=— — — ( in — V a 2 m 2 4-b 2 m 2 —a 2 n 2 j.
n 2 —m 2 \ /



EQUATIONS OF THE SECOND DEGREE. 18?

EXAMPLES IN THE THIRD FORM.

1. What are the values of x in the equation

a: 2 -f-4xzrr— 3 ?
First, by completing the square, we have
x 2 + 4x+4=z — 3 + 4 = 1 ;
and by extracting the square root,

x+2 = ±-y/T=: + l or —1:
hence, x—— 2 + 1 = — 1 ; or x=— 2 — 1 = — 3.
That is, in this form both the roots are negative.

Verification.
If we take the first negative value, the equation

X 2 + 4:X=— 3

gives ( — 1)2+4( — 1) = 1— 4= — 3 ;

and by taking the second value, the equation

x 2 +4x= — 3
gives (-3) 2 +4(-3) = 9-12 = -3:

hence, both values of x satisfy the given equation.
2. What are the values of x in the equation

-^L-5x-l6 = V2 + -l - x 2 +6x.
2 2

By transposing and reducing, we have

-x 2 -l\x=2S\

then since the coefficient of the second power of x is nega*
tive, we change the signs of all the terms which gives

a?+ll*:s— 28,



188 FIRST LESSONS IN ALGEBRA.

then by completing the square

tf 2 +ll*+30,25=2,25,
hence,

x— 5,5 = ±-v/2,25 = + l,5 or —1,5.

consequently,

x= — 4 or x=—7.

3. What are the values of x in the equation

— ^ 2x— 5=— x 2 +5x+5.

8 8

Arcs. <

4. What are the values of x in the equation

2
2x 2 +$x=— 2f — -*.



-Aras.



•i



5. What are the values of x in the equation



3

4x 2 +—x+3x=—Ux—3±—4x 2 .



6. What are the values of x in the equation

q 4/r 2

_ a? 2_ 4 — -»=— +24*+2.

4 ^

Arts. \ ' ~~

7. What are the values of x in the equation

-i-tf 2 +7z+2Q= - ~tf 2 -lltf-60.

9 y



AttS,



(a=-10.

i *= - 8.



EQUATIONS OF THE SECOND DEGREE. 189



8. What are the values of x in the equation

5 2 . l oi 1 2 l
— ar— xA = — 9-i-a: — —x 2 .

6 2 8 6 2

Ans. | ~"~"
t a?=— £.

9. What are the values of x in the equation

A (a?=-10
Ans. <

( *=-tV

1 0. What are the values of x in the equation

x— a 2 — 3 = 6a?+l.

^4ns. < ~~

t*=-l.

1 1 . What are the values of x in the equation

a 2_(_4 a ,_9 _ : __93 #



(*= — 1.



EXAMPLES IN THE FOURTH FORM.

1 . What are the values of x in the equation
x 2 — 8#= — 7.
By completing the square we have

a?— 8a?+16 = — 7+16 = 9 ;
then by extracting the square root

x— 4=± A /9 = + 3 or —3;

hence,

#= + 7 or #= + 1.

That is, in this form, both the roots are positive.



190 FIRST LESSONS IN ALGEBRA

Verification.
If we take the largest root, the equation

x 2 — 8x= — 7 gives 7 2 — 8 x 7 = 49 — 56 = — 7;
and for the smaller, the equation

x 2 ~8x= — 7 gives l 2 — 8 x 1=1— 8 = — 7 :
hence, both of the roots will satisfy the equation.

2. What are the values of x in the equation

40

-lJ-* 2 +3tf-10 = lltf 2 -18;r+— .

2

By clearing the fractions, we have

— Zx 2 +6x— 20 = 3# 2 — 36* + 40;

then by collecting the like terms

— 6* 2 +42;r=60 ;

then by dividing by the coefficient of x 2 , and at the same
time changing the signs of all the terms, we have

x 2 — 7x= — 10.

By completing the square, we have

£ 2 -7*+ 12,25=2,25,

and by extracting the square root of both members,

x— 3,5=± V^25 = + l,5 or —1,5.
hence, *

ar=3,5 + l,5=5, or a?=3,5 — 1,5=2.



EQUATIONS OF THE SECOND DEGREE. 191

Verification.
If we take the larger root, the equation
x 2 — 7x= — 10 gives 5 2 — 7x5=25— 35 = — 10 ;
and if we take the smaller root, the equation

x 2 — 7x= — 10 gives 2 2 — 7x2=4— 14 = — 10.
3. What are the values of x in the equation
— 3*+2tf 2 +l = 17fa:— 2* 2 — 3.
By transposing and collecting the terms, we have

4a? 2 — 20|a= — 4;
then dividing by the coefficient of x 2 we have

x 2 — 5j#= — 1.
By completing the square, we obtain

X 5s * + 25- I+ 25~25'
and by extracting the root



* 2 - 2 ^ ± V / ?T= +



12 12

T or ""IP



hence,



12 « OS 12 1

— =5; or, a!= 2f- 1 -= T



Verification.
If we take the larger root, the equation
x 2 —b\x= — 1 gives 5 2 — 5jx5=25— 26 = — 1 ;
and if we take the smaller root, the equation

rim i • /^Y* ki * * 26

, 2 -5i,= -l gxves (_^5ix-= - -=-l.



192 FIRST LESSONS IN ALGEBRA.



4. What are the values of x in the equation



\

Ans



5. What are the values of x in the equation
1

T



— 4a? 2 — — -a?+ l-y= — 5o? 2 + ^ ?



i a? =8.
Ans.



i x=S
S ' \x^

6. What are the values of x in the equation

^20 40 20 ^40

Ans. < ~f*

7. What are the values of x in the equation

X 2 — l0j\xz=:—l?

, (#=10.
Ans. <
t x



8. What are the values of x in the equation

173,2 Or 2

_27a?+-— -+100=— — +12a?-26?
5 5



1
TXT-



[x=7.
-.6.
9. What are the values of x in the equation



Ans. <

{ a?=(



8a? 2 7a: 2
— 22a?+I5 = — h28o:-30?



Ans. <

( x=l.



10. What are the values of x in the equation

_3_

To



2a? 2 -30a?+3 = -a? 2 +3 T V-T7: ?



Ans. { "«.



EQUATIONS OF THE SECOND DEGREE. 193

Properties of the Roots.

128. We have thus far, only explained the methods of
finding the roots of an equation of the second degree. We
are now going to show some of the properties of these roots.

The first form.

129. The first form

x 2 -\-2px~q



gives 1st root x= — p+\/q-{-p 2 ,

2nd root x~ — p — ^q-\-p 2 ,

and their sum =—2p.

Since, in this form q is supposed positive, the quantity
q-\-p 2 under the radical sign will be greater than p 2 , and
hence its root will be greater than p. Consequently the
first root, which is equal to the difference between p and
the radical, will be positive and less than p. In the second
root, p and the radical have the same sign ; hence, the
second root will be equal to their sum and negative. If we
multiply the two roots together, we have

— p + V$+P 2
-p - Vq+p 2



Product equal to



-\-p 2 — p->Jq-\-p 2

+pVq-{-p 2 —q—p 2



Quest. — 129. In the first form, have the roots the same or contrary-
signs 1 What is the sign of the first root 1 What of the second 1
Which is the greater 1 What is their sum equal to 1 What is their
product equal- to 1

17



194 FIRST LESSONS IN ALGEBRA.

Hence we conclude,

1 st. That in the first form one of the roots is always posi-
tive, and the other negative.

2nd. That the positive root is numerically less than the
negative.

3rd. That the sum of the two roots is equal to the coefficient
of x in the second term, taken with a contrary sign.

4th. That the product of the two roots is equal to the known
term in the second member, taken with a contrary sign.

EXAMPLES.

1. In the equation

# 2 +tf=20,

we find the roots to be 4 and —5. Their sum is — 1,
and their product — 20.

2. In the equation

tf 2 +2#=3,

we find the roots to be 1 and —3. Their sum is equal to
— 2, and their product to —3.

3. The roots of the equation

x 2 +x=§0,

are +9 and —10. Their sum is — 1, and their product
-90.

4. The roots of the equation

a 2 +4tf=60,

are 6 and —10. Their sum is —4, and their product is
-60.



EQUATIONS OF THE SECOND DEGREE. 195

Let these principles be applied to each of the examples
under " examples in the first form."



Second Form.
130. The second form is,

x 2 —2px=q ;
and by resolving the equation we find
1st root, x=+p+i/q+p 2



2nd root, x = -\-p — ^\/q -\-p 2

and their sum = 2p.

In this form, the first root is positive and the second
negative. If we multiply the two roots together, we have

Hence we conclude,

1st. That in the second form one of the roots is positive
and the other negatwe.

2nd. That the positive root is numerically greater than the
negative.

3rd. That the sum of the roots is equal to the coefficient of
X in the second term, taken with a contrary sign.

4th. That the product of the roots is equal to the known
term in the second member, taken with a contrary sign.



Quest. — 130. What is the sign of the first root in the second form 1
What is the sign of the second 1 Which is the greater 1 What is their
sum equal to 1 What is their product equal to 1



196 FIRST LESSONS IN ALGEBRA.

EXAMPLES.

1. The roots of the equation

x 2 — #=12,

are +4 and —3. Their sum is -f-1, and their product
— 12.

2. The roots of the equation

in ' )

are +10 and — — . Their sum is 9^, and their product
is —1.

3. The roots of the equation

x 2 — 6x=l6,

are +8 and —2. Their sum is +6, and their product
is -16.

4. The roots of the equation

X 2 — 11#:=80,

are +16 and —5. Their sum is + 1 1 , and their product
is —80.

Let these principles be applied to each of the examples
under " examples in the second form."

Third Form. *

131. The third form is,

x 2 -\-2pxz=—q;
and by resolving the equation we find,
1st root, x=— p+V - #~hP 2 >



2nd root, x = —p — -\f^q +J° 2

Their sum is =i—2p



EQUATIONS OF THE SECOND DEGREE. 197

In this form, the quantity under the radical being less
than p 2 , its root will be less than p : hence both the roots
will be negative, and the first will be numerically the least.

If we multiply the roots together, we have



(— P+V— q+P z )x{— P— V— 'S r +F 8 ) = +fl f -
Hence we conclude,

1st. That in the third form both the roots are negative.
2nd. That the first root is numerically less than the second.

3rd. That the sum of the two roots is equal to the coefficient
of x in the second term, taken with a contrary sign.

4th. That the product of the roots is equal to the known
term in the second member, taken with a contrary sign.

EXAMPLES.

1. The roots of the equation

x 2 +9x=— 20,

are -—4 and —5. Their sum is —9, and their product
+20.

2. The roots of the equation

X *+I3x=z— 42,

are —6 and —7. Their sum is —13, and their product

+ 42.



Quest. — 131. In the third form, what are the signs of the roots 1
Which root is the least 1 What is the sum of the roots equal to 1
What is their product equal to 1

17*



196 FIRST LESSONS IN ALGEBRA.

3. The roots of the equation

* 2 +2f*=-li,

3

are and ~2. Their sum is — 2f, and their product


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