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ond term, taken with a contrary sign, plus or minus the square root of the second member, increased by the square of half the coefficient of the second term. Ex. 1. Let it be required to solve the equation

22 - 10x=-16. Completing the square by adding to each member the square of half the coefficient of the second term, we have

22 - 10x+25=25-16=9. Extracting the root, 2-5=+3. Whence

x=5+3= 8 or 2, Ans. To verify these values of x, substitute them in the original equation, and we shall bave

82-10 x8=64-803-16. Also,

22-10 x2= 4-20=-16. Ex. 2. Solve the equation 2x2 +82—20=70.

Ans. x=+5 or -9. Ex. 3. Solve the equation 3x2-3x+6=53. Reducing

x2-=Completing the square, x2-x+1=1-4=; Hence

*=*=*=for }, Ans. Second Method of completing the Square. 260. The preceding method of completing the square is always applicable; nevertheless, it sometimes gives rise to inconvenient fractions. In such cases the following method may

be preferred. Let the equation be reduced to the form

ax2 + bx=c, in which a and b are whole numbers, and prime to each other, but c may be either entire or fractional. Multiply each member of this equation by 4a, and it becomes

4a x2 +4abx=4ac. Adding 12 to each member, we have

4a2c +4abu+b2=4ac+52, where the first member is a complete square, and its terms are entire. Extracting the square root, we have

2ax+b=#vac+62.

Transposing b, and dividing by 2a,

- Evac +62

2a which is the same result as would be obtained by the former rule; but by this method we have avoided the introduction of fractions in completing the square. If b is an even number, 5 will be an entire number; and it

2 would have been sufficient to multiply each member by a, and

32 add to each member. Hence we have the following

4

RULE. 1st. Reduce the equation to the form axa + bx=c, where a and b are prime to each other.

2d. If b is an odd number, multiply the equation by four times the coefficient of x?, and add to each member the square of the coefficient of ..

3d. If b is an even number, multiply the equation by the coefficient of a>, and add to each member the square of half the coefficient of x.

Ex. 4. Solve the equation 6x2 — 13x=-6.

Multiplying by 4x6, and adding 132 to each member, we have

144x2 – 312x+169=169-144=25. Extracting the root, 12.–13=$5. Whence

12x=18 or 8, and

x=; or . Ex. 5. Solve the equation 110x2–21x=-1. Multiplying by 440, and adding 21% to each member, we have

48400x2 - 9240x+441=1. Extracting the root, 220x-21=+1. Whence

x=ty or i Ex. 6. Solve the equation 72 – 3x=160.

Ans. x=5 or -37.

261. Modification of the preceding Method.—The preceding method sometimes gives rise to numbers which are unneces

sarily large. When the equation has been reduced to the form ax2 +bx=c, it is sufficient to multiply it by any number which will render the first term a perfect square. Let the resulting equation be

m2c2 +nx=9. The first member will become a complete square by the addition of and the equation will then be

n2 m2 +na+

4m2 This method is expressed in the following

(mm)and the

na

4m2=2+

RULE. Having reduced the equation to the form ax+bx=c, multiply the equation by any quantity (the least possible) which will render the first term a perfect square. Divide the coefficient of w in this new equation by twice the square root of the coefficient of a?, and add the square of this result to both members. Ex. 7. Solve the equation 8x2 +9x=99.

9\2 3249
4

16'
9 57
4x=
44

33
X =3 or Ans.

8'
Ex. 8. Solve the equation 16x2 — 15x=34.

15\2 2401 16

64

17 4x=8 or

162° +18x+ ()

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17
X=2 or Ans.

16
Ex. 9. Solve the equation 12x2=21+2x.
Solve the following equations:-
Ex. 10. 1xc2 — 3x+201=42%.

Ans. x=7 or –61. Ex. 11. 202-2-40=170.

Ans. x=15 or -14. Ex. 12. 3x2+2x-9=76.

=0.

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Ex. 13. 3x2-3x+7f=8.

622-40 3-10 Ex. 14. 3x

-2. 200-1 9-23 This equation reduces to *-154x=-46.

Ans. w=113 or 4. 15 72-63 Ex. 15. =2.

Ans. x=3 or 6. 222 90

90 27 Ex. 16.

2+1, 2+2 Ex. 17. 22-XV3=2-1V3.

V3+3

13-1 Ans. x=

or 2

2 X-1. 2-3 2 Ex. 18.

Ans. x=5 or 1. X-2 2-4 3

atx 5 Ex. 19. +

Ans. x=a or - 2a. ata Ex. 20. 22-(a+)x+ab=0.

Ans. x=a or b. Ex. 21. " (3x—25)(7x+29)=0. Ans. x=8} or —44. 3x-2 2x-5 10

1 Ex. 22. +

Ans. x=

or 23-5' 3x - 2 3

3 Ex. 23. (x-1)(x-2)+(x-2)(x-4)=6(2x–5).

5 Ans. =8 or

2 170 170 51 Ex. 24.

Ans. x=4 or -17. x+1 +2

a2 + ax+x? a?-ax+% ab Ex. 25. ata

3a-4b+

2a2 Ans. x=-3a or 3a.

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Equations which may be solved like Quadratics. 262. There are many equations of a higher degree than the second, which may be solved by methods similar to those employed for quadratics. To this class belong áll equations which contain only two powers of the unknown quantity, and in which the greater exponent is double the less. Such equations are of the form

32n + poc" =9, where n may be either integral or fractional.

For if we assume y=x", then ya=22n, and this equation becomes

y2 +py=9;

P whence

"=y=
2

4 Extracting the nth root of each member, we have

pa q

4 Ex. 1. Solve the equation x*— 13x2 =- F-36. Assuming a?=y, the above becomes

y2-13y=-36; whence

y=9 or 4. But, since x=y, x=#vy. Therefore

x=+V9 or V2. Thus x has four values, viz., +3, -3, +2, -2. To verify these values: 1st value, (+3)* – 13(+3)4=-36, i. e., 81–117=-36. 2d value, (-3)*–13(-3)2=-36, i. e., 81–117=-36. 3d value, (+2)*—13(+2) = -36, i.e., 16– 52=-36. 4th value, (-2)-13(-2)=-36, i. e., 16– 52=-36. Ex. 2. Solve the equation 2 — 35x=-216. Assuming x=y, the above becomes

y? - 35y=-216; whence

y=27 or 8. Hence

x= Vy=3 or 2. This equation has four other roots which can not be de termined by this process.

Ex. 3. Solve the equation +4Væ=21.
Assuming Vã=y, we have

y+4y=21;

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