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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
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
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
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.
15\2 2401 16
17 4x=8 or
162° +18x+ ()
Ans. x=7 or –61. Ex. 11. 202-2-40=170.
Ans. x=15 or -14. Ex. 12. 3x2+2x-9=76.
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.
13-1 Ans. x=
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. +
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
2a2 Ans. x=-3a or 3a.
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
4 Extracting the nth root of each member, we have
4 Ex. 1. Solve the equation x*— 13x2 =- F-36. Assuming a?=y, the above becomes
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.