How do you solve 9x^2+96x+256=09x2+96x+256=0 by completing the square?

1 Answer
Dec 11, 2016

Begin by making the coefficient of x^2x2 equal to 1 and then make the left side fit the pattern, (x +- a)^2 = x^2 +- 2ax + a^2(x±a)2=x2±2ax+a2.

Explanation:

Divide both sides of the equation by 9:

x^2 + 64/3x + 256/9 = 0x2+643x+2569=0

Add a^2 - 256/9a22569 to both sides of the equation:

x^2 + 64/3x + a^2 = a^2 - 256/9x2+643x+a2=a22569

Set the middle term in the right side of the pattern, (x + a)^2 = x^2 + 2ax + a^2(x+a)2=x2+2ax+a2 equal to the middle term in the equation:

2ax = 64/3x2ax=643x

Solve for a:

a = 32/3a=323

Substitute the left side of the pattern into the left side of the equation:

(x + a)^2 = a^2 - 256/9(x+a)2=a22569

Substitute 32/3323 for every "a":

(x + 32/3)^2 = (32/3)^2 - 256/9(x+323)2=(323)22569

Simplify the right side:

(x + 32/3)^2 = 256/3(x+323)2=2563

Use the square root on both sides:

x + 32/3 = +-(16sqrt(3))/3x+323=±1633

Subtract 32/3323 from both sides:

x = (-32 +-16sqrt(3))/3x=32±1633

x = (-32 + 16sqrt(3))/3 and x = (-32 -16sqrt(3))/3x=32+1633andx=321633