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

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Answer 1 · 2016-12-11T20:05:51

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

Divide both sides of the equation by 9:

#x^2 + 64/3x + 256/9 = 0#

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

#x^2 + 64/3x + a^2 = a^2 - 256/9#

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

#2ax = 64/3x#

Solve for a:

#a = 32/3#

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

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

Substitute #32/3# for every "a":

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

Simplify the right side:

#(x + 32/3)^2 = 256/3#

Use the square root on both sides:

#x + 32/3 = +-(16sqrt(3))/3#

Subtract #32/3# from both sides:

#x = (-32 +-16sqrt(3))/3#

#x = (-32 + 16sqrt(3))/3 and x = (-32 -16sqrt(3))/3#