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:

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

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

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