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

Dec 11, 2016

Begin by making the coefficient of ${x}^{2}$ equal to 1 and then make the left side fit the pattern, ${\left(x \pm a\right)}^{2} = {x}^{2} \pm 2 a x + {a}^{2}$.

#### Explanation:

Divide both sides of the equation by 9:

${x}^{2} + \frac{64}{3} x + \frac{256}{9} = 0$

Add ${a}^{2} - \frac{256}{9}$ to both sides of the equation:

${x}^{2} + \frac{64}{3} x + {a}^{2} = {a}^{2} - \frac{256}{9}$

Set the middle term in the right side of the pattern, ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$ equal to the middle term in the equation:

$2 a x = \frac{64}{3} x$

Solve for a:

$a = \frac{32}{3}$

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

${\left(x + a\right)}^{2} = {a}^{2} - \frac{256}{9}$

Substitute $\frac{32}{3}$ for every "a":

${\left(x + \frac{32}{3}\right)}^{2} = {\left(\frac{32}{3}\right)}^{2} - \frac{256}{9}$

Simplify the right side:

${\left(x + \frac{32}{3}\right)}^{2} = \frac{256}{3}$

Use the square root on both sides:

$x + \frac{32}{3} = \pm \frac{16 \sqrt{3}}{3}$

Subtract $\frac{32}{3}$ from both sides:

$x = \frac{- 32 \pm 16 \sqrt{3}}{3}$

$x = \frac{- 32 + 16 \sqrt{3}}{3} \mathmr{and} x = \frac{- 32 - 16 \sqrt{3}}{3}$