# How do you solve the equation 9x^2+30x+25=11 by completing the square?

Sep 14, 2017

Like so:

#### Explanation:

For starters, the ${x}^{2}$ term needs to be a multiple of 1, not 9.
So the first step is to divide both sides by 9, giving:

${x}^{2} + \frac{30}{9} x + \frac{25}{9} = \frac{11}{9}$

For me, it makes some of the subsequent steps easier if we subract $\frac{25}{9}$ from both sides...

${x}^{2} + \frac{30}{9} x = \frac{11}{9} - \frac{25}{9} = - \frac{14}{9}$ <- we'll call this equation 1

Now remember how binomials are squared:

${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$

So, if $\frac{30}{9} = 2 a$, then $a = \frac{1}{2} \cdot \frac{30}{9} = \frac{15}{9}$

...and ${a}^{2} = \frac{225}{81}$

so, what we do is, add and subtract this to the left side of eq. 1:

${x}^{2} + \frac{30}{9} x + \frac{225}{81} - \frac{225}{81} = - \frac{14}{9}$

...and now the first 3 terms of the left side are a perfect square:

${\left(x + \frac{15}{9}\right)}^{2} - \frac{225}{81} = - \frac{14}{9}$

...add $\frac{225}{81}$ to both sides:

${\left(x + \frac{15}{9}\right)}^{2} = - \frac{14}{9} + \frac{225}{81} = \frac{- 126 + 225}{81} = \frac{99}{81}$

...square root of both sides:

$x + \frac{15}{9} = \sqrt{\frac{99}{81}}$

subtract $\frac{15}{9}$ from both sides:

$x = \sqrt{\frac{99}{81}} - \frac{15}{9}$

$x = \frac{\sqrt{99} - 15}{9}$

GOOD LUCK!