# How do you solve the equation x^2-9x=-12 by completing the square?

May 21, 2017

$x = \frac{9}{2} - \frac{\sqrt{33}}{2}$ or $x = \frac{9}{2} + \frac{\sqrt{33}}{2}$

#### Explanation:

As we have to solve ${x}^{2} - 9 x = - 12$ by completing square method

first have alook at LHS, it appears to be of the form

${a}^{2} - 2 a b + \textcolor{red}{{b}^{2}}$, where $a$ is $x$ tp find $b$,

letus write $- 9 x$ as $2 \times x \times \left(- \frac{9}{2}\right)$ $-$ an it is apparent that we can use $- \frac{9}{2}$ as $b$ and therefore for completing square we must add $\textcolor{red}{{b}^{2}}$ i.e. ${\left(- \frac{9}{2}\right)}^{2} = \frac{81}{4}$.

Hence we can write ${x}^{2} - 9 x = - 12$ as

x^2-9x+color(red)(81/4=-12+color(red)(81/4)

i.e. ${\left(x - \frac{9}{2}\right)}^{2} = \frac{- 48 + 81}{4} = \frac{33}{4}$, which can be written as

${\left(x - \frac{9}{2}\right)}^{2} = {\left(\frac{\sqrt{33}}{2}\right)}^{2}$

or ${\left(x - \frac{9}{2}\right)}^{2} - {\left(\frac{\sqrt{33}}{2}\right)}^{2} = 0$

Now, we can write ${a}^{2} - {b}^{2}$ as $\left(a + b\right) \left(a - b\right)$ and our equation becomes

$\left(x - \frac{9}{2} + \frac{\sqrt{33}}{2}\right) \left(x - \frac{9}{2} - \frac{\sqrt{33}}{2}\right) = 0$

i.e. either $x = \frac{9}{2} - \frac{\sqrt{33}}{2}$ or $x = \frac{9}{2} + \frac{\sqrt{33}}{2}$