# How do you solve the equation x^2-8x=-9 by completing the square?

Mar 19, 2017

$x = 6.64575131 \mathmr{and} x = 1.35424868$

#### Explanation:

$\textcolor{red}{C o m m e n c i n g}$ $\textcolor{red}{c o m p l e t i n g}$ $\textcolor{red}{t h e}$ $\textcolor{red}{s q u a r e}$ $\textcolor{red}{m e t h o d}$ $\textcolor{red}{n o w ,}$

1) Know the formula for the perfect quadratic square, which is,

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

2) Figure out your $a \mathmr{and} b$ values,

$a =$ coefficient of ${x}^{2}$, which is $1$.
$\textcolor{red}{b = \frac{- 8}{2 \left(1\right)} = - 4}$

3) Add $\textcolor{red}{{b}^{2}}$ on both sides of the equation, giving you an overall net of 0, hence not affecting the result of the equation,

${x}^{2} - 8 x + \textcolor{red}{{\left(- 4\right)}^{2}} = - 9 + \textcolor{red}{{\left(- 4\right)}^{2}}$
${\left(x - 4\right)}^{2} = 7$

4) Square root both sides,

$x - 4 = \pm \sqrt{7}$

5) Add $4$ on both sides,

$x = \pm \sqrt{7} + 4$

6) Calculate the two values of $x$,

$x = 6.64575131 \mathmr{and} x = 1.35424868$