# How do you solve x^2=24x+10 by completing the square?

Sep 25, 2016

$\textcolor{g r e e n}{x = 12 - \sqrt{154}} \textcolor{w h i t e}{\text{XX}}$ or color(white)("XX")color(green)(x=12+sqrt(154)

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} = 24 x + 10$

Shift all the terms which include the variable $x$ to the left side:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 24 x = 10$

Since, in general, the expansion of a squared binomial has the structure:
$\textcolor{w h i t e}{\text{XXX}} {\left(x + a\right)}^{2} = \underline{{x}^{2} + 2 a x} + {a}^{2}$

If ${x}^{2} - 24 x$ are the first two terms of the expansion of a squared binomial
then the third terms should be ${\left(- \frac{24}{2}\right)}^{2} = {\left(- 12\right)}^{2} = {12}^{2}$

In order to complete the square we will need to add ${12}^{2}$ (to both sides)

$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 24 x + {12}^{2} = 10 + {12}^{2}$

Writing as a squared binomial and simplifying:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 12\right)}^{2} = 154$

Taking the square root of both sides
$\textcolor{w h i t e}{\text{XXX}} x - 12 = \pm \sqrt{154}$
Adding $12$ to both sides
$\textcolor{w h i t e}{\text{XXX}} x = 12 \pm \sqrt{154}$

$x = 12 \pm \sqrt{154}$

#### Explanation:

${x}^{2} - 24 x = 10$
take the factor of x, half it and square, then add both sides
${x}^{2} - 24 x + 144 = 10 + 144$
${\left(x - 12\right)}^{2} = 154$
$\sqrt{{\left(x - 12\right)}^{2}} = \pm \sqrt{154}$
$\left(x - 12\right) = \pm \sqrt{154}$
$x = 12 \pm \sqrt{154}$