# How do you solve x^2 – 10x = 15 using completing the square?

Jun 22, 2015

$x = 5 \pm 2 \sqrt{10}$

#### Explanation:

${x}^{2} - 10 x = 15$
$\textcolor{w h i t e}{\text{XXXX}}$If ${x}^{2} - 10 x$ are the first two terms of a squared binominal
$\textcolor{w h i t e}{\text{XXXX}}$then the third term must be 25, since
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$${\left(x - a\right)}^{2} = {x}^{2} - 2 a x + {a}^{2}$
$\textcolor{w h i t e}{\text{XXXX}}$and, in this case $a = 5$
${x}^{2} - 10 x \textcolor{b l u e}{+ 25} = 15 + \textcolor{b l u e}{25}$

${\left(x - 5\right)}^{2} = 40$

$\textcolor{w h i t e}{\text{XXXX}}$Taking the square root of both sides gives
$x - 5 = \pm 2 \sqrt{10}$

$\textcolor{w h i t e}{\text{XXXX}}$and finally
$x = 5 \pm 2 \sqrt{10}$

Jun 22, 2015

I found:

${x}_{1} = 5 + \sqrt{40}$
${x}_{2} = 5 - \sqrt{40}$

#### Explanation:

You can add and subtract $25$ to get:
${x}^{2} - 10 x \textcolor{red}{+ 25} \textcolor{red}{- 25} = 15$
${x}^{2} - 10 x + 25 = 15 + 25$
${\left(x - 5\right)}^{2} = 40$
$x - 5 = \pm \sqrt{40}$
So:
${x}_{1} = 5 + \sqrt{40}$
${x}_{2} = 5 - \sqrt{40}$