# How do you solve x^2-5x-10=0 by completing the square?

##### 2 Answers
Nov 22, 2017

"By the book":
$\left(x - \frac{5 + \sqrt{65}}{2}\right) \left(x - \frac{5 - \sqrt{65}}{2}\right) = 0$

when:
${x}_{1} = \frac{5 + \sqrt{65}}{2} \approx 6.53$
${x}_{2} = \frac{5 - \sqrt{65}}{2} \approx - 1.53$

#### Explanation:

$a {x}^{2} + b x + c = 0$
${x}^{2} - 5 x - 10 = 0$

$a = 1$
$\implies$

$\left(x + {x}_{1}\right) \left(x + {x}_{2}\right) = 0$
$\implies$
$\left({x}_{1} + {x}_{2}\right) = - 5 = b$
$\left({x}_{1} \cdot {x}_{2}\right) = - 10 = c$

$D \left\{- 10\right\} = \left(- 1 , 10\right) , \left(1 , - 10\right) , \left(- 2 , 5\right) , \left(2 , - 5\right)$

$\left(- 1 , 10\right) :$
$- 1 + 10 = 9 \ne - 5$

$\left(1 , - 10\right) :$
$1 - 10 = - 9 \ne - 5$

$\left(- 2 , 5\right) :$
$- 2 + 5 = 3 \ne - 5$

$\left(2 , - 5\right) :$
$2 - 5 = - 3 \ne - 5$

$\implies$
In that case we don't have any $\mathbb{Z}$ to work with, so we will move to Plan B:

${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 \cdot a}$
$\implies$

${x}_{1 , 2} = \frac{- \left(- 5\right) \pm \sqrt{{\left(- 5\right)}^{2} + 4 \cdot \left(1\right) \cdot \left(- 10\right)}}{2 \cdot \left(1\right)} =$
$= \frac{5 \pm \sqrt{25 + 40}}{2} =$
$= \frac{5 \pm \sqrt{65}}{2} =$
$\implies$
${x}_{1} = \frac{5 + \sqrt{65}}{2} \approx 6.53$
${x}_{2} = \frac{5 - \sqrt{65}}{2} \approx - 1.53$

$\implies$
"By the book":
$\left(x - \frac{5 + \sqrt{65}}{2}\right) \left(x - \frac{5 - \sqrt{65}}{2}\right) = 0$

If we check we can easily see:

$\left(- \frac{5 + \sqrt{65}}{2}\right) \cdot \left(- \frac{5 - \sqrt{65}}{2}\right) = \frac{25 + 5 \sqrt{65} - 5 \sqrt{65} - 65}{4} = - \frac{40}{4} = - 10 = c$

$\left(- \frac{5 + \sqrt{65}}{2}\right) + \left(- \frac{5 - \sqrt{65}}{2}\right) = \frac{- 5 + \sqrt{65} - - 5 - \sqrt{65}}{2} = - \frac{10}{2} = - 5 = b$

Nov 22, 2017

$x = \frac{5}{2} \pm \frac{\sqrt{65}}{2}$

#### Explanation:

$\textcolor{b l u e}{\text{to complete the square}}$

• " ensure the coefficient of the "x^2" term is 1 which it is"

• " add "(1/2"coefficient of the x-term")^2" to both sides"

$\Rightarrow {x}^{2} + 2 \left(- \frac{5}{2}\right) x \textcolor{red}{+ \frac{25}{4}} - 10 = 0 \textcolor{red}{+ \frac{25}{4}}$

$\Rightarrow {\left(x - \frac{5}{2}\right)}^{2} = \frac{25}{4} + 10 = \frac{65}{4}$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

rArr(sqrt((x-5/2)^2)=+-sqrt(65/4)larrcolor(blue)"note plus or minus"

$\Rightarrow x - \frac{5}{2} = \pm \frac{\sqrt{65}}{2}$

$\Rightarrow x = \frac{5}{2} \pm \frac{\sqrt{65}}{2}$