# How do you solve using the completing the square method 0= x^2 + 10x + 5?

May 4, 2016

$\textcolor{b l u e}{x \approx - 9.472 \text{ and "-0.528" to 3 decimal places}}$

#### Explanation:

Let k be the constant of correction

Write as $y = {x}^{2} + 10 x + 5 + k$ At this stage k=0

$y = \left({x}^{\textcolor{red}{2}} + 10 x\right) + 5 + k$

Take the square from x^(color(red)(2) and move it outside the bracket

$y = {\left(x + 10 x\right)}^{\textcolor{red}{2}} + 5 + k$

remove the $x$ from $10 x$

$y = {\left(x + 10\right)}^{2} + 5 + k$

halve the 10

$y = {\left(x + \textcolor{m a \ge n t a}{5}\right)}^{2} + 5 + k$
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Consider $y = a {x}^{2} + b x + c$ written in the form $y = a {\left(x + \textcolor{m a \ge n t a}{\frac{b}{2 a}}\right)}^{2} + c + k$

From within the bracket we have $k = \left(- 1\right) \times a \times {\left(\textcolor{m a \ge n t a}{\frac{b}{2 a}}\right)}^{2}$

The $a \times$ is from the $a$ outside the bracket but in your case $a = 1$

so $k = \left(- 1\right) \times {\left(\textcolor{m a \ge n t a}{+ 5}\right)}^{2} = - 25$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b r o w n}{y = {\left(x + 5\right)}^{2} + 5 + k} \text{ "color(green)(->" } y = {\left(x + 5\right)}^{2} + 5 - 25$

$y = {\left(x + 5\right)}^{2} - 20$
'~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{solving for "x" when } y = 0}$

$\implies 0 = {\left(x + 5\right)}^{2} - 20$

$\sqrt{20} = \sqrt{{\left(x + 5\right)}^{2}}$

$\implies x + 5 = \pm 2 \sqrt{5}$

$\implies x = - 5 \pm 2 \sqrt{5}$

$\textcolor{b l u e}{x \approx - 9.472 \text{ and "-0.528" to 3 decimal places}}$