# How do you solve using the completing the square method x^2 +5x +4=0?

May 3, 2017

$x = - 4 \text{ or } x = - 1$

#### Explanation:

$\text{To solve by "color(blue)"completing the square}$

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

$\text{that is } {\left(\frac{5}{2}\right)}^{2} = \frac{25}{4}$

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

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

$\Rightarrow {\left(x + \frac{5}{2}\right)}^{2} = \frac{25}{4} - 4 = \frac{9}{4}$

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

$\sqrt{{\left(x + \frac{5}{2}\right)}^{2}} = \pm \sqrt{\frac{9}{4}}$

$\Rightarrow x + \frac{5}{2} = \pm \frac{3}{2}$

$\Rightarrow x = - \frac{5}{2} \pm \frac{3}{2}$

$\Rightarrow {x}_{1} = - \frac{5}{2} - \frac{3}{2} = - 4 \text{ and } {x}_{2} = - \frac{5}{2} + \frac{3}{2} = - 1$