# How do you solve 2a^2 + 8a − 5 = 0 by completing the square?

Jul 6, 2015

#### Answer:

$a = \pm \sqrt{\frac{13}{2}} - 2$

#### Explanation:

First, we put the constant in the right-hand side :
$2 {a}^{2} + 8 a - 5 = 0$
$\iff 2 {a}^{2} + 8 a = 5$

Then we want to transform $2 {a}^{2} + 8 a$ to something like ${\left(a + b\right)}^{2}$

Recall : ${\left(\textcolor{b l u e}{x} + \textcolor{red}{y}\right)}^{2} = \textcolor{b l u e}{{x}^{2}} + \textcolor{g r e e n}{2} \cdot \textcolor{b l u e}{x} \textcolor{red}{y} + \textcolor{red}{{y}^{2}}$

Here we have $2 {a}^{2} + 8 a$
$\iff 2 \times \left(\textcolor{b l u e}{{a}^{2}} + 4 a\right)$
$\iff 2 \times \left(\textcolor{b l u e}{{a}^{2}} + \textcolor{g r e e n}{2} \cdot \textcolor{b l u e}{a} \cdot \textcolor{red}{2}\right)$ (Then $\text{y} = 2$)
$\iff 2 \times \left(\textcolor{b l u e}{{a}^{2}} + \textcolor{g r e e n}{2} \cdot \textcolor{b l u e}{a} \cdot \textcolor{red}{2} + \textcolor{red}{{2}^{2}} - {2}^{2}\right)$ (We added and subtract "y^2" for factorize later)

$\iff 2 \times \left({\left(a + 2\right)}^{2} - 4\right)$

Put the last expression inside the equality :

$2 {a}^{2} + 8 a = 5$

$\iff 2 \times \left({\left(a + 2\right)}^{2} - 4\right) = 5$

$\iff 2 \times {\left(a + 2\right)}^{2} - 8 = 5$

$\iff 2 \times {\left(a + 2\right)}^{2} = 13$

$\iff {\left(a + 2\right)}^{2} = \frac{13}{2}$

$\iff a + 2 = \sqrt{\frac{13}{2}} \mathmr{and} a + 2 = - \sqrt{\frac{13}{2}}$

Then $a = \pm \sqrt{\frac{13}{2}} - 2$