# How do you solve 2x^2 + 8x + 7 = 0?

Mar 26, 2016

$x = - 2 \pm \frac{\sqrt{2}}{2}$

#### Explanation:

Completing the Square

Multiply by $2$ to make the leading term into a perfect square, complete the square, then use the difference of square identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(x + 2\right)$ and $b = \sqrt{2}$ as follows:

$0 = 4 {x}^{2} + 16 x + 14$

$= {\left(2 x + 4\right)}^{2} - 16 + 14$

$= {\left(2 x + 4\right)}^{2} - {\left(\sqrt{2}\right)}^{2}$

$= \left(\left(2 x + 4\right) - \sqrt{2}\right) \left(\left(2 x + 4\right) + \sqrt{2}\right)$

$= \left(2 x + 4 - \sqrt{2}\right) \left(2 x + 4 + \sqrt{2}\right)$

$= \left(2 \left(x + 2 - \frac{\sqrt{2}}{2}\right)\right) \left(2 \left(x + 2 + \frac{\sqrt{2}}{2}\right)\right)$

$= 4 \left(x + 2 - \frac{\sqrt{2}}{2}\right) \left(x + 2 + \frac{\sqrt{2}}{2}\right)$

Hence:

$x = - 2 \pm \frac{\sqrt{2}}{2}$

$\textcolor{w h i t e}{}$

$2 {x}^{2} + 8 x + 7$ is in the form $a {x}^{2} + b x + c$ with $a = 2$, $b = 8$ and $c = 7$.

This has zeros given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- 8 \pm \sqrt{{8}^{2} - \left(4 \cdot 2 \cdot 7\right)}}{2 \cdot 2}$

$= \frac{- 8 \pm \sqrt{64 - 56}}{4}$

$= \frac{- 8 \pm \sqrt{8}}{4}$

$= \frac{- 8 \pm \sqrt{{2}^{2} \cdot 2}}{4}$

$= \frac{- 8 \pm 2 \sqrt{2}}{4}$

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