# How do you solve using the completing the square method 2x^2 + 8x + 1 = 0?

Feb 27, 2016

$2 {x}^{2} + 8 x + 1 = 0$
Dividing both sides by 2 we get
$\implies {x}^{2} + 4 x + \frac{1}{2} = 0$
$\implies {x}^{2} + 2 \cdot x \cdot 2 + {2}^{2} - {2}^{2} + \frac{1}{2} = 0$
$\implies {x}^{2} + 2 \cdot x \cdot 2 + {2}^{2} = 4 - \frac{1}{2} = \frac{7}{2}$
$\implies {\left(x + 2\right)}^{2} = \frac{7}{2}$
$\implies x + 2 = \pm \sqrt{\frac{7}{2}}$
$\implies x = - 2 \pm \sqrt{\frac{7}{2}}$

Feb 27, 2016

See solution below.

#### Explanation:

$2 \left({x}^{2} + 4 x + m\right) = - 1$

$m = {\left(\frac{b}{2}\right)}^{2}$

$m = {\left(\frac{4}{2}\right)}^{2}$

$m = 4$

$2 \left({x}^{2} + 4 x + 4 - 4\right) = - 1$

$2 {\left(x + 2\right)}^{2} - 4 \left(2\right) = - 1$

$2 {\left(x + 2\right)}^{2} = - 1 + 8$

${\left(x + 2\right)}^{2} = \frac{7}{2}$

$x + 2 = \pm \sqrt{\frac{7}{2}}$

$x = \pm \sqrt{\frac{7}{2}} - 2$

Don't forget the $\pm$ sign with the square root. Remember: a quadratic equation must always have two solutions.