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

Mar 31, 2015

$2 {x}^{2} + 8 x + 1 = 0$

Reduce the coefficient of ${x}^{2}$ to $1$ by dividing all terms by $2$
x^2+4x +1/2 = 0

Remove the constant $\frac{1}{2}$ from the left-side by subtracting $\frac{1}{2}$ from both sides of the equation.
${x}^{2} + 4 x = - \frac{1}{2}$

To "complete the square" we are looking for a value $a$
${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$

From our equation we know that $2 a x = 4 x$
$\rightarrow a = 2$
and ${a}^{2} = 4$

Add ${a}^{2}$ ($4$) to both sides of the equation to "complete the square"
${x}^{2} + 4 x + 4 = 4 - \frac{1}{2}$
or
${\left(x + 2\right)}^{2} = \frac{7}{2}$

Take the square root of both sides
$x + 2$ = +-sqrt(7/2)

Therefore
$x = - 2 - \sqrt{\frac{7}{2}} = - \left(\sqrt{2} + \sqrt{7}\right)$
or
$x = - 2 + \sqrt{\frac{7}{2}} = \sqrt{7} - \sqrt{2}$