# What is the solution set of the equation x^2+sqrt2x=1/2?

Jul 10, 2017

$x = - \frac{1}{\sqrt{2}} + 1$ or $- \frac{1}{\sqrt{2}} - 1$

#### Explanation:

${x}^{2} + \sqrt{2} x = \frac{1}{2}$ can be written as

$2 {x}^{2} + 2 \sqrt{2} x - 1 = 0$

$x = \frac{- 2 \sqrt{2} \pm \sqrt{{\left(2 \sqrt{2}\right)}^{2} - 4 \times 2 \times \left(- 1\right)}}{2 \times 2}$
$= \frac{- 2 \sqrt{2} \pm \sqrt{8 + 8}}{4}$
$= - \frac{\sqrt{2}}{2} \pm \frac{4}{4}$
$= - \frac{1}{\sqrt{2}} \pm 1$
i.e $- \frac{1}{\sqrt{2}} + 1$ or $- \frac{1}{\sqrt{2}} - 1$