# Question #35aac

Jan 5, 2018

Assuming you meant to find solutions for $\frac{1}{2} {x}^{2} - 4 x + 4 = 0$,

$x = \pm 2 \sqrt{2} + 4$

#### Explanation:

Firstly, take out the $\frac{1}{2}$ term and complete the square as usual

$\frac{1}{2} {x}^{2} - 4 x + 4 = 0$
$\frac{1}{2} \left[{x}^{2} - 8 x + 8\right] = 0$
$\frac{1}{2} \left[{\left(x - 4\right)}^{2} - 8\right] = 0$
$\frac{1}{2} {\left(x - 4\right)}^{2} - 4 = 0$

Now solve for $x$,

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

And simplify the root term

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