# How do you solve the equation 1/2(x-4)^2=8?

Apr 24, 2017

color(blue)(x=8,0

#### Explanation:

Solve:

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

Multiply both sides by $2$.

$\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \times \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} {\left(x - 4\right)}^{2} = 8 \times 2$

Simplify.

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

Take the square root of both sides.

$\sqrt{{\left(x - 4\right)}^{2}} = \pm \sqrt{16}$

Simplify.

$x - 4 = \pm 4$

Add $4$ to both sides and solve for $x$.

$x = 4 \pm 4$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$x = 4 - 4$

$x = 0$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$x = 4 + 4$

$x = 8$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$x = 8 , 0$

Apr 24, 2017

$x = 0$ and $x = 8$
or
$x = 0 , 8$

#### Explanation:

Order of operations.
Exponents first.
$\frac{1}{2} {\left(x - 4\right)}^{2} = 8$ $- - \to$ $\frac{1}{2} \left({x}^{2} - 8 x + 16\right) = 8$

Now divide both sides by $\frac{1}{2}$

$\cancel{\frac{1}{2}} \left({x}^{2} - 8 x + 16\right) = 8 \div \frac{1}{2}$

${x}^{2} - 8 x + 16 = 16$

Subtract 16 from both sides

${x}^{2} - 8 x \cancel{+ 16} = \cancel{16}$
${x}^{2} - 8 x = 0$

Factor "x"
$x \left(x - 8\right) = 0$

So now we have
$x = 0$ and $x - 8 = 0$

Therefore,
$x = 0$ and $x = 8$
or
$x = 0 , 8$