# How do you solve 1/2(x+8)^2=14?

Jan 28, 2017

$x = 2 \sqrt{7} - 8$

#### Explanation:

$\frac{1}{\text{2}} {\left(x + 8\right)}^{2} = 14$

Multiplying by 2 on both sides

cancel(2) × 1/cancel("2")(x + 8)^2 = 2 × 14

${\left(x + 8\right)}^{2} = 28$

Square root both side

sqrt((x + 8)^2 = $\sqrt{28}$

$x + 8 = \pm 2 \sqrt{7}$

$x = 2 \sqrt{7} - 8 \mathmr{and} x = - 2 \sqrt{7} - 8$

$x = - 8 \pm 2 \sqrt{7}$

#### Explanation:

$\frac{1}{2} {\left(x + 8\right)}^{2} = 14$

I want to expand the squared term, so to do that I'll first multiply both sides by 2:

$\frac{1}{2} \textcolor{red}{\times 2} {\left(x + 8\right)}^{2} = 14 \textcolor{red}{\times 2}$

${\left(x + 8\right)}^{2} = 28$

${x}^{2} + 16 x + 64 = 28$

${x}^{2} + 16 x + 64 \textcolor{red}{- 28} = 28 \textcolor{red}{- 28}$

${x}^{2} + 16 x + 36 = 0$

I don't see any easy factors, so let's use the quadratic formula. The general formula is:

$x = \frac{- \textcolor{g r e e n}{b} \pm \sqrt{{\textcolor{g r e e n}{b}}^{2} - 4 \textcolor{b l u e}{a} \textcolor{b r o w n}{c}}}{2 \textcolor{b l u e}{a}}$

and relates to a trinomial this way:

$\textcolor{b l u e}{a} {x}^{2} + \textcolor{g r e e n}{b} x + \textcolor{b r o w n}{c}$

$x = \frac{- 16 \pm \sqrt{{16}^{2} - 4 \left(1\right) \left(36\right)}}{2 \left(1\right)}$

$x = \frac{- 16 \pm \sqrt{256 - 144}}{2}$

$x = \frac{- 16 \pm \sqrt{112}}{2} = \frac{- 16 \pm \sqrt{16 \times 7}}{2} = \frac{- 16 \pm 4 \sqrt{7}}{2} = - 8 \pm 2 \sqrt{7}$

And we can see these two solutions in the graphing of both sides of the original equation:

graph{(y-(1/2)(x+8)^2)(y-0x-14)=0 [-16.69, 3.31, 9.14, 19.136]}