# How do you solve sqrt(2x+8) = x?

May 6, 2018

x=4

#### Explanation:

to get rid of the square root on the right you square both sides so $2 x + 8 = {x}^{2}$
Then bring $2 x + 8$ to the left so you have $0 = \left({x}^{2}\right) - 2 x + 8$
now you can do the quadratic formula to find the zeroes

May 6, 2018

$x = 4$

#### Explanation:

$\textcolor{b l u e}{\text{square both sides}}$

$\text{note that } \sqrt{a} \times \sqrt{a} = {\left(\sqrt{a}\right)}^{2} = a$

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

$\Rightarrow 2 x + 8 = {x}^{2}$

"rearrange into "color(blue)"standard form ";ax^2+bx+c=0

$\text{subtract "2x+8" from both sides}$

$0 = {x}^{2} - 2 x - 8$

$\text{the factors of - 8 which sum to - 2 are - 4 and + 2}$

$\Rightarrow 0 = \left(x - 4\right) \left(x + 2\right)$

$\text{equate each factor to zero and solve for x}$

$x + 2 = 0 \Rightarrow x = - 2$

$x - 4 = 0 \Rightarrow x = 4$

$\textcolor{b l u e}{\text{As a check}}$

Substitute these values into the left side of the equation and if equal to the right side then they are the solution.

$x = - 2 \to \sqrt{- 4 + 8} = \sqrt{4} = 2 \ne - 2$

$\Rightarrow x = - 2 \text{ is an extraneous solution}$

$x = 4 \to \sqrt{8 + 8} = \sqrt{16} = 4 = \text{ right side}$

$\Rightarrow x = 4 \text{ is the solution}$

If you plot it as $y = 0 = x - \sqrt{2 x + 8}$ you get: 