How do you solve sqrt(8x+1)=x+2?

Mar 25, 2018

$x = 1 \text{ or } x = 3$

Explanation:

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

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

$\Rightarrow 8 x + 1 = {x}^{2} + 4 x + 4$

$\text{rearrange into "color(blue)"standard form}$

$\Rightarrow {x}^{2} - 4 x + 3 = 0$

$\text{The factors of + 3 which sum to - 4 are - 1 and - 3}$

$\Rightarrow \left(x - 1\right) \left(x - 3\right) = 0$

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

$x - 1 = 0 \Rightarrow x = 1$

$x - 3 = 0 \Rightarrow x = 3$

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

Substitute these values into the equation and if both sides are equal then they are the solutions.

x=1rArrsqrt9=3" and "1+2=3 color(white)(x)✔︎

x=3rArrsqrt25=5" and "3+2=5color(white)(x)✔︎

$\Rightarrow x = 1 \text{ or "x=3" are the solutions}$