How do you solve sqrt(6x-3) - 1 = x?

May 24, 2018

$x = 2$

Explanation:

$\text{isolate "sqrt(6x-3)" by adding 1 to both sides}$

$\sqrt{6 x - 3} = x + 1$

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

${\left(\sqrt{6 x - 3}\right)}^{2} = {\left(x + 1\right)}^{2}$

$6 x - 3 = {x}^{2} + 2 x + 1$

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

$\text{subtract "6x-3" from both sides}$

$0 = {x}^{2} - 4 x + 4$

$0 = {\left(x - 2\right)}^{2} \Rightarrow x = 2$

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

Substitute this value into the equation and if both sides are equal then it is the solution.

$\text{left "=sqrt(12-3)-1=sqrt9-1=3-1=2=" right}$

$x = 2 \text{ is the solution}$