# How do you solve sqrt(3x^2-11)=x+1 and identify any restrictions?

Mar 29, 2018

$x = 3$

#### Explanation:

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

$\Rightarrow {\left(\sqrt{3 {x}^{2} - 11}\right)}^{2} = {\left(x + 1\right)}^{2}$

$\Rightarrow 3 {x}^{2} - 11 = {x}^{2} + 2 x + 1$

$\Rightarrow 2 {x}^{2} - 2 x - 12 = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\Rightarrow 2 \left({x}^{2} - x - 6\right) = 0$

$\Rightarrow 2 \left(x - 3\right) \left(x + 2\right) = 0$

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

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

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

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

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

$\sqrt{27 - 11} = \sqrt{16} = 4 \text{ and } 3 + 1 = 4$

$\sqrt{12 - 11} = 1 \text{ and } - 2 + 1 = - 1$

$\text{the solution is } x = 3$

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