91Ó°ÊÓ

The given equation can be written as:

$$\begin{gathered} \sqrt{3x^2−32}=x \\ {\left(\sqrt{3x^2−32}\right)}^2=x^2\left(\text{squaringbothsides}\right) \\ 3x^2−32=x^2 \end{gathered}$$

The solution of the given equation is:

$$\begin{gathered} 3x^2−32=x^2 \\ 3x^2−x^2=32 \\ 2x^2=32 \\ {x}^2=16 \\ x=Â\pm \sqrt{16} \\ x=Â\pm \sqrt{4^2} \\ x=Â\pm 4 \end{gathered}$$

Therefore, the values of x are$x=4$ or $x=-4$.

Check the solution $x=4$,

Substitute 4 for x in the given equation $\sqrt{3x^2−32}=x$.

$$\begin{gathered} \sqrt{3x^2−32}=x \\ \sqrt{3{\left(4\right)}^2−32}\underset{¯}{\underset{¯}{?}}4 \\ \sqrt{48−32}\underset{¯}{\underset{¯}{?}}4 \\ \sqrt{16}\underset{¯}{\underset{¯}{?}}4 \\ \text{4}=4 \end{gathered}$$

Therefore,$x=4$ is the solution of the given equation $\sqrt{3x^2−32}=x$.

Check the solution $x=−4$,

Substitute$-4$ for x in the given equation width="101" height="26" role="math">3x2−32=x.

$$\begin{gathered} \sqrt{3x^2−32}=x \\ \sqrt{3{\left(−4\right)}^2−32}\underset{¯}{\underset{¯}{?}}−4 \\ \sqrt{48−32}\underset{¯}{\underset{¯}{?}}−4 \\ \sqrt{16}\underset{¯}{\underset{¯}{?}}−4 \\ \text{4}≠−4 \end{gathered}$$

Therefore,$x=−4$ is not the solution of the given equation $\sqrt{3x^2−32}=x$.

Therefore, the only solution of the given equation$\sqrt{3x^2−32}=x$ is $x=4$.