Q26. - Algebra 1

Question

Solve each equation. Check your solution.

$\sqrt{3x^2-32}=x$

Step-by-Step Solution

Answer

The solution of the given equation is $x=4$.

Step 1. Rewrite the given equation.

The given equation can be written as:

$$\begin{gathered} \sqrt{3x^2-32}=x \\ {\left(\sqrt{3x^2-32}\right)}^2=x^2\left(\text{squaring both sides}\right) \\ 3x^2-32=x^2 \end{gathered}$$

Step 2. Solve the given equation.

The solution of the given equation is:

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

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

Step 3. Check the solutions.

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" style="max-width: none; vertical-align: -4px;" $\sqrt{3x^2-32}=x$.

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

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 $x=4$.