which statement is true about the solution of $sqrt3{x^{2}-12}=sqrt3{4x}$?\n$x = - 2$ is an extraneous…

which statement is true about the solution of $sqrt3{x^{2}-12}=sqrt3{4x}$?\n$x = - 2$ is an extraneous solution, and $x = 6$ is a true solution.\n$x = 6$ is an extraneous solution, and $x = - 2$ is a true solution.\nboth $x = - 2$ and $x = 6$ are extraneous solutions.\nboth $x = - 2$ and $x = 6$ are true solutions.

which statement is true about the solution of $sqrt3{x^{2}-12}=sqrt3{4x}$?\n$x = - 2$ is an extraneous solution, and $x = 6$ is a true solution.\n$x = 6$ is an extraneous solution, and $x = - 2$ is a true solution.\nboth $x = - 2$ and $x = 6$ are extraneous solutions.\nboth $x = - 2$ and $x = 6$ are true solutions.

Answer

Explanation:

Step1: Cube - both sides of the equation

Since $\sqrt[3]{a}=\sqrt[3]{b}$ implies $a = b$, for $\sqrt[3]{x^{2}-12}=\sqrt[3]{4x}$, we get $x^{2}-12 = 4x$.

Step2: Rearrange to a quadratic equation

Rearrange $x^{2}-12 = 4x$ to $x^{2}-4x - 12=0$.

Step3: Factor the quadratic equation

Factor $x^{2}-4x - 12$ as $(x - 6)(x + 2)=0$.

Step4: Solve for x

Set each factor equal to zero: $x-6 = 0$ gives $x = 6$; $x + 2=0$ gives $x=-2$.

Step5: Check for extraneous solutions

Substitute $x = 6$ into the original equation: $\sqrt[3]{6^{2}-12}=\sqrt[3]{36 - 12}=\sqrt[3]{24}$ and $\sqrt[3]{4\times6}=\sqrt[3]{24}$, so $x = 6$ is a true solution. Substitute $x=-2$ into the original equation: $\sqrt[3]{(-2)^{2}-12}=\sqrt[3]{4 - 12}=\sqrt[3]{-8}$ and $\sqrt[3]{4\times(-2)}=\sqrt[3]{-8}$, so $x=-2$ is a true solution.

Answer:

Both $x = -2$ and $x = 6$ are true solutions.