QuestionWhich statement is true about the solution of $\sqrt[3]{x^{2}-12}=\sqrt[3]{4 x}$ ? $x=-2$ is an extraneous solution, and $x=6$ is a true solution. $x=6$ is an extraneous solution, and $x=-2$ is a true solution. Both $x=-2$ and $x=6$ are extraneous solutions. Both $x=-2$ and $x=6$ are true solutions.

Studdy Solution
Check each solution in the original equation to determine if it is true or extraneous.
First, check $x = 6$:
$$\sqrt[3]{6^2 - 12} = \sqrt[3]{4 \cdot 6}$$ $$\sqrt[3]{36 - 12} = \sqrt[3]{24}$$ $$\sqrt[3]{24} = \sqrt[3]{24}$$
This is true.
Next, check $x = -2$:
$$\sqrt[3]{(-2)^2 - 12} = \sqrt[3]{4 \cdot (-2)}$$ $$\sqrt[3]{4 - 12} = \sqrt[3]{-8}$$ $$\sqrt[3]{-8} = \sqrt[3]{-8}$$
This is also true.
Both $x = -2$ and $x = 6$ are true solutions. Therefore, the correct statement is:
Both $x = -2$ and $x = 6$ are true solutions.