Question

what is the solution to the equation (sqrt3{m - 4} - sqrt3{2m + 8} = 0)?
(m = -24)
(m = -12)
(m = -4)
(m = -6)

Explanation:

Step1: Isolate one cube root

Given the equation \(\sqrt[3]{m - 4}-\sqrt[3]{2m + 8}=0\), we can add \(\sqrt[3]{2m + 8}\) to both sides of the equation. So we get \(\sqrt[3]{m - 4}=\sqrt[3]{2m + 8}\).

Step2: Eliminate the cube roots

Since the cube root function is one - to - one (if \(\sqrt[3]{a}=\sqrt[3]{b}\), then \(a = b\) for all real numbers \(a\) and \(b\)), we can set the expressions inside the cube roots equal to each other. That is \(m-4=2m + 8\).

Step3: Solve for \(m\)

Subtract \(m\) from both sides of the equation \(m - 4=2m+8\):
\(m-4 - m=2m + 8-m\)
\(-4=m + 8\)
Then subtract 8 from both sides:
\(-4-8=m+8 - 8\)
\(m=-12\)
We should also check the solution in the original equation.
For \(m = - 12\), \(\sqrt[3]{-12-4}=\sqrt[3]{-16}\), \(\sqrt[3]{2\times(-12)+8}=\sqrt[3]{-24 + 8}=\sqrt[3]{-16}\)
\(\sqrt[3]{-16}-\sqrt[3]{-16}=0\), so \(m=-12\) is a valid solution.

Answer:

\(m = - 12\) (corresponding to the option with \(m=-12\))