Question

60 solve the radical equation. check for extraneous solutions. \\(sqrt3{2x - 8} = 4\\)

Explanation:

Step1: Cube both sides to eliminate the cube root

To solve the equation \(\sqrt[3]{2x - 8}=4\), we cube both sides of the equation. Cubing a cube root will cancel out the radical. So, \((\sqrt[3]{2x - 8})^3 = 4^3\).
Simplifying both sides, we get \(2x - 8=64\) (since \((\sqrt[3]{a})^3=a\) and \(4^3 = 64\)).

Step2: Solve for \(x\)

Now we solve the linear equation \(2x - 8 = 64\). First, add 8 to both sides of the equation: \(2x-8 + 8=64 + 8\).
This simplifies to \(2x=72\). Then, divide both sides by 2: \(\frac{2x}{2}=\frac{72}{2}\).
So, \(x = 36\).

Step3: Check for extraneous solutions

To check if \(x = 36\) is an extraneous solution, we substitute \(x = 36\) back into the original equation \(\sqrt[3]{2x - 8}\).
Substitute \(x = 36\): \(\sqrt[3]{2(36)-8}=\sqrt[3]{72 - 8}=\sqrt[3]{64}\).
Since \(\sqrt[3]{64}=4\) (because \(4^3 = 64\)), which is equal to the right - hand side of the original equation, \(x = 36\) is a valid solution.

Answer:

\(x = 36\)