Question

radical equations and extraneous roots
solving radical equations
solve: $sqrt3{8x - 2} = 4$
to solve this equation, first add 2 to both sides.
x = 2.25

Explanation:

Step1: Cube both sides to eliminate the cube root

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

Step2: Solve for \(x\)

Now, we solve the linear equation \(8x - 2 = 64\). First, we add 2 to both sides of the equation to isolate the term with \(x\). So, \(8x-2 + 2=64 + 2\), which simplifies to \(8x=66\).
Then, we divide both sides by 8 to solve for \(x\). So, \(x=\frac{66}{8}=\frac{33}{4} = 8.25\) (it seems there was a mistake in the given value of \(2.25\) in the original problem's dropdown, the correct value should be \(8.25\) or \(\frac{33}{4}\)).

Answer:

The correct value of \(x\) is \(\frac{33}{4}\) (or \(8.25\)).