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 = 8.25

Explanation:

Step1: Cube both sides

To eliminate the cube root, we cube both sides of the equation \(\sqrt[3]{8x - 2}=4\). Cubing a cube root cancels out the radical. So we have \((\sqrt[3]{8x - 2})^3 = 4^3\), which simplifies to \(8x - 2 = 64\).

Step2: Add 2 to both sides

Now we solve for \(x\). First, add 2 to both sides of the equation \(8x - 2 = 64\). This gives \(8x - 2 + 2 = 64 + 2\), so \(8x = 66\).

Step3: Divide by 8

Then divide both sides by 8: \(x=\frac{66}{8}=\frac{33}{4} = 8.25\).

Answer:

To solve \(\sqrt[3]{8x - 2}=4\), first \(\boldsymbol{\text{cube}}\) both sides (to eliminate the cube root), then add 2 to both sides, and finally divide by 8. The value of \(x\) is \(\boldsymbol{8.25}\).