# How do you solve root3(4x-2) = 2root3(x+6)?

Jul 30, 2016

$x = - \frac{25}{2}$

#### Explanation:

First, we need to get rid of the cube roots. We can do this by cubing, or taking each side to the third power.

$\sqrt[3]{4 x - 2} = 2 \sqrt[3]{x + 6}$

${\left(\sqrt[3]{4 x - 2}\right)}^{3} = {\left(2 \sqrt[3]{x + 6}\right)}^{3}$

Now simplify.

$4 x - 2 = \left({2}^{3}\right) \left({\left(\sqrt[3]{x + 6}\right)}^{3}\right)$

$4 x - 2 = 8 \left(x + 6\right)$

Distribute the $8$ across the parenthesis on the right side of the equal sign.

$4 x - 2 = 8 x + 48$

Now isolate the variable by getting all $x$ terms to one side. Subtract $8 x$ from both sides and add $2$ to both sides.

$4 x - 8 x \cancel{- 2 + 2} = \cancel{8 x - 8 x} + 48 + 2$

$4 x - 8 x = 48 + 2$

Combine like terms and finish isolating the variable.

$- 4 x = 50$

$x = - \frac{50}{4}$

Simplify the fraction.

$x = - \frac{25}{2}$