# How do you solve and check for extraneous solutions in root3(4x+2) -6 = -10?

Aug 1, 2015

I found: $x = - \frac{33}{2}$

#### Explanation:

I would write it as:
$\sqrt[3]{4 x + 2} = - 10 + 6$
$\sqrt[3]{4 x + 2} = - 4$ apply the power of $3$ on both sides:
$4 x + 2 = {\left(- 4\right)}^{3}$
$4 x + 2 = - 64$
$4 x = - 66$
$x = - \frac{66}{4} = - \frac{33}{2}$

Checking the solution we plug it into the original equaton:
$\sqrt[3]{4 \left(- \frac{33}{2}\right) + 2} - 6 = - 10$
$\sqrt[3]{- 66 + 2} = - 4$
$\sqrt[3]{- 64} = - 4$ true,
considering that: $- 4 \cdot - 4 \cdot - 4 = - 64$