# How do you solve (3x+4)^(1/3) = -5 and find any extraneous solutions?

Feb 25, 2017

#### Answer:

$x = - 43$

#### Explanation:

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} {x}^{\frac{1}{3}} = \sqrt[3]{x}$

$\Rightarrow {\left(3 x + 4\right)}^{\frac{1}{3}} = \sqrt[3]{3 x + 4}$

To 'undo' the cube root, $\textcolor{b l u e}{\text{cube both sides}}$ of the equation.

$\Rightarrow {\left({\left(3 x + 4\right)}^{\frac{1}{3}}\right)}^{3} = {\left(- 5\right)}^{3}$

$\Rightarrow 3 x + 4 = - 125$

subtract 4 from both sides.

$3 x \cancel{+ 4} \cancel{- 4} = - 125 - 4$

$\Rightarrow 3 x = - 129$

divide both sides by 3

$\frac{\cancel{3} x}{\cancel{3}} = \frac{- 129}{3}$

$\Rightarrow x = - 43$

$\textcolor{b l u e}{\text{As a check}}$

Substitute this value into the left side of the equation and if equal to the right side then it is the solution.

$\text{left side } = {\left(\left(3 \times - 43\right) + 4\right)}^{\frac{1}{3}} = {\left(- 125\right)}^{\frac{1}{3}} = - 5$

$\Rightarrow x = - 43 \text{ is the only solution}$

There are no extraneous solutions.