# How do you solve 5root3(4x-1)+2=17?

Jun 1, 2016

$x = 7$

#### Explanation:

$5 \times \sqrt[3]{4 x - 1} + 2 = 17$

Here, we try to keep the cube root term on one side.

Subtract $2$ from both sides:

$5 \times \sqrt[3]{4 x - 1} + 2 \textcolor{red}{- 2} = 17 \textcolor{red}{- 2}$

$5 \times \sqrt[3]{4 x - 1} = 15$

Next, divide both sides by $5$:

$\frac{5 \times \sqrt[3]{4 x - 1}}{\textcolor{red}{5}} = \frac{15}{\textcolor{red}{5}}$

$\sqrt[3]{4 x - 1} = 3$

Now, we remove the cube root.
In order to do so, find the cube of both sides:

${\left(\sqrt[3]{4 x - 1}\right)}^{\textcolor{red}{3}} = {3}^{\textcolor{red}{3}}$

$4 x - 1 = 3 \times 3 \times 3$

$4 x - 1 = 27$

$4 x - 1 \textcolor{red}{+ 1} = 27 \textcolor{red}{+ 1}$

$4 x = 28$

$\frac{4 x}{\textcolor{red}{4}} = \frac{28}{\textcolor{red}{4}}$

$x = 7$