Q. 8.125 - Intermediate Algebra

Question

Solve: $\sqrt[3]{5x-4}=\sqrt[3]{2x+5}$

Step-by-Step Solution

Answer

The solution is $x=3$.

Step 1. Remove the radical

Consider the equation $\sqrt[3]{5x-4}=\sqrt[3]{2x+5}$.

We take the cube both sides of the equations to remove the radical.

$\begin{array}{rcl}{\left(\sqrt[3]{5x-4}\right)}^3& =& {\left(\sqrt[3]{2x+5}\right)}^3 \\ 5x-4& =& 2x+5\end{array}$

Step 2. Simplify and solve the new equation

Add $4-2x$ both sides

$\begin{array}{rcl}5x-4+4-2x& =& 2x+5+4-2x\\ 5x-2x& =& 5+4\\ 3x& =& 9\end{array}$

Divide both sides by $3$ and solve for $x$.

$\begin{array}{rcl}\frac{3x}{3}& =& \frac{9}{3}\\ x& =& 3\end{array}$

Step 3. Check the solution

Substitute $3$ for $x$ in the original equation.

$\begin{array}{rcl}\sqrt[3]{5x-4}& =& \sqrt[3]{2x+5}\\ \sqrt[3]{5·3-4}& =& \sqrt[3]{2·3+5}\\ \sqrt[3]{15-4}& =& \sqrt[3]{6+5}\\ \sqrt[3]{11}& =& \sqrt[3]{11}\end{array}$

As $LHS=RHS$, so the solution is correct.