Q. 8.126 - Intermediate Algebra

Question

Solve: $\sqrt[3]{7x+1}=\sqrt[3]{2x-5}$

Step-by-Step Solution

Answer

The solution is $x=-1.2$.

Step 1. Remove the radical

Consider the equation $\sqrt[3]{7x+1}=\sqrt[3]{2x-5}$.

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

$\begin{array}{rcl}{\left(\sqrt[3]{7x+1}\right)}^3& =& {\left(\sqrt[3]{2x-5}\right)}^3 \\ 7x+1& =& 2x-5\end{array}$

Step 2. Simplify and solve the new equation

Add $-2x-1$ both sides.

$\begin{array}{rcl}7x+1-2x-1& =& 2x-5-2x-1\\ 7x-2x& =& -5-1\\ 5x& =& -6\end{array}$

Divide both sides by $5$ and solve for $x$.

$\begin{array}{rcl}\frac{5x}{5}& =& \frac{-6}{5}\\ x& =& -1.2\end{array}$

Step 3. Check the solution

Substitute $-1.2$ for $x$ in the original equation.

$\begin{array}{rcl}\sqrt[3]{7x+1}& =& \sqrt[3]{2x-5}\\ \sqrt[3]{7\times (-1.2)+1}& =& \sqrt[3]{2\times (-1.2)-5}\\ \sqrt[3]{-8.4+1}& =& \sqrt[3]{-2.4-5}\\ \sqrt[3]{-7.4}& =& \sqrt[3]{-7.4}\end{array}$

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