## Intermediate Algebra (12th Edition)

$k=-4$
$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $\sqrt[3]{2k-11}=\sqrt[3]{5k+1} ,$ raise both sides of the equation to the third power. Then use properties of equality to isolate and solve the variable. Finally, do checking of the solution/s with the original equation. $\bf{\text{Solution Details:}}$ Raising both sides of the equation to the third power results to \begin{array}{l}\require{cancel} \left( \sqrt[3]{2k-11} \right)^3=\left( \sqrt[3]{5k+1} \right)^3 \\\\ 2k-11=5k+1 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 2k-5k=1+11 \\\\ -3k=12 \\\\ k=\dfrac{12}{-3} \\\\ k=-4 .\end{array} Upon checking, $k=-4$ satisfies the original equation.