## Intermediate Algebra (12th Edition)

$k=\dfrac{1}{3}$
$\bf{\text{Solution Outline:}}$ To solve the given radical equation, $\sqrt{6k-1}=1 ,$ square both sides of the equation and then isolate the variable. Then, do checking of the solution with the original equation. $\bf{\text{Solution Details:}}$ Squaring both sides of the equation results to \begin{array}{l}\require{cancel} \left( \sqrt{6k-1} \right)^2=(1)^2 \\\\ 6k-1=1 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} 6k=1+1 \\\\ 6k=2 \\\\ k=\dfrac{2}{6} \\\\ k=\dfrac{1}{3} .\end{array} Upon checking, $k=\dfrac{1}{3}$ satisfies the original equation.