## Intermediate Algebra (12th Edition)

$\bf{\text{Solution Outline:}}$ Substitute the given value for $x$ in the given equation, $\sqrt{8x-3}-2x=0 .$ If the left side of the equation becomes equal to the right side of the equation, then the given value of $x$ is a solution to the equation. $\bf{\text{Solution Details:}}$ a) Substituting $x$ with $\dfrac{3}{2}$ in the given equation results to \begin{array}{l}\require{cancel} \sqrt{8\left( \dfrac{3}{2} \right)-3}-2\left( \dfrac{3}{2} \right)=0 \\\\ \sqrt{12-3}-3=0 \\\\ \sqrt{9}-3=0 \\\\ 3-3=0 \\\\ 0=0 \text{ (TRUE)} .\end{array} Hence, $x= \dfrac{3}{2}$ is a solution to the given equation. b) Substituting $x$ with $\dfrac{1}{2}$ in the given equation results to \begin{array}{l}\require{cancel} \sqrt{8\left( \dfrac{1}{2} \right)-3}-2\left( \dfrac{1}{2} \right)=0 \\\\ \sqrt{4-3}-1=0 \\\\ \sqrt{1}-1=0 \\\\ 1-1=0 \\\\ 0=0 \text{ (TRUE)} .\end{array} Hence, $x= \dfrac{1}{2}$ is a solution to the given equation.