## Intermediate Algebra (12th Edition)

$x=\pm\dfrac{5\sqrt{a}}{3}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $9x^2-25a=0 ,$ isolate first the squared variable. Then take the square root of both sides (Square Root Property) and simplify the resulting radical. $\bf{\text{Solution Details:}}$ Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} 9x^2=25a \\\\ x^2=\dfrac{25a}{9} .\end{array} Taking the square root of both sides and then simplifying the radical, the equation above is equivalent to \begin{array}{l}\require{cancel} x=\pm\sqrt{\dfrac{25a}{9}} \\\\ x=\pm\sqrt{\dfrac{25}{9}\cdot a} \\\\ x=\pm\sqrt{\left( \dfrac{5}{3} \right)^2\cdot a} \\\\ x=\pm\dfrac{5}{3}\sqrt{a} \\\\ x=\pm\dfrac{5\sqrt{a}}{3} .\end{array}