Intermediate Algebra (12th Edition)

$x=\pm2\sqrt{b}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $x^2=4b ,$ take the square root of both sides (Square Root Property) and simplify the resulting radical. $\bf{\text{Solution Details:}}$ Taking the square root of both sides (Square Root Property), the equation above is equivalent to \begin{array}{l}\require{cancel} x=\pm\sqrt{4b} \\\\ x=\pm\sqrt{4\cdot b} \\\\ x=\pm\sqrt{(2)^2\cdot b} \\\\ x=\pm2\sqrt{b} .\end{array}