## College Algebra (6th Edition)

The solutions are $x=\pm2$
$4x^{2}-16=0$ Take $16$ to the right side: $4x^{2}=16$ Take the $4$ to divide the right side: $x^{2}=\dfrac{16}{4}$ $x^{2}=4$ Take the square root of both sides of the equation: $\sqrt{x^{2}}=\pm\sqrt{4}$ $x=\pm2$