## Intermediate Algebra for College Students (7th Edition)

$(x - 1)^2 = 16$ Using the Square Root Property, $u^2 = d$, then $u = \sqrt d$ or $u = - \sqrt d$. In this case, $u^2 = (x-1)^2$ and $d = 16$ Thus, $x-1 = ±\sqrt 16$ $x = 1± 4$ $x = 5$ or $x = -3$
$(x - 1)^2 = 16$ Using the Square Root Property, $u^2 = d$, then $u = \sqrt d$ or $u = - \sqrt d$. In this case, $u^2 = (x-1)^2$ and $d = 16$ Thus, $x-1 = ±\sqrt 16$ $x = 1± 4$ $x = 5$ or $x = -3$