## Algebra 1: Common Core (15th Edition)

a) 8 b) $\pm4$ c) -11 d) $\pm\frac{1}{6}$
a) $8^2=64$, so $\sqrt 64 = 8$ b) $4^2=16$, so $\pm\sqrt 16 = \pm4$ a) $11^2=121$, so $-\sqrt {121} = -11$ a) $(\frac{1}{6})^2=\frac{1}{36}$, so $\pm\sqrt \frac{1}{36} = \pm\frac{1}{6}$