Question

algebra 2
simplifying radicals review
simplify; show your work to receive full credit... circle your answers.

1. $sqrt{75}$
2. $sqrt{64}$
3. $sqrt{80}$
4. $sqrt{32}$
5. $sqrt{36}$
6. $sqrt{45}$
7. $sqrt{12}$
8. $sqrt{48}$
9. $sqrt{27}$
10. $sqrt{8}$

Explanation:

Step1: Prime - factor the number inside radical

For $\sqrt{75}$, $75 = 25\times3$.

Step2: Use the property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ ($a = 25$, $b = 3$)

$\sqrt{75}=\sqrt{25\times3}=\sqrt{25}\cdot\sqrt{3}=5\sqrt{3}$

For $\sqrt{64}$, since $64 = 8\times8$, then $\sqrt{64}=8$

For $\sqrt{80}$, $80=16\times5$, so $\sqrt{80}=\sqrt{16\times5}=\sqrt{16}\cdot\sqrt{5}=4\sqrt{5}$

For $\sqrt{32}$, $32 = 16\times2$, so $\sqrt{32}=\sqrt{16\times2}=\sqrt{16}\cdot\sqrt{2}=4\sqrt{2}$

For $\sqrt{36}$, since $36=6\times6$, then $\sqrt{36}=6$

For $\sqrt{45}$, $45 = 9\times5$, so $\sqrt{45}=\sqrt{9\times5}=\sqrt{9}\cdot\sqrt{5}=3\sqrt{5}$

For $\sqrt{12}$, $12 = 4\times3$, so $\sqrt{12}=\sqrt{4\times3}=\sqrt{4}\cdot\sqrt{3}=2\sqrt{3}$

For $\sqrt{48}$, $48=16\times3$, so $\sqrt{48}=\sqrt{16\times3}=\sqrt{16}\cdot\sqrt{3}=4\sqrt{3}$

For $\sqrt{27}$, $27 = 9\times3$, so $\sqrt{27}=\sqrt{9\times3}=\sqrt{9}\cdot\sqrt{3}=3\sqrt{3}$

For $\sqrt{8}$, $8 = 4\times2$, so $\sqrt{8}=\sqrt{4\times2}=\sqrt{4}\cdot\sqrt{2}=2\sqrt{2}$

Answer:

1. $5\sqrt{3}$
2. $8$
3. $4\sqrt{5}$
4. $4\sqrt{2}$
5. $6$
6. $3\sqrt{5}$
7. $2\sqrt{3}$
8. $4\sqrt{3}$
9. $3\sqrt{3}$
10. $2\sqrt{2}$