Simplifying Radicals worksheet with exercises for multiplying and dividing square roots.

Problem: Simplifying Radicals (B)

The task involves simplifying expressions involving square roots. We will solve each part step by step.

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#### Section A: Simplify the following (Multiplication of Radicals)

The general rule for multiplying radicals is:
\[
\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}
\]

1. \(\sqrt{6} \times \sqrt{2}\)
\[
\sqrt{6} \times \sqrt{2} = \sqrt{6 \times 2} = \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}
\]

2. \(\sqrt{20} \times \sqrt{5}\)
\[
\sqrt{20} \times \sqrt{5} = \sqrt{20 \times 5} = \sqrt{100} = 10
\]

3. \(\sqrt{7} \times \sqrt{7}\)
\[
\sqrt{7} \times \sqrt{7} = \sqrt{7 \times 7} = \sqrt{49} = 7
\]

4. \(\sqrt{2} \times \sqrt{72}\)
\[
\sqrt{2} \times \sqrt{72} = \sqrt{2 \times 72} = \sqrt{144} = 12
\]

5. \(\sqrt{3} \times \sqrt{3}\)
\[
\sqrt{3} \times \sqrt{3} = \sqrt{3 \times 3} = \sqrt{9} = 3
\]

6. \(\sqrt{4} \times \sqrt{16}\)
\[
\sqrt{4} \times \sqrt{16} = \sqrt{4 \times 16} = \sqrt{64} = 8
\]

7. \(\sqrt{4} \times \sqrt{5} \times \sqrt{5}\)
\[
\sqrt{4} \times \sqrt{5} \times \sqrt{5} = \sqrt{4 \times 5 \times 5} = \sqrt{4 \times 25} = \sqrt{100} = 10
\]

8. \(\sqrt{2} \times \sqrt{4} \times \sqrt{2}\)
\[
\sqrt{2} \times \sqrt{4} \times \sqrt{2} = \sqrt{2 \times 4 \times 2} = \sqrt{16} = 4
\]

9. \(\sqrt{4} \times \sqrt{4} \times \sqrt{4}\)
\[
\sqrt{4} \times \sqrt{4} \times \sqrt{4} = \sqrt{4 \times 4 \times 4} = \sqrt{64} = 8
\]

10. \(\sqrt{3} \times \sqrt{6} \times \sqrt{3}\)
\[
\sqrt{3} \times \sqrt{6} \times \sqrt{3} = \sqrt{3 \times 6 \times 3} = \sqrt{54} = \sqrt{9 \times 6} = 3\sqrt{6}
\]

11. \(\sqrt{6} \times \sqrt{12} \times \sqrt{6}\)
\[
\sqrt{6} \times \sqrt{12} \times \sqrt{6} = \sqrt{6 \times 12 \times 6} = \sqrt{432} = \sqrt{144 \times 3} = 12\sqrt{3}
\]

12. \(\sqrt{2} \times \sqrt{12} \times \sqrt{3}\)
\[
\sqrt{2} \times \sqrt{12} \times \sqrt{3} = \sqrt{2 \times 12 \times 3} = \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}
\]

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#### Section B: Simplify the following (Division of Radicals)

The general rule for dividing radicals is:
\[
\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}
\]

1. \(\sqrt{3} \div \sqrt{3}\)
\[
\sqrt{3} \div \sqrt{3} = \sqrt{\frac{3}{3}} = \sqrt{1} = 1
\]

2. \(\sqrt{15} \div \sqrt{5}\)
\[
\sqrt{15} \div \sqrt{5} = \sqrt{\frac{15}{5}} = \sqrt{3}
\]

3. \(\sqrt{8} \div \sqrt{2}\)
\[
\sqrt{8} \div \sqrt{2} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2
\]

4. \(\sqrt{27} \div \sqrt{3}\)
\[
\sqrt{27} \div \sqrt{3} = \sqrt{\frac{27}{3}} = \sqrt{9} = 3
\]

5. \(\sqrt{48} \div \sqrt{3}\)
\[
\sqrt{48} \div \sqrt{3} = \sqrt{\frac{48}{3}} = \sqrt{16} = 4
\]

6. \(\sqrt{54} \div \sqrt{9}\)
\[
\sqrt{54} \div \sqrt{9} = \sqrt{\frac{54}{9}} = \sqrt{6}
\]

7. \(\sqrt{96} \div \sqrt{12}\)
\[
\sqrt{96} \div \sqrt{12} = \sqrt{\frac{96}{12}} = \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}
\]

8. \(\sqrt{72} \div \sqrt{18}\)
\[
\sqrt{72} \div \sqrt{18} = \sqrt{\frac{72}{18}} = \sqrt{4} = 2
\]

9. \(\sqrt{48} \div \sqrt{8}\)
\[
\sqrt{48} \div \sqrt{8} = \sqrt{\frac{48}{8}} = \sqrt{6}
\]

10. \(\sqrt{5} \times \sqrt{6} \div \sqrt{2}\)
\[
\sqrt{5} \times \sqrt{6} \div \sqrt{2} = \sqrt{\frac{5 \times 6}{2}} = \sqrt{\frac{30}{2}} = \sqrt{15}
\]

11. \(\sqrt{32} \times \sqrt{10} \div \sqrt{5}\)
\[
\sqrt{32} \times \sqrt{10} \div \sqrt{5} = \sqrt{\frac{32 \times 10}{5}} = \sqrt{\frac{320}{5}} = \sqrt{64} = 8
\]

12. \(\sqrt{24} \times \sqrt{2} \div \sqrt{3}\)
\[
\sqrt{24} \times \sqrt{2} \div \sqrt{3} = \sqrt{\frac{24 \times 2}{3}} = \sqrt{\frac{48}{3}} = \sqrt{16} = 4
\]

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Final Answers:

Section A:
1. \(2\sqrt{3}\)
2. \(10\)
3. \(7\)
4. \(12\)
5. \(3\)
6. \(8\)
7. \(10\)
8. \(4\)
9. \(8\)
10. \(3\sqrt{6}\)
11. \(12\sqrt{3}\)
12. \(6\sqrt{2}\)

Section B:
1. \(1\)
2. \(\sqrt{3}\)
3. \(2\)
4. \(3\)
5. \(4\)
6. \(\sqrt{6}\)
7. \(2\sqrt{2}\)
8. \(2\)
9. \(\sqrt{6}\)
10. \(\sqrt{15}\)
11. \(8\)
12. \(4\)

\boxed{
\begin{array}{ll}
\text{Section A:} & 2\sqrt{3}, 10, 7, 12, 3, 8, 10, 4, 8, 3\sqrt{6}, 12\sqrt{3}, 6\sqrt{2} \\
\text{Section B:} & 1, \sqrt{3}, 2, 3, 4, \sqrt{6}, 2\sqrt{2}, 2, \sqrt{6}, \sqrt{15}, 8, 4
\end{array}
}