Multiplying Radical Expressions worksheet with 32 practice problems for simplifying square root expressions.

Problem: Simplify the given radical expressions.

The task involves simplifying various expressions involving square roots. We will solve each problem step by step, using the properties of radicals and basic algebraic techniques.

---

#### 1. \(\sqrt{5} \times \sqrt{5}\)
Using the property \(\sqrt{a} \times \sqrt{b} = \sqrt{a \cdot b}\):
\[
\sqrt{5} \times \sqrt{5} = \sqrt{5 \cdot 5} = \sqrt{25} = 5
\]

Answer: \(5\)

---

#### 2. \(\sqrt{5} \times \sqrt{10}\)
Using the property \(\sqrt{a} \times \sqrt{b} = \sqrt{a \cdot b}\):
\[
\sqrt{5} \times \sqrt{10} = \sqrt{5 \cdot 10} = \sqrt{50}
\]
Simplify \(\sqrt{50}\):
\[
\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2} = 5\sqrt{2}
\]

Answer: \(5\sqrt{2}\)

---

#### 3. \(\sqrt{2} \times \sqrt{18}\)
Using the property \(\sqrt{a} \times \sqrt{b} = \sqrt{a \cdot b}\):
\[
\sqrt{2} \times \sqrt{18} = \sqrt{2 \cdot 18} = \sqrt{36}
\]
Simplify \(\sqrt{36}\):
\[
\sqrt{36} = 6
\]

Answer: \(6\)

---

#### 4. \(\sqrt{14} \times \sqrt{21}\)
Using the property \(\sqrt{a} \times \sqrt{b} = \sqrt{a \cdot b}\):
\[
\sqrt{14} \times \sqrt{21} = \sqrt{14 \cdot 21} = \sqrt{294}
\]
Simplify \(\sqrt{294}\):
\[
\sqrt{294} = \sqrt{49 \cdot 6} = \sqrt{49} \cdot \sqrt{6} = 7\sqrt{6}
\]

Answer: \(7\sqrt{6}\)

---

#### 5. \(\sqrt{5} \times -4\sqrt{20}\)
First, simplify \(\sqrt{20}\):
\[
\sqrt{20} = \sqrt{4 \cdot 5} = \sqrt{4} \cdot \sqrt{5} = 2\sqrt{5}
\]
Now substitute back:
\[
\sqrt{5} \times -4\sqrt{20} = \sqrt{5} \times -4(2\sqrt{5}) = \sqrt{5} \times -8\sqrt{5}
\]
Using the property \(\sqrt{a} \times \sqrt{b} = \sqrt{a \cdot b}\):
\[
\sqrt{5} \times -8\sqrt{5} = -8(\sqrt{5} \cdot \sqrt{5}) = -8(\sqrt{25}) = -8 \cdot 5 = -40
\]

Answer: \(-40\)

---

#### 6. \(3\sqrt{12} \times \sqrt{6}\)
First, simplify \(\sqrt{12}\):
\[
\sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3}
\]
Now substitute back:
\[
3\sqrt{12} \times \sqrt{6} = 3(2\sqrt{3}) \times \sqrt{6} = 6\sqrt{3} \times \sqrt{6}
\]
Using the property \(\sqrt{a} \times \sqrt{b} = \sqrt{a \cdot b}\):
\[
6\sqrt{3} \times \sqrt{6} = 6\sqrt{3 \cdot 6} = 6\sqrt{18}
\]
Simplify \(\sqrt{18}\):
\[
\sqrt{18} = \sqrt{9 \cdot 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2}
\]
Thus:
\[
6\sqrt{18} = 6 \cdot 3\sqrt{2} = 18\sqrt{2}
\]

Answer: \(18\sqrt{2}\)

---

#### 7. \(5\sqrt{42} \times \sqrt{3}\)
Using the property \(\sqrt{a} \times \sqrt{b} = \sqrt{a \cdot b}\):
\[
5\sqrt{42} \times \sqrt{3} = 5\sqrt{42 \cdot 3} = 5\sqrt{126}
\]
Simplify \(\sqrt{126}\):
\[
\sqrt{126} = \sqrt{9 \cdot 14} = \sqrt{9} \cdot \sqrt{14} = 3\sqrt{14}
\]
Thus:
\[
5\sqrt{126} = 5 \cdot 3\sqrt{14} = 15\sqrt{14}
\]

Answer: \(15\sqrt{14}\)

---

#### 8. \(\sqrt{3} \times -\sqrt{25}\)
Simplify \(\sqrt{25}\):
\[
\sqrt{25} = 5
\]
Now substitute back:
\[
\sqrt{3} \times -\sqrt{25} = \sqrt{3} \times -5 = -5\sqrt{3}
\]

Answer: \(-5\sqrt{3}\)

---

#### 9. \(\sqrt{99} \times \sqrt{48}\)
Using the property \(\sqrt{a} \times \sqrt{b} = \sqrt{a \cdot b}\):
\[
\sqrt{99} \times \sqrt{48} = \sqrt{99 \cdot 48}
\]
Simplify \(99 \cdot 48\):
\[
99 \cdot 48 = 4752
\]
Now simplify \(\sqrt{4752}\):
\[
\sqrt{4752} = \sqrt{144 \cdot 33} = \sqrt{144} \cdot \sqrt{33} = 12\sqrt{33}
\]

Answer: \(12\sqrt{33}\)

---

#### 10. \(5\sqrt{45} \times 3\sqrt{176}\)
First, simplify \(\sqrt{45}\) and \(\sqrt{176}\):
\[
\sqrt{45} = \sqrt{9 \cdot 5} = \sqrt{9} \cdot \sqrt{5} = 3\sqrt{5}
\]
\[
\sqrt{176} = \sqrt{16 \cdot 11} = \sqrt{16} \cdot \sqrt{11} = 4\sqrt{11}
\]
Now substitute back:
\[
5\sqrt{45} \times 3\sqrt{176} = 5(3\sqrt{5}) \times 3(4\sqrt{11}) = 15\sqrt{5} \times 12\sqrt{11}
\]
Using the property \(\sqrt{a} \times \sqrt{b} = \sqrt{a \cdot b}\):
\[
15\sqrt{5} \times 12\sqrt{11} = (15 \cdot 12)\sqrt{5 \cdot 11} = 180\sqrt{55}
\]

Answer: \(180\sqrt{55}\)

---

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \ 5 \\
2. & \ 5\sqrt{2} \\
3. & \ 6 \\
4. & \ 7\sqrt{6} \\
5. & \ -40 \\
6. & \ 18\sqrt{2} \\
7. & \ 15\sqrt{14} \\
8. & \ -5\sqrt{3} \\
9. & \ 12\sqrt{33} \\
10. & \ 180\sqrt{55}
\end{aligned}
}
\]