IMG 7907.jpeg - Kuta Software - Infinite Algebra 1 Name ... - Free Printable

Problem Analysis:

The image contains a series of mathematical expressions and equations related to simplifying radical expressions. The task involves solving these problems step by step. Below, I will solve each problem systematically and explain the solution process.

---

Problem 1:

Simplify \( \sqrt{48} \).

#### Solution:
1. Factorize 48 into its prime factors:
\[
48 = 2^4 \cdot 3
\]

2. Rewrite the square root using the factorization:
\[
\sqrt{48} = \sqrt{2^4 \cdot 3}
\]

3. Use the property of square roots that \( \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b} \):
\[
\sqrt{48} = \sqrt{2^4} \cdot \sqrt{3}
\]

4. Simplify \( \sqrt{2^4} \):
\[
\sqrt{2^4} = 2^{4/2} = 2^2 = 4
\]

5. Combine the results:
\[
\sqrt{48} = 4 \cdot \sqrt{3} = 4\sqrt{3}
\]

#### Final Answer:
\[
\boxed{4\sqrt{3}}
\]

---

Problem 2:

Simplify \( \sqrt{75} \).

#### Solution:
1. Factorize 75 into its prime factors:
\[
75 = 3 \cdot 5^2
\]

2. Rewrite the square root using the factorization:
\[
\sqrt{75} = \sqrt{3 \cdot 5^2}
\]

3. Use the property of square roots that \( \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b} \):
\[
\sqrt{75} = \sqrt{5^2} \cdot \sqrt{3}
\]

4. Simplify \( \sqrt{5^2} \):
\[
\sqrt{5^2} = 5
\]

5. Combine the results:
\[
\sqrt{75} = 5 \cdot \sqrt{3} = 5\sqrt{3}
\]

#### Final Answer:
\[
\boxed{5\sqrt{3}}
\]

---

Problem 3:

Simplify \( \sqrt{128} \).

#### Solution:
1. Factorize 128 into its prime factors:
\[
128 = 2^7
\]

2. Rewrite the square root using the factorization:
\[
\sqrt{128} = \sqrt{2^7}
\]

3. Use the property of square roots that \( \sqrt{a^n} = a^{n/2} \):
\[
\sqrt{2^7} = 2^{7/2}
\]

4. Express \( 2^{7/2} \) as \( 2^3 \cdot 2^{1/2} \):
\[
2^{7/2} = 2^3 \cdot 2^{1/2} = 8 \cdot \sqrt{2}
\]

5. Combine the results:
\[
\sqrt{128} = 8\sqrt{2}
\]

#### Final Answer:
\[
\boxed{8\sqrt{2}}
\]

---

Problem 4:

Simplify \( \sqrt{98} \).

#### Solution:
1. Factorize 98 into its prime factors:
\[
98 = 2 \cdot 7^2
\]

2. Rewrite the square root using the factorization:
\[
\sqrt{98} = \sqrt{2 \cdot 7^2}
\]

3. Use the property of square roots that \( \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b} \):
\[
\sqrt{98} = \sqrt{7^2} \cdot \sqrt{2}
\]

4. Simplify \( \sqrt{7^2} \):
\[
\sqrt{7^2} = 7
\]

5. Combine the results:
\[
\sqrt{98} = 7 \cdot \sqrt{2} = 7\sqrt{2}
\]

#### Final Answer:
\[
\boxed{7\sqrt{2}}
\]

---

Problem 5:

Simplify \( \sqrt{162} \).

#### Solution:
1. Factorize 162 into its prime factors:
\[
162 = 2 \cdot 3^4
\]

2. Rewrite the square root using the factorization:
\[
\sqrt{162} = \sqrt{2 \cdot 3^4}
\]

3. Use the property of square roots that \( \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b} \):
\[
\sqrt{162} = \sqrt{3^4} \cdot \sqrt{2}
\]

4. Simplify \( \sqrt{3^4} \):
\[
\sqrt{3^4} = 3^{4/2} = 3^2 = 9
\]

5. Combine the results:
\[
\sqrt{162} = 9 \cdot \sqrt{2} = 9\sqrt{2}
\]

#### Final Answer:
\[
\boxed{9\sqrt{2}}
\]

---

Problem 6:

Simplify \( \sqrt{200} \).

#### Solution:
1. Factorize 200 into its prime factors:
\[
200 = 2^3 \cdot 5^2
\]

2. Rewrite the square root using the factorization:
\[
\sqrt{200} = \sqrt{2^3 \cdot 5^2}
\]

3. Use the property of square roots that \( \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b} \):
\[
\sqrt{200} = \sqrt{5^2} \cdot \sqrt{2^3}
\]

4. Simplify \( \sqrt{5^2} \):
\[
\sqrt{5^2} = 5
\]

5. Simplify \( \sqrt{2^3} \):
\[
\sqrt{2^3} = \sqrt{2^2 \cdot 2} = \sqrt{2^2} \cdot \sqrt{2} = 2\sqrt{2}
\]

6. Combine the results:
\[
\sqrt{200} = 5 \cdot 2\sqrt{2} = 10\sqrt{2}
\]

#### Final Answer:
\[
\boxed{10\sqrt{2}}
\]

---

Problem 7:

Simplify \( \sqrt{18} \).

#### Solution:
1. Factorize 18 into its prime factors:
\[
18 = 2 \cdot 3^2
\]

2. Rewrite the square root using the factorization:
\[
\sqrt{18} = \sqrt{2 \cdot 3^2}
\]

3. Use the property of square roots that \( \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b} \):
\[
\sqrt{18} = \sqrt{3^2} \cdot \sqrt{2}
\]

4. Simplify \( \sqrt{3^2} \):
\[
\sqrt{3^2} = 3
\]

5. Combine the results:
\[
\sqrt{18} = 3 \cdot \sqrt{2} = 3\sqrt{2}
\]

#### Final Answer:
\[
\boxed{3\sqrt{2}}
\]

---

Problem 8:

Simplify \( \sqrt{50} \).

#### Solution:
1. Factorize 50 into its prime factors:
\[
50 = 2 \cdot 5^2
\]

2. Rewrite the square root using the factorization:
\[
\sqrt{50} = \sqrt{2 \cdot 5^2}
\]

3. Use the property of square roots that \( \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b} \):
\[
\sqrt{50} = \sqrt{5^2} \cdot \sqrt{2}
\]

4. Simplify \( \sqrt{5^2} \):
\[
\sqrt{5^2} = 5
\]

5. Combine the results:
\[
\sqrt{50} = 5 \cdot \sqrt{2} = 5\sqrt{2}
\]

#### Final Answer:
\[
\boxed{5\sqrt{2}}
\]

---

Final Answers:

1. \( \sqrt{48} = \boxed{4\sqrt{3}} \)
2. \( \sqrt{75} = \boxed{5\sqrt{3}} \)
3. \( \sqrt{128} = \boxed{8\sqrt{2}} \)
4. \( \sqrt{98} = \boxed{7\sqrt{2}} \)
5. \( \sqrt{162} = \boxed{9\sqrt{2}} \)
6. \( \sqrt{200} = \boxed{10\sqrt{2}} \)
7. \( \sqrt{18} = \boxed{3\sqrt{2}} \)
8. \( \sqrt{50} = \boxed{5\sqrt{2}} \)

---

Boxed Final Answer:

\[
\boxed{4\sqrt{3}, 5\sqrt{3}, 8\sqrt{2}, 7\sqrt{2}, 9\sqrt{2}, 10\sqrt{2}, 3\sqrt{2}, 5\sqrt{2}}
\]