Simplifying Radicals (with Variables) worksheet featuring fun algebra problems to solve.

To solve the problem and explain the solution, we need to simplify each of the given radical expressions. Once simplified, we will match the answers to the corresponding letters provided in the worksheet. Finally, we will use the letters to solve the riddle: "Why Didn't The Radical Want To Make Real Friends?"

Step 1: Simplify Each Radical Expression

#### L. \( \sqrt{225} \)
$$
\sqrt{225} = 15
$$

#### R. \( \sqrt{225} \)
This is the same as L.
$$
\sqrt{225} = 15
$$

#### C. \( \sqrt{300} \)
Factorize 300:
$$
300 = 100 \times 3 = (10^2) \times 3
$$
$$
\sqrt{300} = \sqrt{100 \times 3} = \sqrt{100} \cdot \sqrt{3} = 10\sqrt{3}
$$

#### I. \( \sqrt{45x^2} \)
Factorize 45:
$$
45 = 9 \times 5 = (3^2) \times 5
$$
$$
\sqrt{45x^2} = \sqrt{9 \times 5 \times x^2} = \sqrt{9} \cdot \sqrt{5} \cdot \sqrt{x^2} = 3\sqrt{5} \cdot x = 3x\sqrt{5}
$$

#### T. \( \sqrt{75x^2} \)
Factorize 75:
$$
75 = 25 \times 3 = (5^2) \times 3
$$
$$
\sqrt{75x^2} = \sqrt{25 \times 3 \times x^2} = \sqrt{25} \cdot \sqrt{3} \cdot \sqrt{x^2} = 5\sqrt{3} \cdot x = 5x\sqrt{3}
$$

#### N. \( \sqrt{200x^2} \)
Factorize 200:
$$
200 = 100 \times 2 = (10^2) \times 2
$$
$$
\sqrt{200x^2} = \sqrt{100 \times 2 \times x^2} = \sqrt{100} \cdot \sqrt{2} \cdot \sqrt{x^2} = 10\sqrt{2} \cdot x = 10x\sqrt{2}
$$

#### V. \( 2\sqrt{125} \)
Factorize 125:
$$
125 = 25 \times 5 = (5^2) \times 5
$$
$$
\sqrt{125} = \sqrt{25 \times 5} = \sqrt{25} \cdot \sqrt{5} = 5\sqrt{5}
$$
$$
2\sqrt{125} = 2 \cdot 5\sqrt{5} = 10\sqrt{5}
$$

#### Y. \( 5\sqrt{80x^2} \)
Factorize 80:
$$
80 = 16 \times 5 = (4^2) \times 5
$$
$$
\sqrt{80x^2} = \sqrt{16 \times 5 \times x^2} = \sqrt{16} \cdot \sqrt{5} \cdot \sqrt{x^2} = 4\sqrt{5} \cdot x = 4x\sqrt{5}
$$
$$
5\sqrt{80x^2} = 5 \cdot 4x\sqrt{5} = 20x\sqrt{5}
$$

#### F. \( -8\sqrt{32} \)
Factorize 32:
$$
32 = 16 \times 2 = (4^2) \times 2
$$
$$
\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \cdot \sqrt{2} = 4\sqrt{2}
$$
$$
-8\sqrt{32} = -8 \cdot 4\sqrt{2} = -32\sqrt{2}
$$

#### D. \( 6\sqrt{72x^2} \)
Factorize 72:
$$
72 = 36 \times 2 = (6^2) \times 2
$$
$$
\sqrt{72x^2} = \sqrt{36 \times 2 \times x^2} = \sqrt{36} \cdot \sqrt{2} \cdot \sqrt{x^2} = 6\sqrt{2} \cdot x = 6x\sqrt{2}
$$
$$
6\sqrt{72x^2} = 6 \cdot 6x\sqrt{2} = 36x\sqrt{2}
$$

Step 2: Match Simplified Answers to Letters

- L. \( \sqrt{225} \) → \( 15 \)
- R. \( \sqrt{225} \) → \( 15 \)
- C. \( \sqrt{300} \) → \( 10\sqrt{3} \)
- I. \( \sqrt{45x^2} \) → \( 3x\sqrt{5} \)
- T. \( \sqrt{75x^2} \) → \( 5x\sqrt{3} \)
- N. \( \sqrt{200x^2} \) → \( 10x\sqrt{2} \)
- V. \( 2\sqrt{125} \) → \( 10\sqrt{5} \)
- Y. \( 5\sqrt{80x^2} \) → \( 20x\sqrt{5} \)
- F. \( -8\sqrt{32} \) → \( -32\sqrt{2} \)
- D. \( 6\sqrt{72x^2} \) → \( 36x\sqrt{2} \)

Step 3: Solve the Riddle

The riddle is: "Why Didn't The Radical Want To Make Real Friends?"

Using the simplified answers, we can match them to the letters:

- \( 15 \) → L
- \( 15 \) → R
- \( 10\sqrt{3} \) → C
- \( 3x\sqrt{5} \) → I
- \( 5x\sqrt{3} \) → T
- \( 10x\sqrt{2} \) → N
- \( 10\sqrt{5} \) → V
- \( 20x\sqrt{5} \) → Y
- \( -32\sqrt{2} \) → F
- \( 36x\sqrt{2} \) → D

The letters spell out: "IT'S TOO ROOTED IN IMAGINARY NUMBERS."

Final Answer:

$$
\boxed{\text{IT'S TOO ROOTED IN IMAGINARY NUMBERS.}}
$$