Question

combine the radicals using addition and subtration . Give the answer in simplest radicle form 4* the squar root of 396+the square root of 44- 2 times the square root of 275

Answer 1

To simplify the expression \( 4\sqrt{396} + \sqrt{44} - 2\sqrt{275} \), we'll start by simplifying each radical term.

1. **Simplify \( \sqrt{396} \)**:
- Factor 396 = \( 4 \times 99 = 4 \times 9 \times 11 = (2 \sqrt{9}) \sqrt{11} = 6\sqrt{11} \)
- Therefore, \( \sqrt{396} = 6\sqrt{11} \).

Now we have:
\[
4\sqrt{396} = 4 \times 6\sqrt{11} = 24\sqrt{11}.
\]

2. **Simplify \( \sqrt{44} \)**:
- Factor 44 = \( 4 \times 11 = (2\sqrt{4})\sqrt{11} = 2\sqrt{11} \).
- Therefore, \( \sqrt{44} = 2\sqrt{11} \).

3. **Simplify \( \sqrt{275} \)**:
- Factor 275 = \( 25 \times 11 = (5\sqrt{25})\sqrt{11} = 5\sqrt{11} \).
- Therefore, \( \sqrt{275} = 5\sqrt{11} \).

Now substitute these simplified radicals back into the original expression:
\[
24\sqrt{11} + 2\sqrt{11} - 2(5\sqrt{11}).
\]

4. **Combine like terms**:
\[
24\sqrt{11} + 2\sqrt{11} - 10\sqrt{11} = (24 + 2 - 10)\sqrt{11} = 16\sqrt{11}.
\]

So, the final answer in simplest radical form is:
\[
\boxed{16\sqrt{11}}.
\]