Question

Add and subtractin radical √24+3√135+√96

Answer 1

To simplify the expression \(\sqrt{24} + 3\sqrt{135} + \sqrt{96}\), we first simplify each radical:

1. **Simplifying \(\sqrt{24}\)**:
\[
\sqrt{24} = \sqrt{4 \times 6} = \sqrt{4} \times \sqrt{6} = 2\sqrt{6}
\]

2. **Simplifying \(3\sqrt{135}\)**:
\[
\sqrt{135} = \sqrt{9 \times 15} = \sqrt{9} \times \sqrt{15} = 3\sqrt{15}
\]
Thus,
\[
3\sqrt{135} = 3 \times 3\sqrt{15} = 9\sqrt{15}
\]

3. **Simplifying \(\sqrt{96}\)**:
\[
\sqrt{96} = \sqrt{16 \times 6} = \sqrt{16} \times \sqrt{6} = 4\sqrt{6}
\]

Now, substituting back into the expression, we have:
\[
\sqrt{24} + 3\sqrt{135} + \sqrt{96} = 2\sqrt{6} + 9\sqrt{15} + 4\sqrt{6}
\]

Next, we can combine like terms:
\[
(2\sqrt{6} + 4\sqrt{6}) + 9\sqrt{15} = 6\sqrt{6} + 9\sqrt{15}
\]

Thus, the simplified expression is:
\[
\boxed{6\sqrt{6} + 9\sqrt{15}}
\]