Question

Simplify √6x(√8x +√12xy) to its simplest form

Answer 1

To simplify the expression \( \sqrt{6x}(\sqrt{8x} + \sqrt{12xy}) \), we will follow these steps:

1. **Simplify each square root:**

- For \( \sqrt{8x} \):
\[
\sqrt{8x} = \sqrt{4 \cdot 2 \cdot x} = \sqrt{4} \cdot \sqrt{2} \cdot \sqrt{x} = 2\sqrt{2x}
\]

- For \( \sqrt{12xy} \):
\[
\sqrt{12xy} = \sqrt{4 \cdot 3 \cdot x \cdot y} = \sqrt{4} \cdot \sqrt{3} \cdot \sqrt{x} \cdot \sqrt{y} = 2\sqrt{3xy}
\]

2. **Substitute the simplified roots back into the expression:**
\[
\sqrt{6x}(\sqrt{8x} + \sqrt{12xy}) = \sqrt{6x}(2\sqrt{2x} + 2\sqrt{3xy})
\]
\[
= \sqrt{6x} \cdot 2(\sqrt{2x} + \sqrt{3xy})
\]

3. **Distribute \( \sqrt{6x} \):**
\[
= 2\sqrt{6x}(\sqrt{2x} + \sqrt{3xy})
\]

4. **Now simplify \( \sqrt{6x} \cdot \sqrt{2x} \) and \( \sqrt{6x} \cdot \sqrt{3xy} \):**
- For \( \sqrt{6x} \cdot \sqrt{2x} \):
\[
\sqrt{6x \cdot 2x} = \sqrt{12x^2} = \sqrt{4 \cdot 3 \cdot x^2} = 2x\sqrt{3}
\]

- For \( \sqrt{6x} \cdot \sqrt{3xy} \):
\[
\sqrt{6x \cdot 3xy} = \sqrt{18x^2y} = \sqrt{9 \cdot 2 \cdot x^2 \cdot y} = 3x\sqrt{2y}
\]

5. **Combining the results:**
\[
2(2x\sqrt{3} + 3x\sqrt{2y}) = 4x\sqrt{3} + 6x\sqrt{2y}
\]

So, the final simplified expression is:
\[
\boxed{4x\sqrt{3} + 6x\sqrt{2y}}
\]