Question
Real Tutor Solution
Tutor-Verified Answer
Answer
The simplified form of the expression is \( 3\sqrt{2} - 2\sqrt{3} \).
Solution
To simplify the expression:
\[ \frac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} \]
we can rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate of \(\sqrt{2} + \sqrt{3}\) is \(\sqrt{2} - \sqrt{3}\).
**Step 1: Multiply by the Conjugate**
\[
\frac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} \times \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \frac{\sqrt{6}(\sqrt{2} - \sqrt{3})}{(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})}
\]
**Step 2: Simplify the Denominator**
\[
(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3}) = (\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1
\]
**Step 3: Distribute in the Numerator**
\[
\sqrt{6}(\sqrt{2} - \sqrt{3}) = \sqrt{6} \cdot \sqrt{2} - \sqrt{6} \cdot \sqrt{3} = \sqrt{12} - \sqrt{18}
\]
Simplify the square roots:
\[
\sqrt{12} = 2\sqrt{3} \quad \text{and} \quad \sqrt{18} = 3\sqrt{2}
\]
So,
\[
\sqrt{12} - \sqrt{18} = 2\sqrt{3} - 3\sqrt{2}
\]
**Step 4: Combine Everything**
\[
\frac{2\sqrt{3} - 3\sqrt{2}}{-1} = -2\sqrt{3} + 3\sqrt{2} = 3\sqrt{2} - 2\sqrt{3}
\]
**Final Answer:**
\[
\boxed{3\sqrt{2} - 2\sqrt{3}}
\]
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Simplify this solution Bonus Knowledge
To simplify \( \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} \), you can rationalize the denominator. Multiply the numerator and denominator by the conjugate of the denominator, which is \( \sqrt{2} - \sqrt{3} \): \[ \frac{\sqrt{6}(\sqrt{2} - \sqrt{3})}{(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})} \] The denominator becomes: \[ (\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1 \] The numerator becomes: \[ \sqrt{6}(\sqrt{2} - \sqrt{3}) = \sqrt{12} - \sqrt{18} = 2\sqrt{3} - 3\sqrt{2} \] Putting it all together gives: \[ \frac{2\sqrt{3} - 3\sqrt{2}}{-1} = -2\sqrt{3} + 3\sqrt{2} \] So, the simplified result is: \[ 3\sqrt{2} - 2\sqrt{3} \]
