Question

write the expression in simplest form.
$\frac{sqrt{6}}{sqrt{3}-sqrt{5}} = \frac{square}{square}$

Explanation:

Step1: Rationalize the denominator

Multiply numerator and denominator by the conjugate $\sqrt{3}+\sqrt{5}$:
$\frac{\sqrt{6}}{\sqrt{3}-\sqrt{5}} \times \frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}} = \frac{\sqrt{6}(\sqrt{3}+\sqrt{5})}{(\sqrt{3})^2 - (\sqrt{5})^2}$

Step2: Simplify the denominator

Calculate the difference of squares:
$(\sqrt{3})^2 - (\sqrt{5})^2 = 3 - 5 = -2$

Step3: Expand the numerator

Distribute $\sqrt{6}$ across the terms:
$\sqrt{6}\times\sqrt{3} + \sqrt{6}\times\sqrt{5} = \sqrt{18} + \sqrt{30}$

Step4: Simplify radical terms

Simplify $\sqrt{18}$ to $3\sqrt{2}$:
$\sqrt{18} + \sqrt{30} = 3\sqrt{2} + \sqrt{30}$

Step5: Combine and simplify the fraction

Divide each term in the numerator by $-2$:
$\frac{3\sqrt{2} + \sqrt{30}}{-2} = -\frac{3\sqrt{2} + \sqrt{30}}{2}$

Answer:

$-\frac{3\sqrt{2} + \sqrt{30}}{2}$