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what is the following quotient? \\(\\frac{\\sqrt{6} + \\sqrt{11}}{\\sqr…

Question

what is the following quotient?
\\(\frac{\sqrt{6} + \sqrt{11}}{\sqrt{5} + \sqrt{3}}\\)
\\(\circ\\) \\(\frac{\sqrt{30} + 3\sqrt{2} + \sqrt{55} + \sqrt{33}}{8}\\)
\\(\circ\\) \\(\frac{\sqrt{30} - 3\sqrt{2} + \sqrt{55} - \sqrt{33}}{2}\\)
\\(\circ\\) \\(\frac{17}{8}\\)
\\(\circ\\) \\(-\frac{5}{2}\\)

Explanation:

Step1: Rationalize the denominator

To rationalize the denominator \(\sqrt{5} + \sqrt{3}\), we multiply the numerator and the denominator by the conjugate of the denominator, which is \(\sqrt{5} - \sqrt{3}\).

$$ \frac{\sqrt{6}+\sqrt{11}}{\sqrt{5}+\sqrt{3}}\times\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{(\sqrt{6}+\sqrt{11})(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})} $$

Step2: Expand the numerator and the denominator

First, expand the numerator using the distributive property (FOIL method):

$$ LATEXBLOCK0 $$

Simplify \(\sqrt{18}\): \(\sqrt{18} = 3\sqrt{2}\), so the numerator becomes \(\sqrt{30}-3\sqrt{2}+\sqrt{55}-\sqrt{33}\)

Next, expand the denominator using the difference of squares formula \((a + b)(a - b)=a^{2}-b^{2}\):

$$ (\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=(\sqrt{5})^{2}-(\sqrt{3})^{2}=5 - 3=2 $$

Step3: Write the final expression

Put the simplified numerator and denominator together:

$$ \frac{\sqrt{30}-3\sqrt{2}+\sqrt{55}-\sqrt{33}}{2} $$

Answer:

\(\frac{\sqrt{30}-3\sqrt{2}+\sqrt{55}-\sqrt{33}}{2}\) (corresponding to the second option: \(\frac{\sqrt{30}-3\sqrt{2}+\sqrt{55}-\sqrt{33}}{2}\))