Question

What is the following quotient? sqr root 6 + sqr root 11/ sqr root 5+ sqr root 3

Answer 1

Answer:

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

Step-by-step explanation:

The given expression is

$\frac{\sqrt{6}+\sqrt{11}}{\sqrt{5}+\sqrt{3} }$

To solve this quotient, we just have to apply a rationalization, which consists in eliminating every root in the denominator. To do so, we multiply and divide the expression by the opposite binomial of the denominator, as follows

$\frac{\sqrt{6}+\sqrt{11}}{\sqrt{5}+\sqrt{3} }=\frac{\sqrt{5}-\sqrt{3} }{\sqrt{5}-\sqrt{3} }\\\\\frac{(\sqrt{6}+\sqrt{11})(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}\\\\\frac{\sqrt{30}-\sqrt{18}+\sqrt{55}-\sqrt{33} }{5-3}\\ \\\frac{\sqrt{30}-3\sqrt{2}+\sqrt{55}-\sqrt{33} }{2}$

Therefore, the right answer is the second option.

Answer 2

The answer is B, the second choice.