Question

(square root of 6+the square root of 11)/(square root of 5+square root of 3) =

Answer 1

It should be 22.880627.

Answer 2

ANSWER

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

EXPLANATION

The given radical expression is

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

We rationalize the denominator to get:

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

This implies that:

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

We expand using the distributive property and also applying difference of two squares on the denominator.

$\frac{( \sqrt{6} \times \sqrt{5} - \sqrt{6} \times \sqrt{3} + \sqrt{5} \times \sqrt{11} - \sqrt{3} \times \sqrt{11} )}{ {( \sqrt{5} )}^{2} - {( \sqrt{3} )}^{2} }$

This will give us,

$\frac{\sqrt{30} - 3\sqrt{2} + \sqrt{55}-\sqrt{33} }{ 5 - 3 }$

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