Question

What is the following quotient? StartFraction StartRoot 6 EndRoot StartRoot 11 EndRoot Over StartRoot 5 EndRoot StartRoot 3 EndRoot EndFraction.

Answer 1

You can use the rationalization method in which we multiply the fraction with conjugate of the denominator.

The quotient of the given fraction is given as

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

How to rationalize a fraction?

Suppose the given fraction is  $\dfrac{a}{b + c}$

Then the conjugate of the denominator is given by b - c

Thus, rationalizing the fraction will give us

$\dfrac{a}{b+c} = \dfrac{a}{b+c} \times \dfrac{b-c}{b-c} = \dfrac{a(b-c)}{b^2 - c^2}\\\\\\(since \: \: (x+y)(x-y) = x^2 - y^2 )$

We actually rationalize just for the use of that later denominator or numerator(if they seem to be helpful).

Remember that $\dfrac{b-c}{b-c} = 1$ thus, multiplying it with the fraction doesn't change its value, and just change the way how it looks. We assume that b-c is not 0

Using above method for getting the quotient of the given fraction

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

Simplifying the fraction:

$\dfrac{(\sqrt{6} + \sqrt{11}) \times ( \sqrt{5} - \sqrt{3})}{(\sqrt{5})^2 - (\sqrt{3})^2} = \dfrac{\sqrt{30} + \sqrt{55} - 3\sqrt{2} - \sqrt{33}}{2}$

Thus,

The quotient of the given fraction is given as

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

Learn more about rationalizing fractions here:

brainly.com/question/14261303