Question 423052
{{{sqrt(11)/(sqrt(33)-sqrt(5))}}}
Rationalizing a two-term denominator takes advantage of the pattern:
{{{(a+b)(a-b) = a^2 - b^2}}}
Notice that the left side is a product of two-term expressions, like your denominator. And the result, the right side, is made up of nothing but perfect squares.<br>
Your denominator has a "-" between the two terms. SO it will play the role fo (a-b) with the "a" being {{{sqrt(33)}}} and the "b" being {{{sqrt(5)}}}. To get the prefect squares, we need to multiply the (a-b) by (a+b). And if we multiply the denominator by something we must also multiply the numerator by the same thing. So we will multiply the numerator and denominator by (a+b), i.e. {{{(sqrt(33)+sqrt(5))}}}:
{{{(sqrt(11)/(sqrt(33)-sqrt(5)))((sqrt(33)+sqrt(5))/(sqrt(33)+sqrt(5)))}}}
We use the distributive Property to multiply the numerators. In the denominators we know how it will multiply because of the pattern we're using:
{{{(sqrt(363)+sqrt(55))/((sqrt(33))^2-(sqrt(5))^2)}}}
which simplifies as follows:
{{{(sqrt(121*3)+sqrt(55))/(33-5)}}}
{{{(sqrt(121)*sqrt(3)+sqrt(55))/28}}}
{{{(11sqrt(3)+sqrt(55))/28}}}