Question 383640
{{{4sqrt(7)/(sqrt(5)-sqrt(3))}}}
Your denominator has two terms. Rationalizing a two term denominator uses the pattern: {{{(a+b)(a-b) = a^2-b^2}}}. The pattern shows us how a binomial, (a+b) or (a-b), can be turned into an expression of perfect squares, {{{a^2-b^2}}}!<br>
Your denominator has a minus between the two terms. So it will plat the role of (a-b) with "a" being {{{sqrt(5)}}} and "b" being {{{sqrt(3)}}}. To rationalize this denominator we will multiply the numerator and denominator by (a+b):
{{{(4sqrt(7)/(sqrt(5)-sqrt(3)))((sqrt(5)+sqrt(3))/(sqrt(5)+sqrt(3)))}}}
In the numerator we will use the distributive Property to multiply. In the denominator we already know we will get {{{a^2=b^2}}}:
{{{(4sqrt(7)*sqrt(5)+4sqrt(7)*sqrt(3))/((sqrt(5))^2-(sqrt(3))^2))}}}
which simplifies as follows:
{{{(4sqrt(35)+4sqrt(21))/(5-3)}}}
You can already see that the denominator is rational.
{{{(4sqrt(35)+4sqrt(21))/2}}}
We can reduce this fraction by factoring out a 2 in the numerator:
{{{(2(2sqrt(35)+2sqrt(21)))/2}}}
{{{(cross(2)(2sqrt(35)+2sqrt(21)))/cross(2)}}}
{{{2sqrt(35)+2sqrt(21)}}}
Not only did we rationalize the denominator but we eliminated the fraction entirely!