Question 390897
{{{(sqrt(3)-sqrt(6))/(sqrt(3)+sqrt(6))}}}
A rational denominator means that the simplified denominator has no square roots. So we have to change the denominator so that it simplifies down to an expression without square roots.<br>
The primary way to change denominators is to multiply both the numerator and denominator by the same expression. Sowe are trying to think of an expression that, when multiplied by the denominator, we get an expression without square roots. With a multiple term denominator like this one, this is more difficult than when there is just one term in the denominator.<br>
The "trick" is to understand the pattern: {{{(a+b)(a-b) = a^2 - b^2}}}. Since the "a" and the "b" can be <i>anything</i>, this pattern shows us how to take any two term expression and turn it into an expression which has only perfect squares ({{{a^2 - b^2}}}).<br>
In your denominator the "a" is {{{sqrt(3)}}} and the "b" is {{{sqrt(6)}}}. Since there is a "+" between them your denominator is a+b. To get the expression of perfect squares we need to multiply a+b by a-b (i.e. {{{sqrt(3)-sqrt(6)}}}):
{{{((sqrt(3)-sqrt(6))/(sqrt(3)+sqrt(6)))((sqrt(3)-sqrt(6))/(sqrt(3)-sqrt(6)))}}}
We already know what the denominator will become: {{{a^2-b^2}}}. To multiplying the numerators we can use another pattrern: {{{(a-b)(a-b) = a^2 - 2ab + b^2}}} (If you don't like the patterns, just use FOIL.)
{{{((sqrt(3))^2 - 2(sqrt(3))*(sqrt(6)) + (sqrt(6))^2)/((sqrt(3))^2 - (sqrt(6))^2)}}}
which simplifes as follows:
{{{(3 -2sqrt(18) + 6)/(3-6)}}}
{{{(9-2sqrt(18))/(-3)}}}
{{{(9-2sqrt(9*2))/(-3)}}}
{{{(9-2sqrt(9)*sqrt(2))/(-3)}}}
{{{(9-2*3*sqrt(2))/(-3)}}}
{{{(9-6*sqrt(2))/(-3)}}}
{{{(-3(-3+2*sqrt(2)))/(-3)}}}
{{{(cross(-3)(-3+2*sqrt(2)))/cross((-3))}}}
{{{-3+2*sqrt(2)}}}
Not only did we rationalize the denominator, we eliminated it entirely!