Question 593848
{{{4 / (6-(2) sqrt (6))}}}
First we can reduce the fraction. Factoring a 2 out of both the numerator and denominator we get:
{{{(2*2) / (2*(3-sqrt (6)))}}}
Canceling the factors of 2 we get:
{{{2 / (3- sqrt (6))}}}<br>
Next we need to rationalize the denominator (i.e. eliminate the square root(s) in the denominator). This denominator has two terms. To rationalize a denominator like this we take advantage of the {{{(a+b)(a-b) = a^2 - b^2}}}. Note that the right side of the pattern is made of of perfect square terms! And the left side shows us how to get these perfect square terms. You denominator, {{{3-sqrt(6)}}}, with its subtraction will play the role of (a-b) in the pattern. So get the perfect squares we just need to multiply by (a+b):
{{{(2 / (3- sqrt (6)))((3+sqrt(6))/(3+sqrt(6)))}}}
Multiplying we get:
{{{(6+2sqrt(6))/(3^2 - (sqrt(6))^2)}}}
which simplifies as follows:
{{{(6+2sqrt(6))/(9 - 6)}}}
{{{(6+2sqrt(6))/3}}}