Question 274869
{{{2/(sqrt(5)+2)}}}
To rationalize a two-term denominator like this you multiply the numerator and denominator by the conjugate of the denominator. The conjugate of {{{sqrt(5)+2}}} is {{{sqrt(5)-2}}}. (Note the minus sign!)<br>
You should be familiar with the pattern: {{{(a+b)(a-b) = a^2 - b^2}}}. a+b and a-b are conjugates of each other. And as you can see, when you multiply conjugates you get two terms which are both perfect squares. This is why we use conjugates when rationalizing denominators. The only terms in the priduct are perfect squares.
{{{(2/(sqrt(5)+2))((sqrt(5)-2)/(sqrt(5)-2))}}}
Multiplying we get:
{{{(2(sqrt(5)-2))/((sqrt(5))^2 - 2^2)}}}
which simplifies to:
{{{(2sqrt(5)-4)/(5-4)}}}
or
{{{(2sqrt(5)-4)/1}}}
or
{{{2sqrt(5)-4}}}
Not only do we no longer have a radical in a denominator, we have no denominator at all!