How do you rationalize the denominator and simplify $4/(6-sqrt5)$ ?

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Answer 1 · 2015-10-11T09:13:06

Multiply both numerator and denominator by $(6+sqrt(5))$ and simplify to find:

$4/(6-sqrt(5)) =(24+4sqrt(5))/31$

$4/(6-sqrt(5)) = (4(6+sqrt(5)))/((6-sqrt(5))(6+sqrt(5))) =$

$=(24+4sqrt(5))/(6^2-sqrt(5)^2) = (24+4sqrt(5))/(36-5)$

$=(24+4sqrt(5))/31$

This uses the difference of squares identity to square any square roots in the binomial denominator.

$(a-b)(a+b) = a^2 - b^2$

In our case $a = 6$ and $b = sqrt(5)$ , but it would work if both $a$ and $b$ were square roots.

How do you rationalize the denominator and simplify 4/(6-sqrt5)4/(6-sqrt5)?