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

1 Answer
Oct 11, 2015

Answer:

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

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

Explanation:

#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.