What is the simplest form of the radical expression of #(sqrt2+sqrt5)/(sqrt2-sqrt5)#?

2 Answers
Apr 23, 2015

Multiply and divide by #sqrt(2)+sqrt(5)# to get:

Feb 5, 2017




Just to add on to the other answers,

We decided to multiply the top and the bottom by #sqrt(2)+sqrt(5)# because this is the conjugate of the denominator, #sqrt(2)-sqrt(5)#.

A conjugate is an expression in which the sign in the middle is reversed. If (A+B) is the denominator, then (A-B) would be the conjugate expression.

When simplifying square roots in the denominators, try multiplying the top and bottom by the conjugate. It will get rid of the square root, because #(A+B)(A-B) = A^2-B^2#, meaning you will be left with the numbers in the denominator squared.