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

2 Answers
Gió
Apr 23, 2015

Multiply and divide by #sqrt(2)+sqrt(5)# to get:
#[sqrt(2)+sqrt(5)]^2/(2-5)=-1/3[2+2sqrt(10)+5]=-1/3[7+2sqrt(10)]#

Elijah T.
Feb 5, 2017

Conjugate

Explanation:

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.

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