How do you rationalize # 3 / (sqrt5 -sqrt2) #?

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Answer 1 · 2015-05-14T19:11:16

You can rationalize the denominator by multiplying top and bottom by the conjugate #sqrt(5) + sqrt(2)# as follows:

#3/(sqrt(5)-sqrt(2))#

#= (3*(sqrt(5)+sqrt(2)))/((sqrt(5)-sqrt(2))(sqrt(5)+sqrt(2))#

#= (3*(sqrt(5)+sqrt(2)))/(sqrt(5)sqrt(5)-sqrt(2)sqrt(2))#

#= (3*(sqrt(5)+sqrt(2)))/(5-2)#

#= (3*(sqrt(5)+sqrt(2)))/3#

#= sqrt(5)+sqrt(2)#