How do you rationalize the denominator #7/(sqrt3-sqrt2) #?

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Answer 1 · 2015-05-26T16:23:12

You can multiply and divide your expression by #sqrt(3)+sqrt(2)# to get:
#7/(sqrt(3)-sqrt(2))*(sqrt(3)+sqrt(2))/(sqrt(3)+sqrt(2))=#

in the denominator you have a notable:
#(a+b)(a-b)=a^2-b^2#

So you get:
#(7(sqrt(3)+sqrt(2)))/(3-2)=7(sqrt(3)+sqrt(2))#

Answer 2 · 2015-05-26T16:27:08

Multiply numerator and denominator by the conjugate #(sqrt(3)+sqrt(2))#:

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

#= (7(sqrt(3)+sqrt(2)))/((sqrt(3)-sqrt(2))(sqrt(3)+sqrt(2)))#

#= (7(sqrt(3)+sqrt(2)))/(sqrt(3)^2-sqrt(2)^2)#

#= (7(sqrt(3)+sqrt(2)))/(3-2)#

#= 7(sqrt(3)+sqrt(2))#