How do you rationalize #(2sqrt3-sqrt2)/ (5sqrt2+sqrt3)#?

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Answer 1 · 2015-05-10T21:45:36

Given #(2sqrt(3)-sqrt(2))/(5sqrt(2)+sqrt(3))#, try multiplying both the top and the bottom by #(5sqrt(2)-sqrt(3))#

The numerator:

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

#=-2sqrt(3)sqrt(3)+9sqrt(2)sqrt(3)-5sqrt(2)sqrt(2)#

#=-2*3+9sqrt(2*3)-5*2#

#=9sqrt(6)-16#

The denominator:

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

#=25sqrt(2)sqrt(2)-sqrt(3)sqrt(3)#

#=50-3#

#=47#.

So #(2sqrt(3)-sqrt(2))/(5sqrt(2)+sqrt(3)) = (9sqrt(6)-16)/47#.