How do you rationalize #(2sqrt5)/ ( 2sqrt5 + 3sqrt2)#?

1 Answer
George C.
May 11, 2018

#(2sqrt(5))/(2sqrt(5)+3sqrt(2)) = 10-3sqrt(10)#

Explanation:

Note that the difference of squares identity tells us that:

#A^2-B^2=(A-B)(A+B)#

Hence we can rationalize the denominator of the given expression by multiplying both numerator and denominator by #2sqrt(5)-3sqrt(2)# ...

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

#color(white)((2sqrt(5))/(2sqrt(5)+3sqrt(2))) = ((2sqrt(5))^2-(2sqrt(5))(3sqrt(2)))/((2sqrt(5))^2-(3sqrt(2))^2)#

#color(white)((2sqrt(5))/(2sqrt(5)+3sqrt(2))) = (20-6sqrt(10))/(20-18)#

#color(white)((2sqrt(5))/(2sqrt(5)+3sqrt(2))) = 10-3sqrt(10)#

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