How do you rationalize the denominator #(8sqrt2)/(2sqrt5-3sqrt2)#?

Question

1246 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-14T22:24:04

Multiply both numerator and denominator by the conjugate #(2sqrt(5)+3sqrt(2))# to get:

#(8sqrt(2))/(2sqrt(5)-3sqrt(2)) = #

#(8sqrt(2))/(2sqrt(5)-3sqrt(2))#

#=(8sqrt(2))/(2sqrt(5)-3sqrt(2))*(2sqrt(5)+3sqrt(2))/(2sqrt(5)+3sqrt(2))#

#=(8sqrt(2)(2sqrt(5)+3sqrt(2)))/(20-18)#

#=4sqrt(2)(2sqrt(5)+3sqrt(2))#

#=8sqrt(2)sqrt(5)+24#

#=8sqrt(10)+24#

#=8(sqrt(10)+3)#