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

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Answer 1 · 2015-06-02T22:36:53

Multiply both numerator (top) and denominator (bottom) by the conjugate #(2+sqrt(2))# ...

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

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

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

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

#=(6+3sqrt(2))/(4-2)#

#=(6+3sqrt(2))/2#

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