How do you rationalize the denominator #(1+sqrt2)/(3+sqrt5)#?

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Answer 1 · 2015-05-28T22:15:43

Multiply both the numerator (top) and denominator (bottom) by the conjugate #3-sqrt(5)#...

#(1+sqrt(2))/(3+sqrt(5))#

#= (1+sqrt(2))/(3+sqrt(5))*(3-sqrt(5))/(3-sqrt(5))#

#= ((1+sqrt(2))(3-sqrt(5)))/((3+sqrt(5))(3-sqrt(5)))#

#= ((1+sqrt(2))(3-sqrt(5)))/(3^2-sqrt(5)^2)#

#= ((1+sqrt(2))(3-sqrt(5)))/(9-5)#

#= ((1+sqrt(2))(3-sqrt(5)))/4#