How do you rationalize #x/(sqrt5-sqrt2)#?

1 Answer
George C.
May 21, 2015

You can rationalize the denominator of this expression, by multiplying the numerator (top) and denominator (bottom) by the conjugate of #sqrt(5)-sqrt(2)#, namely #sqrt(5)+sqrt(2)#.

#x/(sqrt(5)-sqrt(2))#

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

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

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

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

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