How do you divide #(sqrt5+1)/(sqrt5-1)#?

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Answer 1 · 2015-06-06T23:12:51

Multiply both numerator (top) and denominator (bottom) by the conjugate #(sqrt(5)+1)# of the denominator:

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

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

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

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

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

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

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