How do you rationalize #(6-sqrt7)/(5-sqrt7)#?

Question

1376 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-01T21:04:05

Multiply both numerator (top) and denominator (bottom) by the conjugate #5 + sqrt(7)#...

#(6-sqrt(7))/(5-sqrt(7))#

#= (6-sqrt(7))/(5-sqrt(7))*(5+sqrt(7))/(5+sqrt(7))#

#= ((6-sqrt(7))(5+sqrt(7))) / ((5-sqrt(7))(5+sqrt(7)))#

#= (30+sqrt(7)-sqrt(7)^2)/(5^2-sqrt(7)^2)#

#= (30+sqrt(7)-7)/(25-7)#

#=(23+sqrt(7))/18#