How do you simplify #sqrt22/(sqrt11-sqrt66)#?

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Answer 1 · 2015-11-01T14:04:22

Use #sqrt(ab) = sqrt(a)sqrt(b)# to find:

#sqrt(22)/(sqrt(11)-sqrt(66))=-sqrt(2)/5-(2sqrt(3))/5#

If #a, b >= 0# then #sqrt(ab) = sqrt(a)sqrt(b)#

So:

#sqrt(22)/(sqrt(11)-sqrt(66))#

#=(sqrt(11)sqrt(2))/(sqrt(11)-sqrt(11)sqrt(6))#

#=sqrt(2)/(1-sqrt(6))#

#=(sqrt(2)(1+sqrt(6)))/((1-sqrt(6))(1+sqrt(6)))#

#=(sqrt(2)+sqrt(2)^2sqrt(3))/(1-6)#

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

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