How do you divide #(8sqrt6)/(2sqrt3)#?

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Answer 1 · 2018-03-04T14:17:22

#4sqrt(2) #

We must use our knowledge of surds:

#sqrt(a*b) = sqrt(a)*sqrt(b) #

#=> sqrt(6) = sqrt(2*3) = sqrt(2) * sqrt(3) #

#(8sqrt(6))/(2sqrt(3)) = (8sqrt(2)*sqrt(3))/(2sqrt(3) ) #

We can cancel out the #sqrt(3) # as on both numerator and denominator:

#(8sqrt(2)*cancel(sqrt(3)))/(2cancel(sqrt(3) ) ) = (8sqrt(2) )/ 2 #

We know #8/2 = 4 #

#=> 4sqrt(2) #