How do you rationalize the denominator and simplify #sqrt(12x^3)/sqrt(7y^5)#?

Question

1649 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-23T12:07:02

#sqrt(12x^3)/sqrt(7y^5)= 2sqrt(21x^3y^5)/(7y^5)#

Or more simplified:

#sqrt(12x^3)/sqrt(7y^5)=(2x)/(7y^3)*sqrt(21xy)#

To rationalize your expression, the idea is to multiply it by a therm that doesn't change it but cancels the #sqrt()# at the denominator.

Therefore:

#sqrt(12x^3)/sqrt(7y^5)=sqrt(12x^3)/sqrt(7y^5)*sqrt(7y^5)/sqrt(7y^5)=#

#=(sqrt(12x^3)*sqrt(7y^5))/(sqrt(7y^5*7y^5))=#

#=sqrt(12x^3*7y^5)/(sqrt((7y^5)^2))=sqrt(3*2^2*7x^3y^5)/(7y^5)=2sqrt(21x^3y^5)/(7y^5)#

Now you can more simplify the results:

#2sqrt(21x^3y^5)/(7y^5)=2sqrt(21x^2*x*y^4*y)/y^5=2xcancel(y^2)sqrt(21xy)/(7y^(cancel(5)^3))=#
#=(2x)/(7y^3)*sqrt(21xy)#