How do you simplify #sqrt(48x^3) / sqrt(3xy^2)#?

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Answer 1 · 2016-07-01T04:58:22

#sqrt(48x^3)/(3xy^2)=(4x)/y#

Let us first factorize each term in #sqrt(48x^3)/(3xy^2)#

= #sqrt(2xx2xx2xx2xx3xx x xx x xx x)/sqrt(3xx x xxyxxy)#

now cancelling #3xx x# as they occur both in numerator and denominator

= #sqrt(2xx2xx2xx2xxcancel3xxcancelx xx x xx x)/sqrt(cancel3xxcancelx xxyxxy)#

Now pairing the factors to enable us to take square roots, we get

= #sqrt(ul(2xx2)xxul(2xx2)xxul(x xx x))/sqrt(ul(yxxy))#

= #(2xx2xx x)/y=(4x)/y#

Answer 2 · 2016-07-24T16:10:36

#sqrt((16x^2)/y^2) = (4x)/y#

When multiplying or dividing with roots it is possible to combine or separate the given roots.

In its present form, the calculation has no squares under either root. BUt we can also write this as one root:

#sqrt((48x^3)/(3xy)) " which simplifies to " sqrt((16x^2)/y^2)#

Now the numerator and denominator are both perfect squares:

#sqrt((16x^2)/y^2) = (4x)/y#