How do you simplify sqrt(48x^3) / sqrt(3xy^2)√48x3√3xy2?

2 Answers
Shwetank Mauria
Jul 1, 2016

sqrt(48x^3)/(3xy^2)=(4x)/y√48x33xy2=4xy

Explanation:

Let us first factorize each term in sqrt(48x^3)/(3xy^2)√48x33xy2

= sqrt(2xx2xx2xx2xx3xx x xx x xx x)/sqrt(3xx x xxyxxy)√2×2×2×2×3×x×x×x√3×x×y×y

now cancelling 3xx x3×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

EZ as pi
Jul 24, 2016

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

Explanation:

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

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