# How do you simplify (sqrt(48x^3y^2))/(sqrt4xy^3)?

May 16, 2015

Before starting, I must say that I believe there has been a little typing mistake and that your function is $\frac{\sqrt{48 {x}^{3} {y}^{2}}}{\sqrt{4 x {y}^{3}}}$. I will consider this, then.

Note that as both numerator and denominator are square roots, we can then merge them in this way: $\sqrt{\frac{48 {x}^{3} {y}^{2}}{4 x {y}^{3}}}$

Now, let's see what composes both numerator and denominator and cancel the common elements.

$\sqrt{\frac{\cancel{2 \cdot 2} \cdot 2 \cdot 2 \cdot 3 \cdot \cancel{x} \cdot x \cdot x \cdot \cancel{y \cdot y}}{\cancel{2 \cdot 2} \cdot \cancel{x} \cdot \cancel{y \cdot y} \cdot y}}$

Now, let's rewrite all that is left from our cancelling:

$\sqrt{\frac{2 \cdot 2 \cdot 3 \cdot x \cdot x}{y}}$

However, we still have two squared numbers ($2$ and $x$) inside the numerator. As we're dealing with a squared root, then we can take out the square roots of these squared numbers, as follows:

$2 x \sqrt{\frac{3}{y}}$ which is the same as $2 x \frac{\sqrt{3}}{\sqrt{y}}$

Now, we can just rationalize this answer:

$2 x \frac{\sqrt{3}}{\sqrt{y}} \cdot \frac{\sqrt{y}}{\sqrt{y}} = \frac{2 x \sqrt{3} \sqrt{y}}{y}$

The multiplication of square roots is the square root of the multiplication. So, the final simplification is:

$\frac{2 x \sqrt{3 y}}{y}$

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In case your function is just exactly as it is written here , then the same steps will follow:

$\frac{\cancel{2} \cdot 2 \cancel{x} \cancel{y} \sqrt{x}}{\cancel{2} \cancel{x} \cancel{y} \cdot y \cdot y}$

Final answer, in this case, would be: $\frac{2 \sqrt{x}}{y} ^ 2$