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

1 Answer
May 16, 2015

Before starting, I must say that I believe there has been a little typing mistake and that your function is #(sqrt(48x^3y^2))/(sqrt(4xy^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((48x^3y^2)/(4xy^3))#

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

#sqrt((cancel(2*2)*2*2*3*cancel(x)*x*x*cancel(y*y))/(cancel(2*2)*cancel(x)*cancel(y*y)*y))#

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

#sqrt((2*2*3*x*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:

#2xsqrt(3/y)# which is the same as #2xsqrt(3)/sqrt(y)#

Now, we can just rationalize this answer:

#2xsqrt(3)/sqrt(y)*(sqrt(y))/(sqrt(y))=(2xsqrt(3)sqrt(y))/y#

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

#(2xsqrt(3y))/y#

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

#(cancel(2)*2cancel(x)cancel(y)sqrt(x))/(cancel(2)cancel(x)cancel(y)*y*y)#

Final answer, in this case, would be: #(2sqrt(x))/y^2#