Question 405426
{{{sqrt(9x^2y)*sqrt(3x^5y^2)}}}
You are correct in that the square root of the 9 will become a 3 at some point. This can be the first thing you do or it can come later. Since you did this first then that is how I will do it too:
{{{sqrt(9)*sqrt(x^2y)*sqrt(3x^5y^2)}}}
{{{3*sqrt(x^2y)*sqrt(3x^5y^2)}}}<br>
The next thing I would do is multiply the remaining square roots together using the property of radicals, {{{root(a, p)*root(a, q) = root(a, p*q)}}}:
{{{3*sqrt(3x^7y^3)}}}
Next we look for perfect square factors (other than 1) in {{{3x^7y^3}}}. There are no perfect square factors in 3. For the {{{x^7}}} and the {{{y^3}}} you are looking for exponents that are multiples of 2, <i>not perfect squares!</i>  For example, {{{a^6}}}, {{{b^24}}} and {{{c^208}}} are all perfect squares not because their exponents are perfect squares (which they are not) but because the exponents are all divisible by 2!<br>
So {{{x^7}}} and {{{y^3}}} are not perfect squares because their exponents are not divisible by 2. But they both have factors that are perfect squares:
{{{3*sqrt(3*x^2*x^2*x^2*x*y^2*y)}}}
For reasons that will become clear shortly I like to use the Commutative Property to rearrange the order of the factors so that all the perfect square factors are n front:
{{{3*sqrt(x^2*x^2*x^2*y^2*3*x*y)}}}
Next we use the same property as before, only in the other direction, to split up this square root of a product into a product of square roots. We want all the perfect square factors in their own square roots. The factors that are not perfect squares all go into the same square root:
{{{3*sqrt(x^2)*sqrt(x^2)*sqrt(x^2)*sqrt(y^2)*sqrt(3*x*y)}}}
The square roots of the perfect squares will all simplify:
{{{3*x*x*x*y*sqrt(3*x*y)}}}
or
{{{3*x^3*y*sqrt(3*x*y)}}}
This is the simplified answer. (Note how the square root is at the end. This is the normal way to write terms like this and it is the reason I put all the perfect square factors in front earlier.)