SOLUTION: square root 9x^2y square root3x^5y^2 i am stuck. I got the obvious the square root of 9 which is 3 but I don't know what to do next and what do to with my variables. My profess

Algebra ->  Radicals -> SOLUTION: square root 9x^2y square root3x^5y^2 i am stuck. I got the obvious the square root of 9 which is 3 but I don't know what to do next and what do to with my variables. My profess      Log On


   

Question 405426: square root 9x^2y square root3x^5y^2
i am stuck. I got the obvious the square root of 9 which is 3 but I don't know what to do next and what do to with my variables. My professor ran out of time to go over this in class. please help!
Found 2 solutions by jsmallt9, ankor@dixie-net.com:
Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%289x%5E2y%29%2Asqrt%283x%5E5y%5E2%29
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%289%29%2Asqrt%28x%5E2y%29%2Asqrt%283x%5E5y%5E2%29
3%2Asqrt%28x%5E2y%29%2Asqrt%283x%5E5y%5E2%29

The next thing I would do is multiply the remaining square roots together using the property of radicals, root%28a%2C+p%29%2Aroot%28a%2C+q%29+=+root%28a%2C+p%2Aq%29:
3%2Asqrt%283x%5E7y%5E3%29
Next we look for perfect square factors (other than 1) in 3x%5E7y%5E3. There are no perfect square factors in 3. For the x%5E7 and the y%5E3 you are looking for exponents that are multiples of 2, not perfect squares! For example, a%5E6, b%5E24 and c%5E208 are all perfect squares not because their exponents are perfect squares (which they are not) but because the exponents are all divisible by 2!

So x%5E7 and y%5E3 are not perfect squares because their exponents are not divisible by 2. But they both have factors that are perfect squares:
3%2Asqrt%283%2Ax%5E2%2Ax%5E2%2Ax%5E2%2Ax%2Ay%5E2%2Ay%29
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%2Asqrt%28x%5E2%2Ax%5E2%2Ax%5E2%2Ay%5E2%2A3%2Ax%2Ay%29
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:

The square roots of the perfect squares will all simplify:
3%2Ax%2Ax%2Ax%2Ay%2Asqrt%283%2Ax%2Ay%29
or
3%2Ax%5E3%2Ay%2Asqrt%283%2Ax%2Ay%29
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.)

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%289x%5E2y%29sqrt%283x%5E5y%5E2%29
You can combine all this under one radical
sqrt%289x%5E2y%2A3x%5E5y%5E2%29
Combine like terms
sqrt%2827x%5E7%2Ay%5E3%29
factor to reveal more perfect squares
sqrt%283%2A9%2Ax%5E2%2Ax%5E2%2Ax%5E2%2Ax%2Ay%5E2%2Ay%29
extract the square roots of each perfect square
3x%2Ax%2Ax%2Ay%2Asqrt%283xy%29
which is:
3x%5E3y%2Asqrt%283xy%29