SOLUTION: Square root of 63x^4y^7. I broke it down to the square root of 7*9*x^3*x^1*y^6*y so far I have my solution as 3x^2y^2 (square root of 7y^3.... I need to know if I am thinking co

Algebra ->  Radicals -> SOLUTION: Square root of 63x^4y^7. I broke it down to the square root of 7*9*x^3*x^1*y^6*y so far I have my solution as 3x^2y^2 (square root of 7y^3.... I need to know if I am thinking co      Log On


   

Question 856503: Square root of 63x^4y^7. I broke it down to the square root of 7*9*x^3*x^1*y^6*y so far I have my solution as 3x^2y^2 (square root of 7y^3....
I need to know if I am thinking correctly.
Thanks
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
If its square root, why aren't you breaking exponents down to multiples of 2??
sqrt%2863x%5E4y%5E7%29=sqrt%287%2A9%2Ax%5E4%2Ay%2A6%2Ay%29
sqrt%2863x%5E4y%5E7%29=sqrt%283%5E2%2Ax%5E4%2Ay%5E6%29%2Asqrt%287%2Ay%29
sqrt%2863x%5E4y%5E7%29=3%2Ax%5E2%2Ay%5E3%2Asqrt%287%2Ay%29