What is the simplest form of #sqrt(80x7y5)#?

1 Answer
Feb 9, 2018

Answer:

#4x^3y^2sqrt(5xy)#

Explanation:

I'm assuming you meant #sqrt(80x^(7)y^(5))#
If so, you take the square root of 80, shown as..

#sqrt80#
#sqrt(4*4*5)#
#2*2sqrt5#
#4sqrt5#

then we divide the powers by 2, if they're even, or if they're odd, subtract one (which will stay inside the radical) then divide by 2

for #x#:
#sqrt(x^7)#
#sqrt(x^6*x#
#x^3sqrtx#

and for #y#:
#sqrt(y^5)#
#sqrt(y^4*y)#
#y^2sqrty#

then multiply the answers together, and there you have it