Square root

• Jan 9th 2013, 07:13 AM
tahirakw
Square root
for x> 0 and y> 0 > . how do I find out whats equivalent to the square root of 16x^9 y^36?
My answer is 4y^18x^4 square root x
Is that correct? I think I did something wrong
• Jan 9th 2013, 07:19 AM
Plato
Re: Square root
Quote:

Originally Posted by tahirakw
for x> 0 and y> 0 > . how do I find out whats equivalent to the $\sqrt{ 16x^9 y^{36}}$?
My answer is $4y^{18}x^4\sqrt x$
Is that correct? I think I did something wrong

Yes that is correct
• Jan 9th 2013, 07:25 AM
tahirakw
Re: Square root
Thanks