View Full Version : Simplifying radicals
peri42
Aug 22, 2009, 04:24 PM
Simplify with positive exponents..
5x^3y^6 under radical sign
divided by 3a^3
Leidenschaftlich für Wahr
Aug 22, 2009, 04:27 PM
Do your own homework...
galactus
Aug 22, 2009, 05:27 PM
Is this it: \frac{\sqrt{5x^{3}y^{6}}}{3a^{3}}?
If so, there is not much to be done---except look at the y term.
peri42
Aug 22, 2009, 09:21 PM
you so would it be y^3 * y^3?
morgaine300
Aug 22, 2009, 11:24 PM
Yes, and that is (y^3)^2, which is under a radical. You can do something with sqrt{(y^3)^2}
Think about it.
peri42
Aug 23, 2009, 01:52 PM
2 squared radical y?I'm not sure
peri42
Aug 23, 2009, 02:31 PM
no wudnt it just be y^3?
morgaine300
Aug 24, 2009, 12:53 AM
no wudnt it just be y^3?
Yup. :-) The square root of a squared number is just that number.
Remember that you have to pull that out from under the radical since you took the square root of it. But I don't see anything further you can do to the rest of it.
peri42
Aug 24, 2009, 09:07 AM
is xy^2 5 under radical sign over 3a^3 right?
galactus
Aug 24, 2009, 11:35 AM
You could write it as \frac{\sqrt{5x^{3}}y^{3}}{3a^{3}}
Remember that \sqrt{y^{6}}=(y^{6})^{\frac{1}{2}}=y^{3}
Not much can be done with the x^3 inside the radical. That is the same as \sqrt{x^{3}}=x^{\frac{3}{2}}, so just leave it alone.
Actually, if we were to be purists, the actual form would be:
\frac{\sqrt{5x^{3}}\cdot |y^{3}|}{3a^{3}}
But I do not believe it is necessary to introduce the absolute value, though technically, it belongs there.
peri42
Aug 26, 2009, 03:11 PM
OK thnks 4 your help =)