# Help for some (probably easy) radicals!?

• Nov 12th 2005, 01:37 AM
Christina03
I'd be grateful for any helpful hints, explanations, or explained solutions. Thank you for looking and (hopefully) for replying!!
• Nov 12th 2005, 01:39 AM
Christina03
I know I haven't been a member of this forum for very long, but it seems like a nice and helpful place. :)
• Nov 13th 2005, 01:09 PM
hpe
http://i33.photobucket.com/albums/d6...iya03/alg4.gif
Since for each real number has exactly one cube and you get the number back by computing the cube root, this is true for all n. This would be different for squares and square roots.
http://i33.photobucket.com/albums/d6...iya03/alg3.gif
First compute y-x, then the square root, then raise it to the third power, then take the reciprocal.
http://i33.photobucket.com/albums/d6...iya03/alg2.gif
The expression under the square root is an instance of the formula $a^2 -2ab+b^2 = (a-b)^2$ So you need to write this as a square (of a difference). The square root of a square is ... not quite the term you squared in the first place, but the absolute value of this term: $\left({A^2}\right)^{1/2} = |A|$.
http://i33.photobucket.com/albums/d6...iya03/alg1.gif
Here's a simpler case: $\left({x^2y^7}\right)^{1/2} = \left({xy^3xy^3y}\right)^{1/2} = |xy^3|\left(y\right)^{1/2}$ Use the same sort of argument.
• Nov 15th 2005, 03:42 AM
Christina03
THANK YOU! Your information was very useful. I think I understand now :)
• Nov 15th 2005, 03:55 AM
Christina03
I got 7|x – y| for the square root of the perfect-square polynomial, and:
http://i33.photobucket.com/albums/d6...iya03/alg5.gif
for the square root of (a^5b^6). Just wondering if those are correct.... :confused: I think so, but I just want to make sure. Thanks!
• Nov 15th 2005, 04:28 AM
hpe
Quote:

Originally Posted by Christina03
I got 7|x – y| for the square root of the perfect-square polynomial, and:
http://i33.photobucket.com/albums/d6...iya03/alg5.gif
for the square root of (a^5b^6). Just wondering if those are correct.... :confused: I think so, but I just want to make sure. Thanks!

Looks correct to me. :)
• Nov 15th 2005, 04:33 AM
Christina03
ok, thank you for all your help!