I'd be grateful for any helpful hints, explanations, or explained solutions. Thank you for looking and (hopefully) for replying!!

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- Nov 12th 2005, 12:37 AMChristina03Help radicals!?
I'd be grateful for any helpful hints, explanations, or explained solutions. Thank you for looking and (hopefully) for replying!!

- Nov 12th 2005, 12:39 AMChristina03
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, 12:09 PMhpe
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 $\displaystyle a^2 -2ab+b^2 = (a-b)^2$ 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: $\displaystyle \left({A^2}\right)^{1/2} = |A|$.

http://i33.photobucket.com/albums/d6...iya03/alg1.gif

Here's a simpler case: $\displaystyle \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, 02:42 AMChristina03
THANK YOU! Your information was very useful. I think I understand now :)

- Nov 15th 2005, 02:55 AMChristina03
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, 03:28 AMhpeQuote:

Originally Posted by**Christina03**

- Nov 15th 2005, 03:33 AMChristina03
ok, thank you for all your help!