hi! is this correct???

http://img149.imageshack.us/img149/4651/formulary5.jpg

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- Apr 27th 2008, 08:29 AMyuriythebestformula question
hi! is this correct???

http://img149.imageshack.us/img149/4651/formulary5.jpg - Apr 27th 2008, 08:35 AMflyingsquirrel
Hi

No, it's wrong, unless * is the symbol you use for the sum and + the one you use for the product... :D - Apr 27th 2008, 08:36 AMMoo
Hello,

Not it isn't.

$\displaystyle a^{{\color{red}n}*k}$ (I suppose it's n, not a) is equal to $\displaystyle (a^n)^k=(a^k)^n$

The only thing you can do is to factor by $\displaystyle a^k$, if $\displaystyle k<n$ :

$\displaystyle a^n+a^k=a^k \left(a^{n-k}+1 \right)$ - Apr 27th 2008, 08:36 AMMathstud28
- Apr 27th 2008, 08:48 AMyuriythebest
wow thanks for the lighting-quick responses. But it seems I am still doing something terribly wrong:

http://img516.imageshack.us/img516/4...tion001ah6.jpg - Apr 27th 2008, 08:50 AMMoo
I'd make a substitution :

$\displaystyle u=(x^3+8)^{1/4}$

Then the equation is now uČ+u=6 :) - Apr 27th 2008, 09:54 AMyuriythebest
ok I tried it that way but it seems I am going around in circles...

http://img135.imageshack.us/img135/1...tion002gt4.jpg - Apr 27th 2008, 10:03 AMflyingsquirrel
No, you're not "going around in circles". Once you have found the two solutions $\displaystyle u_1,\,u_2$ of $\displaystyle u^2+u-6=0$, you can solve for $\displaystyle x$ : $\displaystyle \sqrt[4]{x^3+8}=u_1 \Rightarrow x^3+8=u_1^4$ and so on...

Be careful about $\displaystyle \Delta=b^2-4ac=1^2-4\times 1\times (-6)\neq 40 $ - Apr 27th 2008, 10:26 AMyuriythebest
ok I tried that and got 2 negative roots:

http://img340.imageshack.us/img340/3...tion003vq8.jpg - Apr 27th 2008, 10:32 AMflyingsquirrel
OK you got the idea, now, read again the post #8. (especially the last sentence "Be careful...") :D

- Apr 27th 2008, 11:20 AMyuriythebest
thank you flyingsquirrel! answer iz correct!

- Apr 27th 2008, 10:56 PMearboth
I'm referring to post #7: I don't understand where you've got the results from:

$\displaystyle u^2+u-6=0~\implies~u = -3~\vee~u=2$

And therefore

$\displaystyle \sqrt[4]{x^3+8} = -3$ that means there doesn't exist a real x to satisfy this equation

or

$\displaystyle \sqrt[4]{x^3+8} = 2~\implies~x^3+8=16~\implies~x^3=8~\implies~\boxed {x=2}$ - Apr 28th 2008, 04:04 AMyuriythebest
yeah I understood that. cause of my spelling I mistook one of my u's for a 4 and therefore the incorrect answer. With your help I redid it and everything is correct.