quadratic equations and parabolas involving strange exponents

Mar 2011
7
0
I am not sure how to even approach these two equations. Both of them have exponents that are a bit odd and I'm not even sure how to begin.

I must find all real solutions for these problems.

x^6-10x^3-96=0

and

x^2/3-2x^1/3-15=0

No decimals allowed.We have to put them in terms with whole numbers, radicals, i, or fractions.
 
Mar 2011
182
42
Awetuouncsygg
Let y be \(\displaystyle x^3\), then \(\displaystyle x^6-10x^3-96=0 \Leftrightarrow y^2-10y-96=0\).

The second equation can be solved using a convenient notation too.
 
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Mar 2011
7
0
I got y= -6, or 16. Do I have to root it by something now to get my final answer? 3rd root?
 
Last edited:
Mar 2011
182
42
Awetuouncsygg
No, it is not the final answer, you have to find x, remember?

If y=-6, then \(\displaystyle x=\sqrt[3]{-6}\)
If y=16, then x=...
 
Mar 2011
182
42
Awetuouncsygg
Uhm, what? o.o More words please. ^^'

If you refer at this "If y=16, then x=...", x=\(\displaystyle 2\sqrt[3]{2}\)
 
Mar 2011
7
0
Sorry >.< Would the final answer then be 2 times the cube root of 2, and cube root of 6i?
(sorry I'm not sure how to add a radical symbol on here.)
 
Mar 2011
182
42
Awetuouncsygg
"(sorry I'm not sure how to add a radical symbol on here.)" - quote one of my posts and see how.

"I must find all real solutions for these problems." Your real solutions are \(\displaystyle 2\sqrt[3]{2}\) and \(\displaystyle \sqrt[3]{-6}\).

Edit: You're welcome ^^
 
Mar 2011
7
0
"(sorry I'm not sure how to add a radical symbol on here.)" - quote one of my posts and see how.

Your real solutions are \(\displaystyle 2\sqrt[3]{2}\) and \(\displaystyle \sqrt[3]{-6}\).
Thank you for your help :)