1. ## quadratic equations and parabolas involving strange exponents

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.

2. Let y be $x^3$, then $x^6-10x^3-96=0 \Leftrightarrow y^2-10y-96=0$.

The second equation can be solved using a convenient notation too.

3. I got y= -6, or 16. Do I have to root it by something now to get my final answer? 3rd root?

4. No, it is not the final answer, you have to find x, remember?

If y=-6, then $x=\sqrt[3]{-6}$
If y=16, then x=...

5. 2cube root 2?

6. Uhm, what? o.o More words please. ^^'

If you refer at this "If y=16, then x=...", x= $2\sqrt[3]{2}$

7. and should I write it as cube root of 6i?

8. 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.)

9. "(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 $2\sqrt[3]{2}$ and $\sqrt[3]{-6}$.

Edit: You're welcome ^^

10. Originally Posted by veileen
"(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 $2\sqrt[3]{2}$ and $\sqrt[3]{-6}$.

11. can anyone help me with the fractional exponent quadratic?

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

12. Originally Posted by Schism462
can anyone help me with the fractional exponent quadratic?

x^2/3-2x^1/3-15=0
Let $\displaystyle~U=x^{\frac{1}{3}}$

Then, your quadratic becomes: $U^2-2U-15=0$

Post your solutions so we can verify them.