quadratic equations and parabolas involving strange exponents

• March 31st 2011, 11:51 AM
Schism462
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.
• March 31st 2011, 11:56 AM
veileen
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.
• March 31st 2011, 12:07 PM
Schism462
I got y= -6, or 16. Do I have to root it by something now to get my final answer? 3rd root?
• March 31st 2011, 12:13 PM
veileen
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=...
• March 31st 2011, 12:16 PM
Schism462
2cube root 2?
• March 31st 2011, 12:19 PM
veileen
Uhm, what? o.o More words please. ^^'

If you refer at this "If y=16, then x=...", x= $2\sqrt[3]{2}$
• March 31st 2011, 12:20 PM
Schism462
and should I write it as cube root of 6i?
• March 31st 2011, 12:22 PM
Schism462
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.)
• March 31st 2011, 12:27 PM
veileen
"(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 ^^
• March 31st 2011, 12:29 PM
Schism462
Quote:

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 $\sqrt[3]{-6}$.

Thank you for your help :)
• March 31st 2011, 01:21 PM
Schism462
can anyone help me with the fractional exponent quadratic?

x^2/3-2x^1/3-15=0
• March 31st 2011, 02:04 PM
Quacky
Quote:

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.