quadratic equations and parabolas involving strange exponents

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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.