# quadratic equations and parabolas involving strange exponents

#### Schism462

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.

#### veileen

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.

HallsofIvy

#### Schism462

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

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#### veileen

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=...

2cube root 2?

#### veileen

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

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

#### Schism462

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

#### 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.)

#### 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 $$\displaystyle 2\sqrt[3]{2}$$ and $$\displaystyle \sqrt[3]{-6}$$.

Edit: You're welcome ^^

#### Schism462

"(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 $$\displaystyle \sqrt[3]{-6}$$.