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Math Help - quadratic equations and parabolas involving strange exponents

  1. #1
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    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.
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  2. #2
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    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.
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  3. #3
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    I got y= -6, or 16. Do I have to root it by something now to get my final answer? 3rd root?
    Last edited by Schism462; March 31st 2011 at 12:10 PM. Reason: add something
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  4. #4
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    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=...
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  5. #5
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    2cube root 2?
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  6. #6
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    Uhm, what? o.o More words please. ^^'

    If you refer at this "If y=16, then x=...", x= 2\sqrt[3]{2}
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  7. #7
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    and should I write it as cube root of 6i?
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  8. #8
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    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.)
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  9. #9
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    "(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 ^^
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  10. #10
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    Quote Originally Posted by veileen View Post
    "(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
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  11. #11
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    can anyone help me with the fractional exponent quadratic?

    x^2/3-2x^1/3-15=0
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  12. #12
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    Quote Originally Posted by Schism462 View Post
    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.
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