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Math Help - non-whole roots

  1. #1
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    non-whole roots

    How could I know that 4x^4 + 3^2 - 3 has two roots?
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  2. #2
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    Re: non-whole roots

    I've noticed it would have a parabola shape and subtracting 3 creates a y-axis crossing point at y=-3.
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  3. #3
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    Re: non-whole roots

    Quote Originally Posted by Stuck Man View Post
    How could I know that 4x^4 + 3^2 - 3 has two roots?
    You don't. It's not true. Or are you asking about real roots?

    Is that really the problem? 4x^4+ 3^2- 3= 4x^4+ 9- 3= 4x^4+ 6= 0 so x^4= -6/4 so that there are four roots, the four fourth roots of -3/2, none of which are real.

    If the equation is 4x^4+ x^2- 3= 0, we can let y= x^2 so we are dealing with the quadratic equation 4y^2+ y- 3= 0. By the quadratic formula,
    y= \frac{-1\pm\sqrt{1+ 48}}{8}= \frac{-1\pm 7}{8}
    which has the two roots y= 6/8= 3/4 and y= -8/8= -1.

    Now, If y= x^2= 3/4, then x= \pm\sqrt{3}/2, two real roots. If y= x^2= -1, then x= \pm i, two imaginary roots.

    The crucial point was that, because this had only even factors, we could reduce to a second order equation. Positive roots of that equation give real roots of the original equation.
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  4. #4
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    Re: non-whole roots

    I typed it wrong, its 4x^4+3x^2-3.
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: non-whole roots

    Use the substitution, let x^2=t so you get a quadratic equation in t which you can solve, afterwards do the back-substitution. Let see how many roots you get.
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    Re: non-whole roots

    Thanks.
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  7. #7
    MHF Contributor Siron's Avatar
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    Re: non-whole roots

    You're welcome! But, what's your result? ...
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  8. #8
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    Re: non-whole roots

    There is 2 real roots. I have exact values but there is quite a lot to type. They are about +/- 0.75.
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