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Thread: Solution for equation

  1. #1
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    Question Solution for equation

    Why is the solution for:

    $$ ax^{4} + bx^{2} + c = 0$$

    $$x_1 = \frac{-b + \sqrt{b^2-4ac}}{2a}$$

    $$x_{final} = \sqrt[4]{x_1}$$?

    Thank you.
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  2. #2
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    Re: Solution for equation

    It isn't.

    $ax^4+bx^2+c=0$
    Let $x_1=x^2$
    $x_1=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} $
    $x=\pm \sqrt{x_1} $

    You will wind up with four solutions, not one.
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  3. #3
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    Re: Solution for equation

    Quote Originally Posted by SlipEternal View Post
    It isn't.

    $ax^4+bx^2+c=0$
    Let $x_1=x^2$
    $x_1=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} $
    $x=\pm \sqrt{x_1} $

    You will wind up with four solutions, not one.
    Why did you choose $$x_1 = x^2$$ and not $$x_1 = x^4$$?
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  4. #4
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    Re: Solution for equation

    Because if $x_1=x^4$, then you have $ax_1+bx^2+c=0$, which is an equation in two variables, so it is not at all helpful. We can also say $ax_1+b\sqrt{x_1}+c=0$, but this is not a quadratic equation. By saying $x_1=x^2$, I wind up with a quadratic equation of $x_1$ that is solvable.
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