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Thread: Quadratic equations

  1. #1
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    Quadratic equations

    Hello all,

    I have never really thought about the following question:

    -when does the quadratic equation have only positive solutions?
    -when does the quadratic equation have only negative solutions?
    -When are the roots in a biquadratic equation positive/negative?

    Thanks
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  2. #2
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    e^(i*pi)'s Avatar
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    Re: Quadratic equations

    You use the discriminant: $\displaystyle \Delta = b^2-4ac$

    If $\displaystyle \Delta > 0$ then there are two real solutions
    If $\displaystyle \Delta = 0$ then there are two, equal real roots[/tex]
    If $\displaystyle \Delta < 0$ then there are no real solutions (2 complex ones)[/tex]


    "only positive solutions" essentially means that any solution is to the right of 0 on the x-axis.
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  3. #3
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    Re: Quadratic equations

    "only positive solutions" essentially means that any solution is to the right of 0 on the x-axis
    that is exactly what I wanted to know. When is this the case, i.e what needs to holds in order for the roots to be only positive/negative. Could one set up some type of criteria?
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    e^(i*pi)'s Avatar
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    Re: Quadratic equations

    I'm not entirely sure. We can rearrange the general form of the quadratic into $\displaystyle (x-h)^2+k$ form. From that we could link "only positive solutions" to h.

    $\displaystyle ax^2+bx+c = 0 \Longleftrightarrow x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0$

    $\displaystyle \left(x+\dfrac{b}{2a}\right)^2 - \dfrac{b^2-4ac}{4a^2} = 0$

    Since the "+k" part is a vertical shift it's value does not make a difference on where the roots lie so we can forget about it for now

    Spoiler:
    If you want to try this out for yourself you're more than welcome to - test $\displaystyle y = (x-1)^2 + 4 \text{ and } y = (x-1)^2 +1$ and see what happens. It's always good to check for yourself anyway


    Due the properties of squaring real expressions we know that $\displaystyle \left(x+\dfrac{b}{2a}\right)^2 \geq 0$

    For positive solutions it follows that $\displaystyle \displaystyle | x + \dfrac{b}{2a}| \geq 0$

    $\displaystyle x \geq -\dfrac{b}{2a}$

    I'm not 100% this is right you may want to try with some examples
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