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Thread: Location of complex roots?

  1. #1
    Super Member fardeen_gen's Avatar
    Jun 2008

    Location of complex roots?

    I have been thinking about this:
    When we graph $\displaystyle y = ax^2 + bx + c$, we get a parabola opening upwards or downwards. It will intersect/touch the x-axis if it has real roots and we know the value of the roots from the points of intersection. When the equation has complex roots, the parabola naturally doesn't intersect with the x-axis. Is it possible for us to locate those complex roots? For eg., if the roots are $\displaystyle 1 \pm \sqrt{5}i$, is it possible to discern from the graph the location of those roots? My professor says Yes. And I fail to see how. (And no, this isn't about the Argand plane. We are talking about the real x-y axes)
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  2. #2
    Jul 2010
    In my opinion of course it's impossible to locate complex roots on a functional graph. But you can find complex roots by performing some calculations on some specfic points on the parabola. There are probably many methods and this one requires you to locate the vertex of the parabola. Let $\displaystyle y=ax^2+bx+c$ and the discriminant is less than zero.

    Let the coordinate of the vertex be $\displaystyle (p, q)$, then by doing some calculations:
    $\displaystyle \displaystyle p=-\frac{b}{2a}$

    $\displaystyle \displaystyle q=\frac{-b^2+4ac}{4a}$

    Now you can see that the quadratic formula can be rewritten in terms of the coordinates of the vertex, which is easy to derive:
    $\displaystyle x=p\pm \sqrt{-\frac{q}{a}}$.
    (Actually this generates all the roots.)
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  3. #3
    Senior Member
    Jul 2010
    [ignore]I think you can start thinking about it by making some assumptions and putting some constraints.

    For example, since you want to find an x for which y(x) = 0, then you can assume a = 1 since you can divide the equation out by a if it isn't 1. With proper scaling and expansion/contraction you can just look at the very simple case y = x^2+ c where c is a constant. That should help a lot in simplifying the problem.[/ignore]

    Terence has a great explanation.
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