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Homework Help: Quadratic Formula

  1. Jul 17, 2006 #1
    Umm... I guess I don't remember this from algebra, but I have a rather basic question.

    Let's say you have the expression,
    [tex] f(x) = 4x^2 + 5x + 1 [/tex]

    And we want to find the roots.


    [tex] \frac{-5 \pm \sqrt{5-4(4)}}{2(4)}= \{ \frac{1}{2},-1 \} [/tex]

    Now, if we plug these into [itex] f(x) [/itex] we have:
    [tex] f(1/2) = 9/2 [/tex]
    [tex] f(-1) = 0 [/tex]

    So obviously the quadratic equation does not yield the roots. But if we let [itex] a=1 [/itex].

    [tex] 4(x^2+\frac{5}{4}x+\frac{1}{4}) [/tex]
    Again, using the quadratic:
    [tex] x = \{ -1,-\frac{1}{4} \} [/tex]

    So now plugging this into [itex] f(x) [/itex]

    [tex] f(x) = 4(x^2+\frac{5}{4}x+\frac{1}{4}) [/tex]

    [tex] f(-1) = 0 [/tex]
    [tex] f(\frac{-1}{4}) = 0 [/tex]

    So what gives. I don't think I was ever taught that "a" has to be equal to 1 for it to hold.

    If someone could clarify why this is that would be awesome. I seem to remember something about completing the square, but the thing is, is I have an engineering test on Thursday, and the last thing I want to be using my time on is figuring out why this is. But I'm curious :smile:

    Last edited: Jul 17, 2006
  2. jcsd
  3. Jul 17, 2006 #2
    Recheck your quadratic equation. It should be:
    [tex] \frac{-b \pm \sqrt{b^2 -4ac}}{2a} [/tex]

    You forgot to take the sqrt of b^2 - 4ac :smile:
    Last edited: Jul 17, 2006
  4. Jul 17, 2006 #3
    You can also try to complete the square or find the numbers m, n such as m*n = c and m+n = b when a = 1 .
    Last edited: Jul 17, 2006
  5. Jul 17, 2006 #4

    Ok, cool... so I'm not crazy. It works fine for the reals. Now when I get complex conjugates, it seems to be a different story. Please tell me I'm making a simple mistake again ! :cry:

    [tex] f(s) = 10s^2 + 10s + 40 [/tex]
    [tex] \frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-10 \pm \sqrt{100-4(10)(40)}}{2(10)}=\frac{-10 \pm\sqrt{-1500}}{2(10)} = \ldots [/tex]
    [tex] \ldots = \left\{ \frac{-1}{2} + \frac{\sqrt{15}}{2}j \,\, , \,\, \frac{-1}{2} - \frac{\sqrt{15}}{2}j \right\} = \{ \lambda_1 \,\, , \,\, \lambda_2 \} [/tex]
    [tex] f(\lambda_1) = f(\lambda_2) = 0 [/tex]
    So that's cool. But what about,

    [tex] (s-\lambda_1)(s-\lambda_2)=s^2+s+4 [/tex]

    If the roots were real then wouldn't that equal the original function. Maybe I'm doing drugs, because I can't believe I'm getting caught up with this right now :surprised
  6. Jul 17, 2006 #5


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    It works fine for complex roots.

    In your examples of s^2+s+4 and 10s^2+10s+40, remember if u and v are roots of a polynomial ax^2+bx+c, then


    it's not just a product over factors corresponding to the roots, you need to take care of the leading coefficient as well.
  7. Jul 17, 2006 #6

    Ahh, that's what I was missing! Thanks man. I can't believe I forgot that :mad:
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