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Math Help - quadratic equation - discrimination

  1. #1
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    quadratic equation - discrimination

    quadratic equation - discrimination
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  2. #2
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    mr fantastic's Avatar
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    Your expression for the discriminant can be re-arranged into
    D = 4(a^2 - a(b+c) + (b^2 - bc + c^2)).

    For unequal roots (ie. two distinct roots) you want a^2 - a(b+c) + (b^2 - bc + c^2) \neq 0.

    NB (an acronym for the latin phrase nota bene which translates as note well, that is, what follows is very important so pay close attention, chum): We don't give a fetid dingos kidney whether the roots are real or not, as long as they're distinct. So the requirement is D \neq 0, NOT D > 0.

    Let's assume a^2 - a(b+c) + (b^2 - bc + c^2) = 0. Since a is real, then (IF the assumption is true) you can solve this quadratic to find real values of a. Therefore (IF the assumption is true) the discriminant of this quadratic is greater than zero.

    From here, I'll let you fill in the steps that lead to (b - c)^2 < 0, an impossibility for real values of b and c.

    Hence the assumption a^2 - a(b+c) + (b^2 - bc + c^2) = 0 is false. So a^2 - a(b+c) + (b^2 - bc + c^2) \neq 0. Q.E.D.
    Last edited by mr fantastic; December 19th 2007 at 12:17 AM. Reason: Corrected a small typo (and you know how worried the kiddies get if there's a small mistake, even when it's an obvious typo)
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  3. #3
    Senior Member JaneBennet's Avatar
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    You can also write

    D\ =\ (2a-b-c)^2+3(b-c)^2

    Since b\neq c,\ 3(b-c)^2\ >\ 0.

    Hence D is (strictly) positive, so the equation has real and unequal roots.
    Last edited by JaneBennet; December 19th 2007 at 04:41 AM.
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