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Math Help - A nice quadratic ^^

  1. #1
    Moo
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    A nice quadratic ^^

    Hello,

    What's so special about this polynomial ? :

    P_m(x)=(m-3)x^2+(2m-1)x+(m+2), m \in \mathbb{R}

    The discriminant is \Delta=(2m-1)^2-4(m-3)(m+2)=4m^2-4m+1-4(m^2-m-6)=\boxed{25}

    One can see that the discriminant is independent from m !

    The solutions are :

    x_1=\frac{-(2m-1)-5}{2(m-3)} and x_2=\frac{-(2m-1)+5}{2(m-3)}

    Hey, what else ??
    Let's see x_1 :

    x_1=\frac{-2m+2-5}{2(m-3)}=\frac{-2(m-3)}{2(m-3)}=\boxed{-1} !!!

    So -1 is always a root of this quadratic polynomial, whatever m is

    This, however, can be explained : (2m-1)-\bigg[(m-3)+(m+2)\bigg]=0, so -1 is a root.



    Isn't it nice ?
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  2. #2
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    Quote Originally Posted by Moo View Post
    Hey, what else ??
    Hey

    Let's see x_1 :

    x_1=\frac{-2m+2-5}{2(m-3)}=\frac{-2(m-3)}{2(m-3)}=\boxed{-1} !!!
    Why?
    ---
    By the way, m\not = 3
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