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Thread: common root

  1. #1
    Junior Member
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    common root

    For which values of b do the equations:

    x^3 + bx^2 + 2bx - 1 = 0 and x^2 + (b-1)x + b = 0

    have a common root?
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  2. #2
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
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    I believe there's a little trick involved here:

    Let
    (1) $\displaystyle x^3+bx^2+2bx-1=0$
    (2)$\displaystyle x^2+(b-1)x+b=0$

    Observe that $\displaystyle x=-1$ is not a root of (2) for any choice of b. Hence this can not be a common root.
    So we may replace the problem of finding b such that (3),(4) have a common root

    (3) $\displaystyle x^3+bx^2+2bx-1=0$
    (4) $\displaystyle (x+1)(x^2+(b-1)x+b)=0 $

    Then a common root of (3),(4) must also be a root of the difference of (3),(4)

    $\displaystyle (3)-(4)= x^3+bx^2+2bx-1 - (x+1)(x^2+(b-1)x+b) = x-(1+b)$

    This implies that $\displaystyle x=1+b$ is the only possible common root they can have.

    substituting $\displaystyle x= 1+b$ in (2) gives 2 possible values for b.
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