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Math Help - Roots of quadratics.

  1. #1
    Member grgrsanjay's Avatar
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    Roots of quadratics.

    if the greater root of the equation x^2 -(a^2 +a+1)x+a(a^2+1)=0 lies between the roots of the equation x^2 -a^2 x-2(a^2 -2) =0 then a lies in
    (a)[1,2]
    (b)[-2,-1]
    (c)[-8,-root5]U[root5,I]
    (d)none of these
    Last edited by mr fantastic; August 8th 2010 at 06:59 PM. Reason: Re-titled.
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  2. #2
    Master Of Puppets
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    What are the roots of these equations?

    Do you know how to find them?

    Hint: x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}
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  3. #3
    Member grgrsanjay's Avatar
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    i know to find them
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  4. #4
    Senior Member eumyang's Avatar
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    The solutions of the quadratic equation
    x^2 -a^2 x-2(a^2 -2) = 0 are
    \begin{aligned}<br />
x &= \dfrac{a^2 \pm \sqrt{a^4 - 4(1)(2)(a^2 - 2)}}{2} \\<br />
&= \dfrac{a^2 \pm \sqrt{a^4 - 8a^2 + 16}}{2} \\<br />
&= \dfrac{a^2 \pm (a^2 - 4)}{2} \\<br />
&= 2,\:a^2 - 2<br />
\end{aligned}

    The solutions of the quadratic equation
    x^2 -(a^2 +a+1)x+a(a^2+1) = 0 are
    \begin{aligned}<br />
x &= \dfrac{a^2 + a + 1 \pm \sqrt{(-(a^2 +a + 1))^2 - 4(1)(a)(a^2 + 1)}}{2} \\<br />
&= \dfrac{a^2 + a + 1 \pm \sqrt{a^4 + 2a^3 + 3a^2 + 2a + 1 -4a^3 - 4a}}{2} \\<br />
&= \dfrac{a^2 + a + 1 \pm \sqrt{a^4 - 2a^3 + 3a^2 - 2a + 1}}{2} \\<br />
&=  \dfrac{a^2 + a + 1 \pm (a^2 - a + 1)}{2} \\<br />
&= a,\:a^2 + 1<br />
\end{aligned}
    The greater root here is a^2 + 1.
    ...

    Jeez, I could have "factored" these...
    Can you take it from here?
    Last edited by eumyang; August 8th 2010 at 07:37 PM.
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  5. #5
    Senior Member eumyang's Avatar
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    (Duplicate post).
    Last edited by eumyang; August 8th 2010 at 07:36 PM.
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