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Math Help - Location of roots problem.

  1. #1
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    Location of roots problem.

    Let
    4x^2-4(α-2)x+α-2=0
    (α is real number)

    be a quadratic equation, then find the value of α for which both the roots are positive.


    the conditions will be
    1) Discriminant D≥0
    by this condition i got α (-∞,2][3,∞)
    2) f(0) greater than or equal to 0
    by this we get α (2,∞)

    3) now should i use -b/2a(point exactly between both roots)
    and equate as -b/2a greater than 0


    if-3rd point is right then what will be the final answer
    α (?,?)union(?,?)
    please provide help
    Attached Thumbnails Attached Thumbnails Location of roots problem.-iag94.png  
    Last edited by sumedh; August 29th 2011 at 06:45 AM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Location of roots problem.

    Quote Originally Posted by sumedh View Post
    Let  4x^2-4(α-2)xα-2=0   (α\epsilon R)
    What does this mean?
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  3. #3
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    Re: Location of roots problem.

    sorry the alpha are not seen α=alpha
    4x^2-4(α-2)x+α-2=0
    where alpha is real.
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Location of roots problem.

    What you've done is fine. The condition that the equation has two roots can be satisfied by D\geq 0 and so
    \alpha \in ]-\infty, 2]\cup [3,+\infty[

    To continue I would say, let \delta, \gamma be the two positive roots of the function and therefore:
    \delta=\frac{-b+\sqrt{D}}{2a}=\frac{4(\alpha-2)+\sqrt{16(\alpha-2)^2-16(\alpha-2)}}{8}\geq 0
    \gamma =\frac{-b-\sqrt{D}}{2a}=\frac{4(\alpha-2)-\sqrt{16(\alpha-2)^2-16(\alpha-2)}}{8}\geq 0

    Now, You have to solve two irrational inequality's in function of \alpha and so you'll find out the values of \alpha wherefore \delta\geq 0 and \gamma \geq 0

    This is the way I would try to solve this question.
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  5. #5
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    Re: Location of roots problem.

    Thanks i got the answer,
    But in a different way:-
    This method is based on the graph of quadratic equation that is a parabola

    In third point ' 3)...... '

    As (-b/2a ,-D/4a) is the vertex of parabola
    it's x component lies exactly in the middle of the two roots( in case of one roots they coincide)

    So it(-b/2a) will be greater than the smallest root,
    and hence it will be greater than zero.
    As zero is the point on x axis from where positive numbers start,
    so taking it as reference point.
    We get that -b/2a will be greater than 0
    equating this we will get an interval (2,∞).

    From this and other two interval in question
    (-∞,2][3,∞)-------------(1)
    (2,∞) ------------(2)
    We get final interval as
    [3,∞)

    To continue I would say, let \delta, \gamma be the two positive roots of the function and therefore:
    \delta=\frac{-b+\sqrt{D}}{2a}=\frac{4(\alpha-2)+\sqrt{16(\alpha-2)^2-16(\alpha-2)}}{8}\geq 0
    \gamma =\frac{-b-\sqrt{D}}{2a}=\frac{4(\alpha-2)-\sqrt{16(\alpha-2)^2-16(\alpha-2)}}{8}\geq 0

    Now, You have to solve two irrational inequality's in function of \alpha and so you'll find out the values of \alpha wherefore \delta\geq 0 and \gamma \geq 0
    Thank you for your method.
    I will try this
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  6. #6
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    Re: Location of roots problem.

    Quote Originally Posted by sumedh View Post
    Let

    4x^2-4(a-2)x+a-2=0

    (a is real number)

    be a quadratic equation, then find the value of a for which both the roots are positive.


    the conditions will be
    1) Discriminant D\ge 0 by this condition i got a: (-\infty,2][3,\infty)
    Now Descartes rule of signs tells us under the conditions in force here that neither of the roots are negative iff (a-2)\ge 0, or rather a\ge 2, so we are now restricted to a \in [3,\infty).

    CB
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  7. #7
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    Re: Location of roots problem.

    Thank you very much
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