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Finding for k in quadratic equation.

  1. Aug 23, 2011 #1
    1. The problem statement, all variables and given/known data


    Find the least integral value of k for which the quadratic polynomial
    (k-2)x2 + 8x + k+4 > 0 where x is real.



    3. The attempt at a solution

    i am trying to solve the discriminant by equating it to>0
    D>0
    but i don't think it is correct.
    Please provide hints for this solutions.
     
  2. jcsd
  3. Aug 23, 2011 #2

    SteamKing

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    Show us what you have done.
     
  4. Aug 23, 2011 #3
    D=b2-4ac
    =64-4(k-2)(k+4)

    as x is real
    D>0

    .'. 64-4(k-2)(k+4)>0

    on solving
    i got
    (k-4)(k+6)<0

    after that what should I do

    should i find the values of k from this inequality???
     
  5. Aug 23, 2011 #4

    eumyang

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    No, it's not... at least the way I'm interpreting the question.

    You are looking for the smallest integer k such that, if you plug in ANY REAL NUMBER for x, the quadratic becomes positive. That means that the graph of the parabola would be entirely above the x-axis. What does that say about the discriminant?
     
  6. Aug 23, 2011 #5

    SteamKing

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    You might want to check your algebra on trying to solve for k in the inequality. What happened to the constant 64?
     
  7. Aug 23, 2011 #6

    eumyang

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    The algebra is actually correct. The OP multiplied the binomials, combined like terms, and then divided both sides by -4. The inequality symbol used in the beginning is wrong, however.
     
  8. Aug 23, 2011 #7
    the smallest value of the function:[tex]f(x)=ax^2+bx+c=0[/tex]

    here[tex]f(x)=(k-2)x^2+8x+k+4>0[/tex]
    is

    [tex]\frac {-D}{4a}[/tex]

    to get this value positive D should be negative-------------(I)
    and solving according this we get
    64-4(k-2)(k+4) < 0
    as this may result in correct answer

    but as x is real .'. D should be positive or zero--------------(II)
    [tex](I)\neq(II)[/tex]

    I am confused with (I) (II)
    ????
     
    Last edited: Aug 23, 2011
  9. Aug 23, 2011 #8
    i got the answer
    by using this

    [tex]\frac{-D}{4a}[/tex]

    before i was running on the wrong concept:confused:

    now i got the answer
    thank you very much for your valuable suggestion and valuable time for me

    thank you once again:smile::smile:
     
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