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Math Help - Finding for k in quadratic equation.

  1. #1
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    Finding for k in quadratic equation.

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



    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.
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  2. #2
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    Re: Finding for k in quadratic equation.

    Quote Originally Posted by sumedh View Post
    Find the least integral value of k for which the quadratic polynomial
    (k - 2)x^2 + 8x + k+4 > 0 where x is real.



    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.
    I would complete the square...
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  3. #3
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    Re: Finding for k in quadratic equation.

    Quote Originally Posted by sumedh View Post
    Find the least integral value of k for which the quadratic polynomial
    (k - 2)x^2 + 8x + k+4 > 0 where x is real.
    If \Delta is the discriminant make \Delta <0 so there are no real roots. Then look at the graph to see where it is positive.
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  4. #4
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    Re: Finding for k in quadratic equation.

    i did in this way

    d=discriminant


    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???
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    Re: Finding for k in quadratic equation.

    Quote Originally Posted by sumedh View Post
    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
    So k<-6\text{ or }k>4.
    I suggest that in each of those cases you make a graph of an example.
    You want the graph to be above the x-axis.
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    Re: Finding for k in quadratic equation.

    but the final answer is k=5
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    Re: Finding for k in quadratic equation.

    Quote Originally Posted by sumedh View Post
    but the final answer is k=5
    The question clearly asks for the least integer that works.
    What is the least integer greater than 4?
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  8. #8
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    Re: Finding for k in quadratic equation.

    but why it cannot be -7 or - infinity or some integer less than -6
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    Re: Finding for k in quadratic equation.

    Quote Originally Posted by sumedh View Post
    but why it cannot be -7 or - infinity or some integer less than -6
    GRAPH the function for k=-7. Is it positive everywhere?
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  10. #10
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    Re: Finding for k in quadratic equation.

    i think that on solving (k-4)(k+6)<0
    we get -6<k<4

    ???
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  11. #11
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    Re: Finding for k in quadratic equation.

    Quote Originally Posted by sumedh View Post
    i think that on solving (k-4)(k+6)<0
    we get -6<k<4
    You have made some algebra errors somewhere.

    Solve 8^2-4(k-2)(k+4)<0 and you get k<-6\text{ or }4<k~.
    Typo is fixed.
    Last edited by Plato; August 23rd 2011 at 10:57 AM.
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  12. #12
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    Re: Finding for k in quadratic equation.

    no it is not ...(k+1)
    it is ...(k+4)




    i think that i have to first find the zeros of the following

    (k-4)(k+6)<0

    that is k=-6 and k=4

    then try putting any value less than -6 then value between -6 and 4
    and at last value greater than 4
    thus we will get the solution as -6<k<4 ???
    Last edited by sumedh; August 23rd 2011 at 11:03 AM. Reason: edit
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  13. #13
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    Re: Finding for k in quadratic equation.

    Quote Originally Posted by sumedh View Post
    no it is not ...(k+1)
    it is ...(k+4)
    That was a TYPO.
    It is now fixed. See reply #11
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  14. #14
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    Re: Finding for k in quadratic equation.

    the smallest value of the function: f(x)=ax^2+bx+c=0

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

    \frac {-D}{4a}

    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)
    (I)\neq(II)

    I am confused with (I) (II)
    ????
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  15. #15
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    Re: Finding for k in quadratic equation.

    i got the answer
    by using this

    \frac{-D}{4a}

    before i was running on the wrong concept

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

    thank you once again
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