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Math Help - value of k and roots

  1. #1
    rcs
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    value of k and roots

    Can anybody help... how is it possible to find the root of 16x^2-24x+k = 0 ? when one root is 1/2 and find k?

    it is possible to find the value of k if the sum and product of the roots are equal?

    if im going to solve i understand that in the k is the c part of quadratic equation.
    so,
    k = c/a = k/16 then in this part i am stuck.
    it made me solve for hours but still can't get the answer correctly.

    thank you so much and more power.
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    Re: value of k and roots

    Since 1/2 is a root then you know that (2x -1)(Ax+B) = 16x^2-24x+k where A and B are constants.

    Comparing coefficients

    x^2 \rightarrow 2A = 16 \text{   and   } x^0 \rightarrow -B = k


    To find the value of B you can compare the coefficients of x. Once you have B it should be easy to find k
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  3. #3
    rcs
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    Re: value of k and roots

    -B = k
    and that is -24 = k is this correct sir? or should i have to make it -24 = r1 + r2; - 24 = 1/2 + r2 then -24 -1/2 = r2, then r2 = -24 1/2 which the other root now. Oh my gosh please hope this is right.
    Last edited by mr fantastic; July 10th 2011 at 04:19 PM.
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    Re: value of k and roots

    Comparing the coefficients of x gives 2B-A = -24 and since we know A we have 2B-8 = -24 which makes B = -8. Therefore I get k=8.

    If we sub this back we get 16x^2-24x+8 = 8(2x-1)(x-1) = 0
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  5. #5
    rcs
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    Re: value of k and roots

    thank you sir... hope i can learn this more... i am not so good in mathematics

    more power
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    Re: value of k and roots

    Hello RCS,
    given one root substitute it into your equation and find k =8
    factor the new equation after simplification and find the other root ( 1)



    bjh
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