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    Question:

    find the value of k (\neq 1) such that the quadratic function of x

    k(x+2)^2 - (x-1)(x-2)
    is equal to zero for only one value of x
    Find also, (a) th range of vaules of k for which the function possess a minimum value, (b) the range of values of k for which the value of the function never exceeds 12.5

    sketch the graph of the function for  k=\frac{1}{2} and  k = 2\frac{1}{2}


    My attempt:

    let
      f(x) =  k(x+2)^2 - (x-1)(x-2)
      k(x^2+4x+4)-(x^2-3x+2)

    i skip the a couple of steps ( please note it is only to save time)

    i got  x^2(k-1)+ x(4k+3)+4k-2

    since we need to find the value of k when x = 0
    in other for one of the root to be 0, the quadratic function has to be in the form;  x^2 + bx where c = 0

    since c = 0
     c = 4k-2
    thus  4k-2 =0
    k = 0.5


    for (a)
    using   x^2(k-1)+ x(4k+3)+4k-2
    complete the square

     (k-1)\left [ x^2 + \frac{x(4k+3)}{(k-1)} + \frac{4k-2}{k-1}\right ]

     (k-1)\left [ \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + \frac{4k-2}{k-1} - \left ( \frac{(4k+3)}{2(k-1)} \right )^2\right ]

     (k-1) \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + 4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2

    completing the square produces;
     (k-1) \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + 4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2

    since f(x) has a minimum function a > 0, thus it has a least value of  \frac{4ac-b^2}{4a} when x =  -\frac{b}{2a}
    therefore if x =  -\frac{ 4k+3}{2(k-1)}
    then use  \left ( x + \frac{4k+3}{2(k-1)} \right ) = 0  to find k
    x =  -\frac{ 4k+3}{2(k-1)}
     -\frac{ 4k+3}{2(k-1)} = 0 k =  - \frac{3}{4}
    sub k = -(3/4) into   4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2
    since  4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2  =  \frac{4ac-b^2}{4a}

    therefore sub in  k = - \frac{3}{4} into  4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2

    .k > 175/4


    part b

    using  (k-1) \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + 4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2

    note the maximum value = 12.5

    therefore   4k-2 -  \frac{(4k+3)^2}{4(k-1)}  =  12.5

     \frac{(4k-2)(4k-4) - (4k+3)^2}{4(k-1)} = 12.5

     (4k-2)(4k-4) - (4k+3)^2 = (4)12.5(k-1)
     (16^2 -24k + 8) - (16k^2 + 24k + 9) = 50k - 50
     -48k -1 = 50k -50
     49 = 98k
     k = \frac{1}{2}



    In the bostock book 'The core Course for A-level' The answer was k = -(1/48) , a) k > 1, b)  k \leq  \frac{1}{2}


    please help me, point out where i went wrong
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    (Original post by bigmansouf)
    Question:

    find the value of k (\neq 1) such that the quadratic function of x

    k(x+2)^2 - (x-1)(x-2)
    is equal to zero for only one value of x
    Find also, (a) th range of vaules of k for which the function possess a minimum value, (b) the range of values of k for which the value of the function never exceeds 12.5

    sketch the graph of the function for  k=\frac{1}{2} and  k = 2\frac{1}{2}

    please help me, point out where i went wrong
     (k-1)\left [ \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + \frac{4k-2}{k-1} - \left ( \frac{(4k+3)}{2(k-1)} \right )^2\right ]

    (k-1) \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + 4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2

    Are you sure you expanded the square bracket correctly?
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    X doesn't equal 0. That's your mistake, I think. (This seems to have a carry on effect for the rest of the question)
    I think the first part is asking for a value for k such that the equation has only one solution.

    EDIT: On doing the maths on paper, this is what the question is asking for. I get k=-1/48, which matches your book
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    (Original post by an_atheist)
    X doesn't equal 0. That's your mistake, I think. (This seems to have a carry on effect for the rest of the question)
    I think the first part is asking for a value for k such that the equation has only one solution.

    EDIT: On doing the maths on paper, this is what the question is asking for. I get k=-1/48, which matches your book
    Good spot.

    At first glance I interpreted the question in the same way as the OP.
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    (Original post by an_atheist)
    X doesn't equal 0. That's your mistake, I think. (This seems to have a carry on effect for the rest of the question)
    I think the first part is asking for a value for k such that the equation has only one solution.

    EDIT: On doing the maths on paper, this is what the question is asking for. I get k=-1/48, which matches your book
    thanks for correcting me
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    (Original post by notnek)
     (k-1)\left [ \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + \frac{4k-2}{k-1} - \left ( \frac{(4k+3)}{2(k-1)} \right )^2\right ]

    (k-1) \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + 4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2

    Are you sure you expanded the square bracket correctly?
     (k-1)\left [ \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + \frac{4k-2}{k-1} - \left ( \frac{(4k+3)}{2(k-1)} \right )^2\right ]

    (k-1) \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + 4k-2 - \left ( \frac{(4k+3)^2}{4(k-1)} \right )

    im sorry but i am stuck on this question
    normally to find the range of value of k you use the examples stated in Chandler's book as a guide to find it

    I understand why it k > 1. We know that a quadratic function hasa a minimum value if a >0. therefore since for part a it has a minimum value for the quadratic function k(x+2)^2 - (x-1)(x-2)
    rearranging the function gives

     (k-1)x^2 + (4k+3)x +4k-2
    since a = k-1 and a > 0 thus k must be k> 1 in order to have minimum value

    The problem is now although i understand the question and have the answer how do i write a mathematically structured answer to prove that k > 1
 
 
 
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