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Math Help - Need help: Find for the values....

  1. #1
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    Smile Need help: Find for the values....

    Hi, everyone..I'm a new recruit and just joined this forum several weeks ago. So, hope u all can help me through every difficulties I face..yeah, this is my first ever question and I hope there's somebody can help me out...I'm not sure its the question problems or its mine problems..anyways have a look, thanks..







    a) Find the set of values of m for which the graph of y = (3 - m)x^2 + 4x - m lies above the x-axis for all real values of x.




















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  2. #2
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    Quote Originally Posted by Ah Chan View Post
    Hi, everyone..I'm a new recruit and just joined this forum several weeks ago. So, hope u all can help me through every difficulties I face..yeah, this is my first ever question and I hope there's somebody can help me out...I'm not sure its the question problems or its mine problems..anyways have a look, thanks..







    a) Find the set of values of m for which the graph of y = (3 - m)x^2 + 4x - m lies above the x-axis for all real values of x.

















    Hello fellow Malaysian ,

    note that it has no real roots since it is floating above the x-axis and this implies that its discriminant ie b^2-4ac<0

    where a=3-m , b=4 and c=-m .

    Then this is a positive graph , where a>0 , so you will need to pick the correct ranges of m at the end .
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  3. #3
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    Hello, Ah Chan!

    Find the set of values of m for which the graph of . y \:=\: (3 - m)x^2 + 4x - m
    lies above the x-axis for all real values of x.
    First of all, the parabola must open upward.
    . . Hence, the leading coefficient must be positive: . 3-m \:>\:0 \quad\Rightarrow\quad m \:<\:3 .[1]

    Second, there are no x-intercepts.
    . . That is, the equation: . (3-m)x^2 + 4x - m \:=\:0 . has no real roots.

    Quadratic Formula: . x \;=\;\frac{-4 \pm\sqrt{4^2 - 4(3-m)(-m)}}{2(3-m)} \;=\;\frac{-2 \pm\sqrt{(1+m)(4-m)}}{3-m}

    To have no real roots, the discriminant must be negative: . (1+m)(4-m) \:<\:0


    There are two cases: . \begin{array}{ccc}(1) & \text{(pos)}\times\text{(neg.)} \\ \\[-4mm] (2) & \text{(neg)}\times\text{(pos)} \end{array}


    (1)\;\begin{array}{ccc}1+m \:>\:0 & \Longrightarrow & m \:>\:\text{-}1 \\ 4-m \:<\:0 & \Longrightarrow & m \:>\:4 \end{array} \quad\Longrightarrow\quad m > 4\quad\hdots . which contradicts [1].


    (2)\:\begin{array}{ccc}1+m \:<\:0 & \Longrightarrow & m \:<\:\text{-}1 \\ 4-m \:>\:0 & \Longrightarrow & m \:<\:4\end{array}\quad\Longrightarrow\quad m \:<\:-1 \quad\hdots . which agrees with [1].


    . . . . Therefore: . m \;<\;-1

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