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Math Help - Simple inequality

  1. #1
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    Simple inequality

    The question is:

    Given that the equation kx^2 - 4x + (k-3) = 0 has real roots, show that k^2 - 3k - 4 < 0.
    Find the range of values of k satisfying this inequality.

    b^2 - 4ac = 0

    (-4)^2 - 4(k)(k-3) = 0
    16 - 4k^2 + 12k = 0
    (/4 4 - k^2 + 3k = 0
    (Rearranged) k^2 - 3k - 4 = 0
    (Factorized) (k-4)(k+1) = 0
    k = 4, -1


    I'm guessing that the rest involves the solution of an inequality but what would I do here?
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  2. #2
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    Hello, db5vry!

    You had a right approach, but we want an inequality . . .


    Given that the equation kx^2 - 4x + (k-3) \:= \:0 has real roots,

    (a) show that: k^2 - 3k - 4 \:< \:0

    (b) Find the range of values of k satisfying this inequality.
    (a) A quadratic has two real roots if its discriminant is positive: . b^2 - 4ac \:> \:0

    . . So we have: . (\text{-}4)^2 - 4(k)(k-3) \:>\:0 \quad\Rightarrow\quad -4k^2 + 12k + 16 \:>\:0

    . . Divide by -4: . \boxed{k^2 - 3k - 4 \:<\:0}



    (b) We have: . k^2 - 3k - 4 \:<\:0

    . . Factor: . (k+1)(k-4) \:<\:0


    The product of two factors is negative if they have opposite signs.
    So there are two cases:

    [1]\;\;\begin{array}{ccc}k+1 \:< 0 & \Rightarrow & k \:<\:\text{-}1 \\ k-4\:>\:0 & \Rightarrow & k \:>\:4 \end{array} \quad\Rightarrow\quad k  < \text{-1}\,\text{ and }\,k > 4\quad\hdots\;\;\text{Impossible}

    [2]\;\;\begin{array}{ccc}k+1 \:>\:0 & \Rightarrow & k \:>\:\text{-}1 \\ k-4 \:<\:0 & \Rightarrow & k \:<\:4 \end{array}\quad\Rightarrow\quad k > \text{-}1\,\text{ and }\,k < 4\quad\hdots\;\;\text{Yes!}


    Answer: . \boxed{\text{-}1 \:<\:k\:<\:4}

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