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Math Help - inequality question

  1. #1
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    Question inequality question

    Find the range of values of k if kx^2+8x>6-k for all real values of x

    What does for all the real values of x mean?

    The answer is k>8 and I can't seem to get it
    this is my working by the way:

    Kx^2 +8x+k-6 >0
    a=k b=8 c=k-6
    sub into b^2-4ac
    4k^2-24k-64<0
    k^2-6k-16<0
    (k-8)(k+2)<0

    thank you!
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  2. #2
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    Re: inequality question

    Quote Originally Posted by pumbaa213 View Post
    Find the range of values of k if kx^2+8x>6-k for all real values of x
    The answer is k>8 and I can't seem to get it
    this is my working by the way:
    Kx^2 +8x+k-6 >0
    a=k b=8 c=k-6
    sub into b^2-4ac
    4k^2-24k-64<0
    k^2-6k-16<0
    (k-8)(k+2)<0
    It should be clear to you the inequaity is not true if k<0. Example: k=-1~\&~x=10
    The parabola opens down if k<0.

    On the other hand, we do need b^2-4ac>0. WHY?
    Thanks from topsquark and pumbaa213
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  3. #3
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    Re: inequality question

    I am sorry I still haven't managed to grasp it yet. Does real values of x means that the parabola has to be a positive and u shaped one?
    And I'm not sure why we need b^2-4ac >0
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  4. #4
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    Re: inequality question

    Quote Originally Posted by pumbaa213 View Post
    Find the range of values of k if kx^2+8x>6-k for all real values of x

    What does for all the real values of x mean?

    The answer is k>8 and I can't seem to get it
    this is my working by the way:

    Kx^2 +8x+k-6 >0
    a=k b=8 c=k-6
    sub into b^2-4ac
    4k^2-24k-64<0
    k^2-6k-16<0
    (k-8)(k+2)<0

    thank you!
    \displaystyle \begin{align*} k\,x^2 + 8x &> 6 - k \\ k\,x^2 + 8x + k - 6 &> 0 \\ k \left( x^2 + \frac{8}{k}\,x + 1 -\frac{6}{k} \right) &> 0 \\ k \left[ x^2 + \frac{8}{k} \, x + \left( \frac{4}{k} \right) ^2 - \left( \frac{4}{k} \right) ^2 + 1 - \frac{6}{k} \right] &> 0 \\ k \left[ \left( x + \frac{4}{k} \right) ^2 + 1 - \frac{6}{k} - \frac{16}{k^2} \right] &> 0 \end{align*}

    Notice you have the product of two things being positive. This can only happen if both things are positive or both things are negative.

    Case 1: \displaystyle \begin{align*} k > 0 \end{align*} and \displaystyle \begin{align*} \left( x + \frac{4}{k} \right) ^2 + 1 - \frac{6}{k} - \frac{16}{k^2} > 0 \end{align*}. Working on the second inequality gives

    \displaystyle \begin{align*} \left( x + \frac{4}{k} \right) ^2 + 1 - \frac{6}{k} - \frac{16}{k^2} &> 0 \\ \left( x + \frac{4}{k} \right) ^2 &> \frac{16}{k^2} + \frac{6}{k} - 1 \\ \left( x + \frac{4}{k} \right) ^2 &> \frac{16 + 6k - k^2}{k^2} \\ \left( x + \frac{4}{k} \right) ^2 &> \frac{-\left( k^2 - 6k - 16 \right) }{k^2} \\ \left( x + \frac{4}{k }\right) ^2 &> \frac{- \left[ k^2 - 6k + (-3) ^2 - (-3 ) ^2 - 16 \right] }{k^2} \\ \left( x + \frac{4}{k} \right) ^2 &> \frac{- \left[ \left( k - 3 \right) ^2 - 25 \right] }{k^2} \\ \left( x + \frac{4}{k } \right) ^2 &> \frac{25 - \left( k - 3 \right) ^2 }{k^2} \\ \left| x + \frac{4}{k} \right| &> \frac{\sqrt{25 - \left( k - 3 \right) ^2 }}{k}  \end{align*}

    Now clearly this can only have solutions if \displaystyle \begin{align*} 25 - \left( k - 3 \right) ^2 > 0 \end{align*}, so

    \displaystyle \begin{align*} 25 - \left( k - 3 \right) ^2 &> 0 \\ 25 &> \left( k - 3 \right) ^2 \\ \left( k - 3 \right) ^2 &< 25 \\ | k - 3 | &< 5 \\ -5 < k - 3 &< 5 \\ -2 < k &< 8 \end{align*}

    So putting the inequalities together, we find that \displaystyle \begin{align*} 0 < k < 8 \end{align*}.

    Now try Case 2, the situation where both quantities have to be negative...
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  5. #5
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    Re: inequality question

    Quote Originally Posted by pumbaa213 View Post
    I am sorry I still haven't managed to grasp it yet. Does real values of x means that the parabola has to be a positive and u shaped one? And I'm not sure why we need b^2-4ac >0
    For the quadratic equation ax^2+bx+c=0 we do call \Delta=b^2-4ac is the discriminant.
    The name comes from the fact it discriminates between real and complex roots.

    If \Delta <0 the quadratic has no real roots. That means the entire graph is either above the x-axis or below the x-axis.

    Let's review the OP
    Quote Originally Posted by pumbaa213 View Post
    Find the range of values of k if kx^2+8x>6-k for all real values of x
    kx^2 +8x+k-6 >0
    Here \Delta=64-4(k)(k-6). If k<0 we see 6-k>0 so there are no real roots.

    So this leaves one possible solution set: k>8
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    Re: inequality question

    Thanks everyone for the help!
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