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Math Help - Quadratic Inequality

  1. #1
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    Question Quadratic Inequality

    Given that y = tx^2 + 8x + 10 - t find the range of values of t for which y is always positive.
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  2. #2
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    The answer depends on both t and x.

    tx^2 + 8x + 10 - t > 0

    Isolating the variable t produces

    tx^2 - t > -8x - 10

    \iff t(x^2 - 1) > -8x - 10

    \iff t > \dfrac{-8x - 10}{x^2 - 1} if x^2 - 1 > 0 or t < \dfrac{-8x - 10}{x^2 - 1} if x^2 - 1 < 0
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Ilsa View Post
    Given that y = tx^2 + 8x + 10 - t find the range of values of t for which y is always positive.

    For any t\neq 0 express

    y(x)=\ldots=t\left(x+\dfrac{4}{t}\right)^2-\dfrac{16}{t}+10-t

    We have to find only values of t>0 (why?). For t>0 , y(x)> 0 for all x\in\mathbb{R} iff

    y(-4/t)=-\dfrac{16}{t}+10-t>0

    Now, solve the last inequality on t .
    Last edited by FernandoRevilla; April 7th 2011 at 09:13 AM.
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by Ilsa View Post
    Given that y = tx^2 + 8x + 10 - t find the range of values of t for which y is always positive.

    First for y to always be positive the parabola must open up this means that
    y=ax^2+bx+c

    a =t > 0 so we can divide by t

    Now we know that the vertex will be the lowest point and it occurs at

    x=-\frac{b}{2a}=\frac{-8}{2t}=\frac{-4}{t}

    Now if we put this back into the equation we want to solve

    \displaystyle y > 0 \iff t\left( \frac{-4}{t}\right)^2+8\left( \frac{-4}{t}\right)+10-t >0 \iff t^2-10t+16 <0

    We can reduce to this because we already know that t > 0

    so the inequality is preserved.

    Here is a fun animation

    Quadratic Inequality-parabola.gifClick image for larger version. 

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    Last edited by TheEmptySet; April 7th 2011 at 09:08 AM. Reason: too slow
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  5. #5
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    How dull of me not to realize y > 0 \iff t > 0 (or clever of you to realize so).
    Last edited by NOX Andrew; April 7th 2011 at 09:14 AM.
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by NOX Andrew View Post
    The answer depends on both t and x.
    It seems clear that t is a parameter and x the variable i.e. the question should be:

    Find all values of t\in\mathbb{R} such that y_t(x)=tx^2+8x+10-t> 0 for all x\in\mathbb{R} .
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