Results 1 to 6 of 6

Thread: Quadratic Inequality

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    66

    Question Quadratic Inequality

    Given that $\displaystyle y = tx^2 + 8x + 10 - t$ find the range of values of $\displaystyle t $ for which $\displaystyle y$ is always positive.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Dec 2009
    Posts
    226
    The answer depends on both $\displaystyle t$ and $\displaystyle x$.

    $\displaystyle tx^2 + 8x + 10 - t > 0$

    Isolating the variable $\displaystyle t$ produces

    $\displaystyle tx^2 - t > -8x - 10$

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

    $\displaystyle \iff t > \dfrac{-8x - 10}{x^2 - 1}$ if $\displaystyle x^2 - 1 > 0$ or $\displaystyle t < \dfrac{-8x - 10}{x^2 - 1}$ if $\displaystyle x^2 - 1 < 0$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46
    Quote Originally Posted by Ilsa View Post
    Given that $\displaystyle y = tx^2 + 8x + 10 - t$ find the range of values of $\displaystyle t $ for which $\displaystyle y$ is always positive.

    For any $\displaystyle t\neq 0$ express

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

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

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

    Now, solve the last inequality on $\displaystyle t$ .
    Last edited by FernandoRevilla; Apr 7th 2011 at 09:13 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Ilsa View Post
    Given that $\displaystyle y = tx^2 + 8x + 10 - t$ find the range of values of $\displaystyle t $ for which $\displaystyle y$ is always positive.

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

    $\displaystyle a =t > 0$ so we can divide by $\displaystyle t$

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

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

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

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

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

    so the inequality is preserved.

    Here is a fun animation

    Quadratic Inequality-parabola.gifClick image for larger version. 

Name:	parabola.gif 
Views:	14 
Size:	214.2 KB 
ID:	21394


    Last edited by TheEmptySet; Apr 7th 2011 at 09:08 AM. Reason: too slow
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2009
    Posts
    226
    How dull of me not to realize $\displaystyle y > 0 \iff t > 0$ (or clever of you to realize so).
    Last edited by NOX Andrew; Apr 7th 2011 at 09:14 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46
    Quote Originally Posted by NOX Andrew View Post
    The answer depends on both $\displaystyle t$ and $\displaystyle x$.
    It seems clear that $\displaystyle t$ is a parameter and $\displaystyle x$ the variable i.e. the question should be:

    Find all values of $\displaystyle t\in\mathbb{R}$ such that $\displaystyle y_t(x)=tx^2+8x+10-t> 0$ for all $\displaystyle x\in\mathbb{R}$ .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Quadratic Inequality
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Apr 8th 2011, 09:27 AM
  2. Quadratic Inequality
    Posted in the Algebra Forum
    Replies: 2
    Last Post: May 12th 2009, 10:51 PM
  3. quadratic inequality
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Nov 27th 2008, 11:56 PM
  4. Quadratic Inequality
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: Nov 17th 2008, 07:42 AM
  5. inequality/quadratic!!!
    Posted in the Algebra Forum
    Replies: 0
    Last Post: May 19th 2008, 07:57 AM

Search Tags


/mathhelpforum @mathhelpforum