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Math Help - Inequality solution

  1. #1
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    Inequality solution

    Here's the quadratic I need to solve: 2x^2 -7x > 4
    Now I know the solutions are 4, -\frac {1}{2} but my book shows a number line that essentially has  x > 4 and  x < -\frac {1}{2} Why is this the case? Why aren't they all "greater than" inequalities?
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  2. #2
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    Quote Originally Posted by donnagirl View Post
    Here's the quadratic I need to solve: 2x^2 -7x > 4
    Now I know the solutions are 4, -\frac {1}{2} but my book shows a number line that essentially has  x > 4 and  x < -\frac {1}{2} Why is this the case? Why aren't they all "greater than" inequalities?
    The inequality is equivalent to 2x^2 - 7x - 4 > 0.

    Draw a graph of y = 2x^2 - 7x - 4 and be sure to label the x-intercepts. For what values of x is y > 0 ....?
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  3. #3
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    But all y values are greater than 0 (where they are not is part of the interval I don't need..)
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  4. #4
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    Quote Originally Posted by donnagirl View Post
    But all y values are greater than 0 (where they are not is part of the interval I don't need..)
    Draw the graph!!
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  5. #5
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    I DID draw the graph, it does work--if x < -\frac{1}{2} all values of y are positive. I don't need the part of the interval where the y values are negative. BUT, why can't I manipulate it algebraically? Was I supposed to multiply by a negative or something??
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  6. #6
    Super Member Bacterius's Avatar
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    Consider your quadratic equation :

    2x^2 - 7x - 4

    This can be factorized as :

    (2x + 1)(x - 4)

    Therefore your inequality becomes :

    (2x + 1)(x - 4) > 0

    In order for this inequality to hold, the factors need to have the same sign. Can you follow on ?
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  7. #7
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    Quote Originally Posted by Bacterius View Post
    ...
    In order for this inequality to hold, the factors need to have the same sign. Can you follow on ?
    Not sure what you mean?? That's the same setup I have..yet I got  x > 4, x > -\frac{1}{2}. Why is it failing algebraically?
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  8. #8
    Super Member Bacterius's Avatar
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    It is not failing, you are just not going to the end of things. I would draw a sign table :

    \begin{array}{|l|l|}<br />
  \hline x & -\infty \ \ \ \frac{-1}{2} \ \ \ \ \ \ 4 \ \ \ +\infty \\ \hline<br />
   2x + 1 & \ \ - \ \ \ \ \ 0 \ \ + \ \ | \ \ \ \ + \ \ \ \ \ \\ \hline<br />
   x - 4 & \ \ - \ \ \ \ \ | \ \ \ - \ \ 0 \ \ \ + \ \ \ \ \\ \hline<br />
   (2x + 1)(x - 4) & \ \ + \ \ \ \ \ 0 \ \ - \ \ 0 \ \ \ + \ \ \ \ \\ \hline<br />
\end{array}

    And you get the correct answer x < \frac{-1}{2} and x > 4.

    This is how I was taught to do it, makes it a lot easier to visualize

    That table was a pain to build.

    *PS* : you can achieve the same result using algebra but I forgot how
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  9. #9
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    Now THAT makes sense! Thanks for your hard work on it Bacterius I appreciate it!
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  10. #10
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    Quote Originally Posted by donnagirl View Post
    Now THAT makes sense! Thanks for your hard work on it Bacterius I appreciate it!
    It was my pleasure
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  11. #11
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    Quote Originally Posted by donnagirl View Post
    Now THAT makes sense! Thanks for your hard work on it Bacterius I appreciate it!
    If you had said in the first place that you wanted an algebraic approach rather than a graphical approach (the latter is superior in my opinion, by the way) less time and effort would have been wasted here.
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  12. #12
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    first link in my signature.
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