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Math Help - A Quick Question of Inequalities

  1. #1
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    Question A Quick Question of Inequalities

    I was a bit confused on something.
    It's inequalities.

    The question is
    Find the set values of x for which:
    2x^2 - 7x + 3 > 0

    So, I did this.
    (2x - 1) (x - 3) > 0
    And after quickly sketching it onto a graph, chose the 'inside' regions.
    x > 1/2
    x < 3

    My question is, Is it actually supposed to be the 'outside' regions? ie.
    x < 1/2
    x > 3

    Thanks.
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  2. #2
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    Re: A Quick Question of Inequalities

    Quote Originally Posted by dementedsquirrel View Post
    I was a bit confused on something.
    It's inequalities.

    The question is
    Find the set values of x for which:
    2x^2 - 7x + 3 > 0

    So, I did this.
    (2x - 1) (x - 3) > 0
    And after quickly sketching it onto a graph, chose the 'inside' regions.
    x > 1/2
    x < 3

    My question is, Is it actually supposed to be the 'outside' regions? ie.
    x < 1/2
    x > 3

    Thanks.
    outside ... note the graph of y = 2x^2 - 7x + 3 is a parabola that opens upward and has roots at x = 1/2 and x = 3. the y-values of the parabola are negative (i.e. less than 0) between those two roots.
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  3. #3
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    Re: A Quick Question of Inequalities

    Thanks a lot for your help.
    I drew a complete blank for a while.
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  4. #4
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    Re: A Quick Question of Inequalities

    Quote Originally Posted by dementedsquirrel View Post
    I was a bit confused on something.
    It's inequalities.

    The question is
    Find the set values of x for which:
    2x^2 - 7x + 3 > 0

    So, I did this.
    (2x - 1) (x - 3) > 0
    And after quickly sketching it onto a graph, chose the 'inside' regions.
    x > 1/2
    x < 3

    My question is, Is it actually supposed to be the 'outside' regions? ie.
    x < 1/2
    x > 3

    Thanks.
    The most direct method to solve quadratic inequalities is to complete the square and make use of the absolute value function.

    \displaystyle \begin{align*} 2x^2 - 7x + 3 &> 0 \\ 2\left(x^2 - \frac{7}{2}x + \frac{3}{2}\right) &> 0 \\ 2\left[x^2 - \frac{7}{2} + \left(-\frac{7}{4}\right)^2 - \left(-\frac{7}{4}\right)^2 + \frac{3}{2}\right] &> 0 \\ 2\left[\left(x - \frac{7}{4}\right)^2 - \frac{49}{16} + \frac{24}{16}\right] &> 0 \\ 2\left[\left(x - \frac{7}{4}\right)^2 - \frac{25}{16}\right] &> 0 \\ \left(x - \frac{7}{4}\right)^2 - \frac{25}{16} &> 0 \\ \left(x - \frac{7}{4}\right)^2 &> \frac{25}{16} \\ \left|x - \frac{7}{4}\right| &> \frac{5}{4} \\ x - \frac{7}{4} < -\frac{5}{4} \textrm{ or }x - \frac{7}{4} &> \frac{5}{4} \\ x < \frac{1}{2} \textrm{ or } x &> 3 \end{align*}
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  5. #5
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    Smile Re: A Quick Question of Inequalities

    Thanks.
    This way is good, but the best way for me (during an exam - for a couple of marks) personally is to factorise and work it out accordingly.
    Though thanks for the tips!
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