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Thread: Different methods of solving Inequalities

  1. #1
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    Different methods of solving Inequalities

    Hi folks,


    When asked to find the range of values x can take given an inequality, there are at least three methods:

    Sketching the function, completing the square or checking the sign of the factors. All three should agree, naturally.


    So, given $y = 4x^4 - 17x^2 + 4 < 0 $


    let $y = x^2$ then


    $4y^2 - 17y + 4 < 0$


    $y^2 -\frac{17}{4}y + 1 < 0$ ...................................(1)


    $(4y - 1)(y - 4) < 0 $


    so, examining the factors:


    $
    \begin{array}{lccc}
    & y < \frac{1}{4} & \frac{1}{4} < y < 4 & y > 4 \\
    4y - 1 & - & + & + \\
    y - 4 & - & - & + \\
    f(y) & + & - & +
    \end{array}
    $


    we get $ \frac{1}{4} < y < 4$


    resubstituting, $\frac{1}{4} < x^2 < 4$


    and a little simplification yields the range:


    $-2 < x < -\frac{1}{2} , \frac{1}{2} < x < 2$


    Now,
    if I decided to complete the square:


    $y^2 -\frac{17}{4}y + 1 < 0$ then from (1)


    we get


    $(y - \frac{17}{8})^2 -(\frac{17}{8})^2 + 1 < 0$


    i.e.


    $(y -\frac{17}{8})^2 < \frac{289}{64} -\frac{64}{64} $


    which gives


    $(y - \frac{17}{8}) < \frac{15}{8}$


    whence


    $y < 4$ or $x^2 < 4$


    giving the range


    $ -2 < x < 2$ which does not match the result above.

    clearly there is an arithmetic error somewhere. I can't see it, can anyone else?
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  2. #2
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    Re: Different methods of solving Inequalities

    $y = 4x^4-17x^2+4<0$

    $(4x^2-1)(x^2-4) < 0$

    critical values at $x = \pm \dfrac{1}{2}$ and $x=\pm 2$

    $x \in (-\infty,-2) \implies y > 0$

    $x \in \left(-2, -\dfrac{1}{2}\right) \implies y < 0$

    $x \in \left(-\dfrac{1}{2}, \dfrac{1}{2} \right) \implies y > 0$

    $x \in \left(\dfrac{1}{2}, 2 \right) \implies y < 0$

    $x \in (2,\infty) \implies y > 0$



    $x^4-\dfrac{17}{4}x^2+ 1 < 0$

    $x^4-\dfrac{17}{4}x^2+ \dfrac{289}{64} - \dfrac{289}{64}+1 < 0$

    $\left(x^2-\dfrac{17}{8}\right)^2 -\dfrac{225}{64} < 0$

    $\bigg[\left(x^2-\dfrac{17}{8}\right)-\dfrac{15}{8}\bigg] \bigg[\left(x^2-\dfrac{17}{8}\right)+\dfrac{15}{8}\bigg] < 0$

    $(x^2-4)\left(x^2-\dfrac{1}{4}\right) < 0$ ... which leads to the same intervals above
    Thanks from s_ingram
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    Re: Different methods of solving Inequalities

    yes, I see the mistake.

    I said since $x^2 < b^2$ then $ x < b $ only considering the positive root.

    thanks.
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  4. #4
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    Re: Different methods of solving Inequalities

    Quote Originally Posted by s_ingram View Post
    yes, I see the mistake.

    I said since $x^2 < b^2$ then $ x < b $ only considering the positive root.

    thanks.
    Yes it should be |x| < |b|.
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