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Thread: Getting the signs right with inequalities

  1. #1
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    Getting the signs right with inequalities

    Hi folks,


    I am getting the wrong range of values to the following question:

    calculate the range of x for which:


    \[ \left | \frac{2x + 5}{x^2 - 4} \right | \ge \frac{1}{5} \]


    the modulus means that both numerator and denominator are positive, therefore:


    \[ 2x + 5 \ge \frac{1}{5}(x^2 - 4) \]


    $ -x^2 + 10x + 29 \ge 0 $ ....................................(1)


    \[ x = \frac{-10 \pm \sqrt{100 + 4.1.29}}{-2} \Rightarrow 5 \pm \sqrt{216} \]


    \[ x = 5 \pm 3\sqrt{6} \]


    \[ (x - (5 - 3\sqrt{6})) (x - (5 + 3\sqrt{6})) \ge 0 \]


    \[ (x - 5 + 3\sqrt{6}) (x - 5 - 3\sqrt{6}) \ge 0 \]


    for this inequality to be true:


    \[ x - 5 + 3\sqrt{6} \ge 0 \ and \ x - 5 - 3\sqrt{6} \ge 0 \]


    \[ x \ge 5 - 3\sqrt{6} \ and \ x \ge 5 + 3\sqrt{6} \]


    for both to be true simultaneously x must be $ \ge 5 + 3\sqrt{6} $


    or


    \[ x - 5 + 3\sqrt{6} \le 0 \ and \ x - 5 - 3\sqrt{6} \le 0 \]


    \[ x \le 5 - 3\sqrt{6} \ and \ x \le 5 + 3\sqrt{6} \]


    for both to be true simultaneously x must be $ \le 5 - 3\sqrt{6} $


    so the range is


    \[ 5 + 3\sqrt{6} \le x \le 5 - 3\sqrt{6} \]


    but the answer given is


    \[ 5 - 3\sqrt{6} \le x \le 5 + 3\sqrt{6} \]


    I think the reason for the problem is that (1) can be framed as shown above or written as


    $ x^2 -10x -29 \le 0$ ....................................(2)


    in both cases the roots are the same $x = 5 \pm 3\sqrt{6}$


    so my question to the forum is should I use equation (1) or (2) ? (2) gives the right answer, but I should be able to use either. Is it the roots that should be different? I was a bit unsure when I was taking the negative of $\pm$ !
    Last edited by s_ingram; Aug 2nd 2017 at 04:13 AM.
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  2. #2
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    Re: Getting the signs right with inequalities

    I do not agree with the given answer. Attached is the graph of the equivalent inequality

    $\bigg| \dfrac{2x+5}{x^2-4} \bigg| -\dfrac{1}{5} \ge 0$

    As depicted in the graph, the inequality is true over the intervals $[-7,-3]$, $[5-3\sqrt{6},-2)$, $(-2,2)$, $(2,5+3\sqrt{6}]$
    Attached Thumbnails Attached Thumbnails Getting the signs right with inequalities-img_1644.png  
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    Re: Getting the signs right with inequalities

    Quote Originally Posted by s_ingram View Post


    the modulus means that both numerator and denominator are positive, or both are negative, or the numerator is zero
    .
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    Re: Getting the signs right with inequalities

    The solutions of |A|\geq B where B is positive are given by the solutions of

    A\geq B

    together with (Union) the solutions of

    A\leq -B

    First, solve \frac{2x+5}{x^2-4}\geq \frac{1}{5}
    solution:

    5-3 \sqrt{6}\leq x<-2 or 2<x\leq 5+3 \sqrt{6}

    Next, solve \frac{2x+5}{x^2-4}\leq  - 1/5
    solution:

    -7\leq x\leq -3 or -2<x<2
    Last edited by Idea; Aug 2nd 2017 at 06:57 AM.
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    Re: Getting the signs right with inequalities

    Good Idea (I guess everyone says that!)

    Yes, it seems I am only doing part of the question. Let me give it a try!
    Last edited by s_ingram; Aug 2nd 2017 at 08:14 AM. Reason: applied reply to wrong poster.
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    Re: Getting the signs right with inequalities

    Quote Originally Posted by s_ingram View Post
    Not sure I understand this: the modulus sign makes a value positive so |-3| = 3, so the numerator and denominator are independently positive. If the numerator is zero, then the function would not be defined.
    $\dfrac{0}{4} = 0$ is definitely defined.
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    Re: Getting the signs right with inequalities

    Thanks Skeeter. The full answer given is $-7 \le x \le -3, (5 - 3\sqrt{6}) \le x \le (5 + 3\sqrt{6}) \ with \ x \ne \pm 2$
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    Re: Getting the signs right with inequalities

    Quote Originally Posted by s_ingram View Post
    Not sure I understand this: the modulus sign makes a value positive so |-3| = 3, so the numerator and denominator are independently positive. If the numerator is zero, then the function would not be defined.
    absolute value of a rational function ...

    $\bigg|\dfrac{f(x)}{g(x)}\bigg| = \dfrac{f(x)}{g(x)}$ if $\dfrac{f(x)}{g(x)} \ge 0$ ... $f(x) = 0$ or $f(x)$ and $g(x)$ have the same sign.

    $\bigg|\dfrac{f(x)}{g(x)}\bigg| = -\dfrac{f(x)}{g(x)}$ if $\dfrac{f(x)}{g(x)} < 0$ ... $f(x)$ and $g(x)$ have opposite signs.

    in either case, $g(x) \ne 0$ ... if the denominator equals zero, the function is undefined.
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    Re: Getting the signs right with inequalities

    Hi SlipEternal,

    I meant if $x = \pm 2$ then the modulus function is undefined.
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    Re: Getting the signs right with inequalities

    Hi Skeeter,

    I agree with everything you say in the above. I was using these methods to get to my answer. Thanks to the replies I have so far, I see that I was only doing part of the question, but even there I seem to be getting into a mess. The issue is that I get different answers, using the logic you just laid out, when I try to solve $x^2 -10x -29 \le 0$ and $-x^2 +10x +29 \ge 0$.
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    Re: Getting the signs right with inequalities

    Hi Idea,

    I followed your suggestion and it certainly helps!
    When I solve \[ \frac{2x + 5}{x^2 - 4} \le \frac{-1}{5} \ I get \ -7 \le x \le -3 \] which is great.
    But I still have a problem with \[ \frac{2x + 5}{x^2 - 4} \ge \frac{1}{5} \] which gives me $(5 + 3\sqrt{6}) \le x \le (5 - 3\sqrt{6})$ instead of $(5 - 3\sqrt{6}) \le x \le (5 + 3\sqrt{6})$, using the reasoning shown in the original post. No one, so far, has explained where I am going wrong. Any ideas?
    Last edited by s_ingram; Aug 2nd 2017 at 09:10 AM.
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    Re: Getting the signs right with inequalities

    Quote Originally Posted by s_ingram View Post
    Hi Idea,

    I followed your suggestion and it certainly helps!
    When I solve \[ \frac{2x + 5}{x^2 - 4} \le \frac{-1}{5} \ I get \ -7 \le x \le -3 \] which is great.
    But I still have a problem with \[ \frac{2x + 5}{x^2 - 4} \ge \frac{1}{5} \] which gives me $(5 + 3\sqrt{6}) \le x \le (5 - 3\sqrt{6})$ instead of $(5 - 3\sqrt{6}) \le x \le (5 + 3\sqrt{6})$, using the reasoning shown in the original post. No one, so far, has explained where I am going wrong. Any ideas?
    Let us solve the inequality

    \frac{2x+5}{x^2-4}\leq  \frac{-1}{5}

    \frac{2x+5}{x^2-4}+ \frac{1}{5}\leq 0

    \frac{x^2+10 x+21}{5 \left(x^2-4\right)}\leq 0

    \frac{(x+7)(x+3)}{5 (x-2)(x+2)}\leq 0

    There are four factors on the left that affect the sign of the expression giving rise to 5 intervals on the real line

    (-\infty ,-7,-3,-2,2,+\infty )

    We check each of those 5 intervals to see which make the expression less or equal to zero

    and we find not only -7\leq x\leq -3 but also -2<x<2
    Last edited by Idea; Aug 2nd 2017 at 10:00 AM.
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    Re: Getting the signs right with inequalities

    another method ...

    note $|u| = \sqrt{u^2}$

    $\left|\dfrac{2x+5}{x^2-4}\right| \ge \dfrac{1}{5}$

    $\sqrt{\left(\dfrac{2x+5}{x^2-4}\right)^2} \ge \dfrac{1}{5}$

    $\dfrac{(2x+5)^2}{(x^2-4)^2} \ge \dfrac{1}{25}$ ... keep in mind that $x \ne \pm 2$

    $25(2x+5)^2 \ge (x^2-4)^2$

    $25(2x+5)^2 - (x^2-4)^2 \ge 0$

    $\left[5(2x+5) - (x^2-4)\right] \cdot \left[5(2x+5) + (x^2-4)\right] \ge 0$

    $(29+10x-x^2)(x^2+10x+21) \ge 0$

    $(29+10x-x^2)(x+7)(x+3) \ge 0$

    critical values are $x=5\pm3\sqrt{6}$, $x=-7$, $x=-3$, and $x=\pm 2$

    plot each critical value on a number line and "test" a value within each interval in the last inequality (see attached diagram) ...
    Attached Thumbnails Attached Thumbnails Getting the signs right with inequalities-intervals.jpg  
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    Re: Getting the signs right with inequalities

    Thanks skeeter. You put all the pieces together. There was certainly more to this than I thought! Many thanks.
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    Re: Getting the signs right with inequalities

    Hi Idea, thanks very much for your detailed reply.
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