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Thread: Quadratic Compound Inequality Question

  1. #1
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    Quadratic Compound Inequality Question

    Good morning all!

    I am studying to prepare myself for a PreCalc/Trig class in the Fall and I am stuck on a quadratic compound inequality with no examples to use and no professor to ask. I have attached a copy of the work I have done on this exercise so far. Would any of you guys mind letting me know where I went wrong ?

    Questions I had while solving:

    When I derived the two possible values for the absolute value, should I have eliminated part of the domain since it has both positive and negative 3? If so, why?

    Is there a better way I can analyze the domains I found to make it easier to find the intersections?


    Any help would be really appreciated.

    LenoraQuadratic Compound Inequality Question-quadratic-compound-inequality-question-forum_page1.jpgQuadratic Compound Inequality Question-quadratic-compound-inequality-question-forum_page2.jpgQuadratic Compound Inequality Question-quadratic-compound-inequality-question-forum_page3.jpg
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  2. #2
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    Re: Quadratic Compound Inequality Question

    Beer soaked suggestion follows.
    Quote Originally Posted by lenoradgray View Post
    Good morning all!

    I am studying to prepare myself for a PreCalc/Trig class in the Fall and I am stuck on a quadratic compound inequality with no examples to use and no professor to ask. I have attached a copy of the work I have done on this exercise so far. Would any of you guys mind letting me know where I went wrong ?

    Questions I had while solving:

    When I derived the two possible values for the absolute value, should I have eliminated part of the domain since it has both positive and negative 3? If so, why?

    Is there a better way I can analyze the domains I found to make it easier to find the intersections?


    Any help would be really appreciated.

    LenoraClick image for larger version. 

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    One easy way is to plot all 3 functions on the same window; you can easily see the solution set to this quadratic compound inequality at
    plot |x^2-9|, 2, 9, x=-5 to 5 - Wolfram|Alpha Results
    Too hammered and rusty at the moment but I'm sure you can verify this easily. Check your results at
    solve 2<=|x^2-9|<9 - Wolfram|Alpha Results
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  3. #3
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    Re: Quadratic Compound Inequality Question

    well, you picked a rather tedious one to solve ...

    $2 \le |x^2-9| < 9$

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

    part (1) ...

    $|x^2-9| \ge 2$

    $\sqrt{(x^2-9)^2} \ge 2$

    $(x^2-9)^2 \ge 4$

    $(x^2-9)^2 - 4 \ge 0$

    $[(x^2-9)-2][(x^2-9)+2] \ge 0$

    $(x^2-11)(x^2-7) \ge 0$

    for the inequality to be true, both factors must be non-negative or both must be strictly negative.

    both factors are non-negative for $x \ge \sqrt{11}$ or $x \le -\sqrt{11}$

    both factors are negative for $-\sqrt{7} < x < \sqrt{7}$

    solution for this part of the inequality is the union of the above intervals.


    part (2) ...

    $|x^2-9| < 9$

    $\sqrt{(x^2-9)^2} < 9$

    $(x^2-9)^2 < 81$

    $(x^2-9)^2 - 81 < 0$

    $[(x^2-9)-9][(x^2-9)+9] < 0$

    $(x^2-18)(x^2) < 0$

    since $x^2 \ge 0$ for all $x$, the above inequality is true for $(-\sqrt{18} < x < 0) \cup (0 < x < \sqrt{18})$


    so, the solution set of the inequality $2 \le |x^2-9| < 9$ is the intersection of the intervals

    $(x \ge \sqrt{11}) \cup (x \le -\sqrt{11})$

    $-\sqrt{7} < x < \sqrt{7}$

    $(-\sqrt{18} < x < 0) \cup (0 < x < \sqrt{18})$

    which is ...

    $(-\sqrt{18} < x \le -\sqrt{11}) \cup (-\sqrt{7} \le x < 0) \cup (0 < x \le \sqrt{7}) \cup (\sqrt{11} \le x < \sqrt{18})$
    Attached Thumbnails Attached Thumbnails Quadratic Compound Inequality Question-abs_inequality2.jpg  
    Thanks from lenoradgray
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  4. #4
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    Re: Quadratic Compound Inequality Question

    Thanks so much for your help!
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