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Question:

Find the maximum integral value which satisfies: $$\frac{x-2}{x^2-9}<0$$

I know that this means either of the following:

#1. $x-2<0$ and $x^2-9>0$. Implies that $x \in (3, 2)$, which is false. [1]

#2. $x-2>0$ and $x^2-9<0$. Implies that $x \in (2, 3)$, which is true. [1]

However, this does not give the correct answer. Of course, I can solve this by checking values, but I want a proper method of solving.

I, thus, need help in figuring out the method to get the correct answer

[1] - $x^2-9>0 \Rightarrow x^2>9 \Rightarrow x>3$

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  • $\begingroup$ Note: $x<2$ and $x^2> 9$ imply $x < -3$, not what you mention. $\endgroup$
    – Macavity
    May 17, 2014 at 17:39
  • $\begingroup$ @Macavity I don't know how you got $x<-3$. I have elaborated on my method, please elaborate on your method. Thanks. $\endgroup$ May 17, 2014 at 17:41
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    $\begingroup$ $x^2> 9 \implies x > 3$ OR $x < -3$. Draw the quadratic or try a few values to convince yourself what you're missing. $\endgroup$
    – Macavity
    May 17, 2014 at 17:42
  • $\begingroup$ @Macavity Well, I never knew this. Thanks for that. $\endgroup$ May 17, 2014 at 17:46

3 Answers 3

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What about "standard" approach to solving such inequalities in real variable $x$? $$ \frac{x-2}{x^2-9} < 0\qquad\qquad |x|\neq3 $$ Now multiply both sides by $(x^2-9)^2$ which is always positive and does not change the inequality sign. $$ (x-2)(x^2-9) < 0 $$ $$ (x-2)(x-3)(x+3) < 0 $$ We got a fairly easy polynomial inequality. Given $|x|\neq3$ the solution is: $$ x\in (-\infty,-3)\cup(2,3) $$ The maximum integer value from this set is obviously $-4$.

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  • $\begingroup$ I had thought about this, but this method has three bracket terms, so it does take a lot of time. $\endgroup$ May 17, 2014 at 17:52
  • $\begingroup$ I solved it in my head. It can't be that hard. ;-) $\endgroup$
    – Piotr Miś
    May 17, 2014 at 17:54
  • $\begingroup$ I am new to all this, so well, that is indeed tough for me :) $\endgroup$ May 17, 2014 at 17:54
  • $\begingroup$ The trick is to sketch the polynomial. After some practice it really becomes automatic. $\endgroup$
    – Piotr Miś
    May 17, 2014 at 17:56
  • $\begingroup$ Well, do you have an example site where I can learn this method ?? $\endgroup$ May 17, 2014 at 17:59
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Because of the denominator, $|x|=3$ is forbidden. If $|x|<3$, we need $x-2>0$, which is impossible for integer $x$. If $|x|>3$, we need $x-2<0$, which together with $|x|>3$ implies that $x\le -4$ (integer!), so $x=-4$ is the answer.

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  • $\begingroup$ Nice! Interesting method. Thanks for that! $\endgroup$ May 17, 2014 at 17:47
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To solve this quick, look at your function $\frac{x-2}{x^2-9}<0$ and find all the roots of its polynomial. Here it is very easy: it equals to $\frac{x-2}{(x-3)(x+3)}<0$

So its numerator/denominator zeroes go in the following order: $-3, 2, 3$. At each point exactly one multiplier changes sign, so the sign of the whole expression changes to the opposite.

As there is no square here, this means that the function's value is negative before -3 (where it does not exist), then positive before 2, crosses zero in 2, goes negative until 3, then it turns positive again. So the inequality is true in $(-\infty,-3) \cup (2,3) $, and in here the maximum integer value is, obviously, 4.

Look here for the method. http://www.bymath.com/studyguide/alg/sec/alg28.html . That's someting we used back at school.

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