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Okay so I did a lot of research online and on stackexchange on solving absolute value inequalities.

I read in an answer here that on having different set of values in different cases, we take union when the inequality is greater than and take intersection when the inequality is less than. This logic solved almost all of my problems correctly however, I came across a question which seems to be an exception to this rule.

$\frac{( x^2-7|x|+10)}{(x^2-6x+9)}<0$

As we see, the denominator is necessarily always positive except for $x=3$, so it won't affect the inequality in general, on solving the two cases, one while opening the mod normally and the other while opening it up with a negative sign, I got two cases as follows.

$x\in (-2,-5)$ and $x\in (2,3) \cup (3,5)$

Now according to the rule the answer should be the intersection of the two cases since the inequality sign in the original problem is less than. However, the answer in my book arrives on taking the union.

Someone just please let me know what is going wrong and if this rules won't work everytime, is there any solid and purely logical rule that I can use blindfoldy for every problem without having to consider checking the values to figure out whether to take union or intersection? Please help me guys, I'm literally going mad over it.

At last, Thanks a lot for showing patience and interest in this querry, thanks for giving me your precious time.

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    $\begingroup$ I suppose you're looking at $-7|x|< -x^2-10$; this is not an "intersection" (because of the negative on the left). Put it in the form $7|x|> x^2+10$ to see a "union" should be used (specifically, $7x>x^2+10$ or $7x<-x^2-10$). $\endgroup$ – David Mitra Aug 12 '17 at 9:41
  • $\begingroup$ Yea I see it now, but the rule said observe the inequality signs in the original form of the problem. $\endgroup$ – Tanuj Aug 12 '17 at 9:49
  • $\begingroup$ @David Mitra so what you're actually saying is that to observe the inequality signs, I should always check and make sure that my modulus term has a positive sign? Is it so? $\endgroup$ – Tanuj Aug 12 '17 at 9:51
  • $\begingroup$ Yes; you want, e.g., either $|x|<\rm something$, or $|x|>\rm something$, before writing the equivalent form. $\endgroup$ – David Mitra Aug 12 '17 at 10:09
  • $\begingroup$ Thanks David. One last querry to be resolved. $\endgroup$ – Tanuj Aug 12 '17 at 10:10
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We can use the intervals method.

We need to solve $$\frac{(|x|-5)(|x|-2)}{(x-3)^2}<0.$$ $(|x|-5)(|x|-2)=0$ for $x\in\{\pm5,\pm2\}$ and $(x-3)^2=0$ for $x=3$.

Now, you can make the following.

  1. Draw the $x$-axes and put there the points: $-5$, $-2$, $2$, $3$ and $5$;

  2. Take the right interval $(5,+\infty)$ and define the sign of our function on this interval.

Easy to see that the sign is $+$.

  1. Now, if a degree of the point is odd then the sign is changed, while if the degree of the point is even then the sign is not changed.

Thus, we get the following signs: $$+,-,+,-,-,+,$$ which gives, the answer: $$(-5,-2)\cup(2,3)\cup(3,5).$$

This method gives possibility to write the answer immediately without distinction between cases.

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  • $\begingroup$ How do you do it? Can I learn this technique from somewhere? Please let me know. $\endgroup$ – Tanuj Aug 12 '17 at 12:15
  • $\begingroup$ Also, I've edited that previous question of mine. I've put up an image showing my effort. Would you please let me know what I'm doing wrong there? $\endgroup$ – Tanuj Aug 12 '17 at 12:16
  • $\begingroup$ Yes, of course. I'll try to explain. It's very useful and easy method, but hard enough to explain this in the net. $\endgroup$ – Michael Rozenberg Aug 12 '17 at 12:17

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