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$$|x-4|^2 -5|x-4| +6 > 0$$

How can I get rid of the absolute value? Does it work the same way equations with absolute value work?

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    $\begingroup$ Set $y=|x-4|$ and solve for $y\ge 0$? $\endgroup$ – Mark Bennet Nov 21 '14 at 22:44
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Hint:
Substitute $u=|x-4|$, and you get the quadratic inequality $$u^2-5u+6\gt0.$$ Do you know how to proceed further?

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  • $\begingroup$ That seems easier. But no idea. Should I just factor or what should I do next? $\endgroup$ – FranklinS Nov 21 '14 at 22:56
  • $\begingroup$ @FranklinS You factor the polynomial and you should get something of the form $(u-r_1)(u-r_2)$ where $r_1,r_2$ are the roots of that polynomial. So your inequality simplifies to $(u-r_1)(u-r_2)\gt0$, now ask yourself: when a product of two numbers is strictly positive? $\endgroup$ – Hakim Nov 21 '14 at 22:59
  • $\begingroup$ Yes, I got (u-3)(u-2)>0. When they have the same sign? $\endgroup$ – FranklinS Nov 21 '14 at 23:01
  • $\begingroup$ @FranklinS Exactly. $\endgroup$ – Hakim Nov 22 '14 at 13:42
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You will probably want to factor first. That is $$|x-4|^2 -5|x-4| +6 = (|x-4|-2)(|x-4|-3)>0.$$ From this, we can see $|x-4|-2>0,$ and $|x-4|-3>0,$ $\mathbf{or}$ $|x-4|-2<0$ and $|x-4|-3<0.$

Thus $x<1$, or $x>7$ (for our first case); but also $x>2$ or $x<6$, that is $2<x<6$ (for the second).

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