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Tried to solving $|x^2-5x+5|<1$ using the square method, but I don't know what I did wrong:

$$-1<x^2-5x+5<1$$ $$-6<x^2-5x<-4$$ $$-6+\frac{25}{4}<x^2-5x+\frac{25}{4}<-4+\frac{25}{4}$$ $$\frac{25-24}{4}<\left(x-\frac{5}{2}\right)^2<\frac{25-16}{4}$$ $$\frac{\pm\sqrt1}{\sqrt4}<\sqrt{\left(x-\frac{5}{2}\right)^2}<\frac{\pm\sqrt9}{\sqrt4}$$ $$\frac{\pm1}{2}<x-\frac{5}{2}<\frac{\pm3}{2}$$ $$\frac{5\pm1}{2}<x<\frac{5\pm3}{2}$$

Possible solutions:

$2<x<1$ (not valid)

$2<x<4$ (ok)

$3<x<1$ (not valid)

$3<x<4$ (ok, but is a subset of solution 2.

Therefore $S=\{2<x<4\}$

The only problem is that the correct solution is $S=\{1<x<2\text{ or }3<x<4\}$. Where am I wrong?

Brian M. Scott
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Luiz Borges
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    Hint: what happens when you take square roots in an inequality and consider both the positive and negative root. For example, $1<4<9$, but taking square roots, $\pm 1<2<\pm 3$ gives something fishy. Afterall, if say $-1<0$ but squaring both sides does not preserve the inequality: $1<0$. – Alex R. May 03 '12 at 20:25
  • The double inequality $$\frac{25-24}{4}<\left(x-\frac{5}{2}\right)^2<\frac{25-16}{4}$$ is not equivalent to $$\frac{\pm\sqrt1}{\sqrt4}<\sqrt{\left(x-\frac{5}{2}\right)^2}<\frac{\pm\sqrt9}{\sqrt4}$$ – Américo Tavares May 03 '12 at 20:29
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    Why not? If I had them separated (like equation 1 and 2) wouldn't it be valid? I'm not questioning your math, I'm just trying to grasp it. – Luiz Borges May 03 '12 at 20:32
  • @Luiz Borges Well, $\sqrt{\left( x-\frac{5}{2}\right) ^{2}}\geq 0$ and $\frac{-\sqrt{9}}{\sqrt{4}}<0$; and similarly for the first inequality. – Américo Tavares May 03 '12 at 21:07
  • @LuizBorges The solution is $1<x<2$ or $3<x<4$. Proof. The solution of the first inequality $x^{2}-5x+4<0$ is $1<x<4$, because the roots of $x^{2}-5x+4=0$ are $x_{1}=1$ and $x_{2}=4$, and the coefficient of $x^{2}$ is positive. The solution of the second inequality $x^{2}-5x+6>0$ is $x<2$ or $x>3$, because the roots of $x^{2}-5x+6=0$ are $x_{1}=2$ and $x_{2}=3$, and the coefficient of $x^{2}$ is positive. – Américo Tavares May 03 '12 at 21:15
  • @LuizBorges The correct equivalence is

    $$\frac{1}{4}<\left( x-\frac{5}{2}\right) ^{2}<\frac{9}{4}\Leftrightarrow \frac{1}{2}<\left\vert x-\frac{5}{2}\right\vert <\frac{3}{2}.$$

     – Américo Tavares May 03 '12 at 21:54

2 Answers2

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When you take the root of an inequality, you have to make sure that everything is positive and then take positive roots.

So after taking the roots, you get:

$\frac 12 < |x-\frac 52| < \frac 32$.

Now, you can regard the two cases $x> \frac 52$ and $x < \frac 52$ to eliminate the absolute value.

Phira
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  • (For Luiz) - When $x > \dfrac{5}{2}$, the absolute value can be dropped. This gives the first inequality in my answer. When $x < \dfrac{5}{2}$, you must multiply everything by $-1$, giving the second inequality in my answer. – The Chaz 2.0 May 03 '12 at 20:33
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You have a $\pm$ on each side of the inequality, but you need to change the direction of inequality for the "minus".

So you would have $$\dfrac{5 + 1}{2} < x < \dfrac{5+3}{2}$$ (The $+$'s go together), or $$\dfrac{5 - 1}{2} > x > \dfrac{5 - 3}{2}$$ (The $-$'s go together)

The Chaz 2.0
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  • I did reached that by trial analysis, but why is that? I'm trying to run away from "magic rules"... :( – Luiz Borges May 03 '12 at 20:31
  • See Phira's answer, which I will now comment on. – The Chaz 2.0 May 03 '12 at 20:32
  • I got it know. I will accept your answer, but it could be Phira as well. I got choose one. Thanks. – Luiz Borges May 03 '12 at 20:54
  • Anytime. Maybe I'll have a look at her other answers and find a few I haven't voted on :) – The Chaz 2.0 May 03 '12 at 21:00
  • @LuizBorges: It's not a "magic rule", it's the fact that $\sqrt{x^2}=|x|$, not $\pm x$. – Arturo Magidin May 03 '12 at 21:31
  • @ArturoMagidin, I refered to: change the direction of inequality for the "minus". What I had missed, is that $|x|=|y|$ is equivalent of saying just $|x|=y$ (the other cases are implied). But when I have a situation like $|x|=|y|=|z|$ I can't imply just two cases. This is where I was wrong. – Luiz Borges May 03 '12 at 21:37
  • @Luiz: I actually disagree. Once you had $$\frac{25-24}{4}<\left(x-\frac{5}{2}\right)^2<\frac{25-16}{4}$$ when you take square roots you get $$\frac{1}{2}\lt \left|x-\frac{5}{2}\right|\lt \frac{3}{2}.$$ That's what Americo was saying above. This triple inequality is equivalent to the triple inequality before it. You don't have three (or two) absolute values, you just have one. Incorrectly identifying your error is just going to lead to more errors. – Arturo Magidin May 04 '12 at 00:33
  • Hum... I think I understand it better now. I'm not at home now, so tomorrow I will sit with pencil and paper and try it out separating the two equations to visualize what you're saying, but I guess I'm almost there. Square roots and absolute values are much worser and counterintuive than I thought. – Luiz Borges May 04 '12 at 01:17