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up vote 6 down vote favorite

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?

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2  
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 May 3 '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 3 '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 3 '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 3 '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 3 '12 at 21:15
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2 Answers

up vote 3 down vote accepted

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)

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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 3 '12 at 20:31
See Phira's answer, which I will now comment on. – The Chaz 2.0 May 3 '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 3 '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 3 '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 3 '12 at 21:31
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up vote 6 down vote

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.

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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 3 '12 at 20:33

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