6
$\begingroup$

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

$$-1 $$-6 $$-6+\frac{25}{4} $$\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} $$\frac{5\pm1}{2}

Possible solutions:

$2 (not valid)

$2 (ok)

$3 (not valid)

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

Therefore $S=\{2

The only problem is that the correct solution is $S=\{1. Where am I wrong?

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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: $1<0$.2012-05-03
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    The double inequality $$\frac{25-24}{4}<\left(x-\frac{5}{2}\right)^2<\frac{25-16}{4}$$ is not equivalent to $$\frac{\pm\sqrt1}{\sqrt4}<\sqrt{\left(x-\frac{5}{2}\right)^2}<\frac{\pm\sqrt9}{\sqrt4}$$2012-05-03
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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.2012-05-03
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    @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.2012-05-03
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    @LuizBorges The solution is $1 or $3. Proof. The solution of the first inequality $x^{2}-5x+4<0$ is $1, 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.2012-05-03
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    @LuizBorges The correct equivalence is $$\frac{1}{4}<\left( x-\frac{5}{2}\right) ^{2}<\frac{9}{4}\Leftrightarrow \frac{1}{2}<\left\vert x-\frac{5}{2}\right\vert <\frac{3}{2}.$$2012-05-03

2 Answers 2

3

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"... :(2012-05-03
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    See Phira's answer, which I will now comment on.2012-05-03
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    I got it know. I will accept your answer, but it could be Phira as well. I got choose one. Thanks.2012-05-03
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    Anytime. Maybe I'll have a look at her other answers and find a few I haven't voted on :)2012-05-03
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    @LuizBorges: It's not a "magic rule", it's the fact that $\sqrt{x^2}=|x|$, not $\pm x$.2012-05-03
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    @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.2012-05-03
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    @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.2012-05-04
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    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.2012-05-04
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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.

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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.2012-05-03