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I'm asked to solve $$|x+1|>x^2-5$$

My attempt,

For my basic inequality skill, this is a very easy question.

Since $$|x|=\left\{\begin{matrix} x, x \geq0 & \\ -x,x<0& \end{matrix}\right.$$

Then for $x<-1$

$x^2+x-4<0$

$\frac{-1-\sqrt{17}}{2}<x<\frac{-1+\sqrt{17}}{2}$

Then for $x\geq-1$

$x+1 >x^2-5$

$x^2-x-6 <0$

$-2 < x<3$

So combine the ranges,

I got $\frac{-1-\sqrt{17}}{2}<x<3$

So, a senior told me that basically my solution is correct, but instead of $x< -1$, I should write $x \leq -1$. Why? According to the definition of modulus function which shows that I'm correct. But why he said that actually the = sign can be for both. So, he asked me don't be bothered too much by this trivial issue. I really don't understand why he said so. Can anyone explain it for me? Thanks in advance.

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  • $\begingroup$ Your solution looks good. $\endgroup$ – quasi Jun 3 '17 at 8:11
  • $\begingroup$ The final result is correct but the intermediate step $x+1 \geq x^2-5$ is wrong. $\endgroup$ – Yves Daoust Jun 3 '17 at 8:29
  • $\begingroup$ Just a typo for it. Fixed ! @YvesDaoust $\endgroup$ – Mathxx Jun 3 '17 at 8:30
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The senior person is both right and wrong. Right because $|x|=-x$ is indeed true for $x\le0$, but wrong because the case of $x=0$ is already handled (by $x\ge0$) and needn't be repeated.


The discussion can be made as follows:

  • $x+1\ge0\to x+1>x^2-5\to-2<x<3\to-1\le x<3$,

  • $x+1<0\to -(x+1)>x^2-5\to-\dfrac{\sqrt{17}+1}2<x<\dfrac{\sqrt{17}-1}2\to-\dfrac{\sqrt{17}+1}2<x<-1$

This is essentially what you did.

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  • $\begingroup$ Can you give me more example of a statement in math, which right and wrong simultaneously? The solution of Mathxx is right and the senior must be shut up. $\endgroup$ – Michael Rozenberg Jun 3 '17 at 8:40
  • $\begingroup$ @MichaelRozenberg "the senior must be shut up": why ?? $\endgroup$ – Yves Daoust Jun 3 '17 at 8:47
  • $\begingroup$ Because the solution of Mathxx is right. What does he want else? $\endgroup$ – Michael Rozenberg Jun 3 '17 at 8:58
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It's $x+1>x^2-5$ or $x+1<-x^2+5$ without any additional cases.

because if $x^2-5<0$ then the inequality is obviously true.

We get $-2<x<3$ or $\frac{-1-\sqrt{17}}{2}<x<\frac{-1+\sqrt{17}}{2}$, which gives your answer: $$\frac{-1-\sqrt{17}}{2}<x<3$$ I think your solution is right, but your way is bad.

For example, solve the following inequality: $$|x^3+x-1|>2-x$$

By your way we need to solve two inequalities: $x^3+x-1\geq0$ and $x^3+x-1<0$.

I don't say that it's hard (sometimes it's just impossible!), I say that it's not necessary.

Indeed, $|x|>a\Leftrightarrow x>a$ or $x<-a$ for all $a\in\mathbb R$

because for $a<0$ the inequality $|x|>a$ is obvious.

Thus, $|x^3+x-1|>2-x$ gives $x^3+x-1>2-x$ or $x^3+x-1<-2+x$,

which is $x^3+2x-3>0$ or $x^3+1<0$, which is

$x^3-x^2+x^2-x+3x-3>0$ or $(x+1)(x^2-x+1)<0$, which is

$(x-1)(x^2+x+3)>0$ or $x<-1$, which gives the answer: $$(-\infty,-1)\cup(1,+\infty)$$

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  • $\begingroup$ This recasting has a variable expression on the right side which might be positive or negative depending on $x.$ $\endgroup$ – coffeemath Jun 3 '17 at 8:09
  • $\begingroup$ @coffeemath I explained this point for you. See now, please. $\endgroup$ – Michael Rozenberg Jun 3 '17 at 8:23
  • $\begingroup$ Very confuse explanation. $\endgroup$ – Yves Daoust Jun 3 '17 at 8:36
  • $\begingroup$ @Yves Daoust I fixed my post for you and explained, what I mean. See now, please. $\endgroup$ – Michael Rozenberg Jun 3 '17 at 8:46
  • $\begingroup$ I can't decifer "without any cases else", I don't know what "the inequality" refer to, and the obviousness is not obvious to me. $\endgroup$ – Yves Daoust Jun 3 '17 at 8:46

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