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Please help me with this! $$x^3+|x| = 0$$ Now one solution is clearly $0.$ We have to find the other solution (i.e, $-1$)

$$Solution:$$ CASE $1$: If $x<0,~|x| = -x$, we can write $x^3+|x| = 0$ as $-x^3-x=0$ $$x^3+x=0$$ $$x(x^2+1)=0$$ $$\Longrightarrow x=0, or, x=\sqrt{-1}$$ Please tell me where I've gone wrong. Many thanks!

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    $\begingroup$ I don't think there should be a minus in front of the $x^3$ $\endgroup$ – James May 18 '15 at 11:02
  • $\begingroup$ $\;|x|=-\;$ , for $\;x<0\;$ ...but $\;x^3\;$ remains the same. $\endgroup$ – Timbuc May 18 '15 at 11:04
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    $\begingroup$ The conclusion of case 1 doesn't make sense: The hypothesis of that case is $x < 0$, but one of the solutions isn't even real. (In fact, if one is counting complex solutions, one should have both of $\pm i$.) $\endgroup$ – Travis Willse May 18 '15 at 11:05
  • $\begingroup$ @Timbuc But why does $x^3$ remain the same? $\endgroup$ – Ishan May 18 '15 at 11:08
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    $\begingroup$ Because it is given $\;x^3\;$ !! The only thing you do is to apply the definition of $\;|x|\;$ , all the rest remains exactly the same as it was given... $\endgroup$ – Timbuc May 18 '15 at 11:09
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Hints:

Suppose

$$\begin{align}&x\ge 0\implies 0= x^3+x=x(x^2+1)\;\ldots\\{}\\&x<0\implies0= x^3-x=x(x^2-1)=\ldots\end{align}$$

All in all, there are two different real solutions.

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  • $\begingroup$ But for the second case, why didn't you put a minus sign before $x^3$ (since x<0$\Rightarrow x^3<0$? $\endgroup$ – Ishan May 18 '15 at 11:05
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    $\begingroup$ @BetterWorld Who told that you must put a minus sign in front of an element to make it negative?? $\endgroup$ – Timbuc May 18 '15 at 11:06
  • $\begingroup$ Why do we get 1 as a solution as well? $\endgroup$ – Gummy bears May 18 '15 at 11:11
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    $\begingroup$ We get -1 as a solution, not 1. $\endgroup$ – Ishan May 18 '15 at 11:11
  • $\begingroup$ @Timbuc I had always learnt this... Sorry $\endgroup$ – Ishan May 18 '15 at 11:13
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the function $$f(x) = \begin{cases} x^3 - x & if \, x \le 0,\\x^3+x &if \, 0 \le x. \end{cases} $$

has one negative zero at $x = -1$ and a zero at $x = 0.$ you can see from the function definition that $f(x) > 0$ for $x > 0.$ the graph of $y = f(x)$ has a cusp at $(0,0)$ which is also a local minimum.

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