3
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Sum of real roots of the equation $x^2 + 5|x| +6 = 0$

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  • $\begingroup$ What is the question? $\endgroup$ – Riccardo.Alestra Nov 7 '13 at 13:16
  • $\begingroup$ You should try to explain what you've tried and why it failed. $\endgroup$ – user37238 Nov 7 '13 at 13:18
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Hint: It $r$ is a solution then so is $-r$

Note: After seeing Ryan's answer, I realized that the solution set is empty. Thus, Ryan's answer is the correct answer.

Now if we are looking for solutions inside $\mathbb{C}$, my hint can be used to deduce that either the solution set is infinite (hence the sum is undefined) or the sum is zero

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  • $\begingroup$ Sir, will you explain why so? $\endgroup$ – Silent Nov 7 '13 at 13:28
  • $\begingroup$ that means sum does not exist $\endgroup$ – Deiknymi Nov 7 '13 at 13:33
  • $\begingroup$ @Akash, for what are you saying:"that means sum does not exixt "? I can't understand. Could you please be precise? $\endgroup$ – Silent Nov 7 '13 at 13:35
  • $\begingroup$ my text book says $x^2, 5|x| and 6 $ are positive so the equation does not have any real root, Therefor sum does not exist $\endgroup$ – Deiknymi Nov 7 '13 at 13:37
  • $\begingroup$ Oh my god! It is really amazing!! Didn't think that way! $\endgroup$ – Silent Nov 7 '13 at 13:39
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Another hint: all the terms are non-negative, and the constant term is actually positive.

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  • $\begingroup$ One should agree about what the sum of the empty set is. ;-) $\endgroup$ – egreg Nov 7 '13 at 13:51
1
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Hint

$|x|=x\forall x\ge0$

$|x|=-x\forall x<0$

Case 1 $x\ge 0$

$x^2+5x+6=(x+2)(x+3)=0\Rightarrow x=-2,-3$ but we already assumed $x\ge 0$ so $(\Leftrightarrow)$

Case 2 $x<0$ then the equation becomes according to the definition of $|x|$

$x^2-5x+6=(x-2)(x-3)=0\Rightarrow x=2,3$ again $(\Leftrightarrow)$

$(\Leftrightarrow)$ is the sign of contradiction

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    $\begingroup$ Are you sure? My impression is that the stated equation has no solution in the real numbers. $\endgroup$ – egreg Nov 7 '13 at 13:41
  • $\begingroup$ Well It was just a hint to show that it has really no real root as I defined $|x|$ $\endgroup$ – Marso Nov 7 '13 at 13:45
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    $\begingroup$ @egreg Then that would be full answer! not a hint :D $\endgroup$ – Marso Nov 7 '13 at 13:47
  • $\begingroup$ @Sade Just add that something else has to be done (I removed my previous comment). $\endgroup$ – egreg Nov 7 '13 at 13:50
  • $\begingroup$ @Sade, Please help me. We should check your hint by putting the value of $x$ back in the original equation, right? $\endgroup$ – Silent Nov 7 '13 at 13:54

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