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

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

• What is the question? – Riccardo.Alestra Nov 7 '13 at 13:16
• You should try to explain what you've tried and why it failed. – user37238 Nov 7 '13 at 13:18

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

• Sir, will you explain why so? – Silent Nov 7 '13 at 13:28
• that means sum does not exist – Deiknymi Nov 7 '13 at 13:33
• @Akash, for what are you saying:"that means sum does not exixt "? I can't understand. Could you please be precise? – Silent Nov 7 '13 at 13:35
• 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 – Deiknymi Nov 7 '13 at 13:37
• Oh my god! It is really amazing!! Didn't think that way! – Silent Nov 7 '13 at 13:39

Another hint: all the terms are non-negative, and the constant term is actually positive.

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

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

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