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

Question

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

You should try to explain what you've tried and why it failed. — user37238, Nov 07 '13 at 13:18

score 8 · Answer 1 · answered Nov 07 '13 at 13:25

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

answered Nov 07 '13 at 13:25

Ryan Reich

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One should agree about what the sum of the empty set is. ;-) – egreg Nov 07 '13 at 13:51

Amr · Accepted Answer · 2013-11-07T13:50:14.877

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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

edited Nov 07 '13 at 13:50

answered Nov 07 '13 at 13:18

Amr

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Sir, will you explain why so? – Silent Nov 07 '13 at 13:28
that means sum does not exist – Deiknymi Nov 07 '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 07 '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 07 '13 at 13:37
Oh my god! It is really amazing!! Didn't think that way! – Silent Nov 07 '13 at 13:39
sir i am little confused by the statement of my text book can you explain it to me – Deiknymi Nov 07 '13 at 13:43
@Akash The correct answer is that the solution set of this equation is empty as noted by RyanReich. It is a common convention to define the sum of elements of an empty set to be zero, thus the sum is zero. When I answered the question, I somehow assumed the existence of solutions. Since if $r$ is a root, then so is its additive inverse. Hence the sum of roots can be shown to be zero, if the solution set were non-empty – Amr Nov 07 '13 at 13:46
@Akash, I think you don't understand your book's explanation. Note that for any real $x$, we have $x^2\geq0$ and $|x|\geq0$. Now, note that $x^2 + 5|x| +6 \geq6> 0$. So, for no real $x$ we have this equation. Hope now you can understand the book's explanation. By the way which book are you reading? Is it ML Khanna or else? – Silent Nov 07 '13 at 14:39
thanks got it. The book is written and provided by our coaching center FIITJEE – Deiknymi Nov 07 '13 at 14:49

Myshkin · Answer 3 · 2013-11-07T13:59:43.230

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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

edited Nov 07 '13 at 13:59

answered Nov 07 '13 at 13:19

Myshkin

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Are you sure? My impression is that the stated equation has no solution in the real numbers. – egreg Nov 07 '13 at 13:41
Well It was just a hint to show that it has really no real root as I defined $|x|$ – Myshkin Nov 07 '13 at 13:45
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@egreg Then that would be full answer! not a hint :D – Myshkin Nov 07 '13 at 13:47
@Sade Just add that something else has to be done (I removed my previous comment). – egreg Nov 07 '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 07 '13 at 13:54
@Sush Is it clear now? – Myshkin Nov 07 '13 at 14:00
@Sade, Many many thanks. Yours is the answer which is most useful for other such problems, too. (And that's why I upvoted.) Sorry for being late, I went for walk. – Silent Nov 07 '13 at 14:33
@Sush How was the walk? you are most welcome – Myshkin Nov 07 '13 at 14:35

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

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