Find the sum of all real solutions

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Discussion Overview

The discussion revolves around finding the sum of all real solutions for the equation $\large (x^2+4x+6)^{{(x^2+4x+6)}^{(x^2+4x+6)}}=2014$. Participants explore the nature of the problem, its solvability, and the characteristics of the functions involved.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses enjoyment in sharing the problem, suggesting it is straightforward.
  • Another participant, Dan, mentions that such problems typically cannot be solved in closed form, indicating skepticism about finding explicit solutions.
  • There is a proposal to share a solution via private message, highlighting a desire to keep the discussion open for others.
  • A participant describes the function $f(y) = y^{y^y}$ as increasing and provides specific values to demonstrate that there is a unique solution for $y$ between $2$ and $3$ that satisfies $f(y) = 2014$.
  • The quadratic function $y = x^2 + 4x + 6$ is discussed, noting that it takes values greater than $2$ exactly twice, leading to a sum of $-4$ for the corresponding $x$ values.
  • Another participant agrees with the enjoyment of the problem and acknowledges the reasoning behind the solution approach, while also reflecting on the initial claim of the problem's simplicity.

Areas of Agreement / Disagreement

Participants express differing views on the problem's difficulty and solvability. While some find it enjoyable and straightforward, others, like Dan, maintain that it cannot be solved in closed form. The discussion remains unresolved regarding the nature of the solutions.

Contextual Notes

The discussion includes assumptions about the behavior of the function $f(y)$ and the properties of the quadratic function, but these are not fully explored or settled, leaving room for further inquiry.

anemone
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Find the sum of all real solutions for $x$ to the equation $\large (x^2+4x+6)^{{(x^2+4x+6)}^{(x^2+4x+6)}}=2014$.

P.S. I know this doesn't count as a challenge(no matter how you slice it) because it's quite obvious and rather a very straightforward sort of problem but I'd like to share it because I enjoyed this problem much!
 
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(sighs) You know I just got done telling someone on another Forum that you can't solve one of these in closed form...

-Dan
 
Hey Dan,(Wave)

In this case, do you want me to PM you how I solved the problem while I let other members to have a little fun with this problem?
 
anemone said:
Hey Dan,(Wave)

In this case, do you want me to PM you how I solved the problem while I let other members to have a little fun with this problem?
Sorry for the misinterpretation...I was a bit tired when I wrote that. I was't implying you were wrong, I was mentioning the irony that I had just told someone else it was impossible and your problem was implying that it could be done. My statement was badly worded, my apologies.

And I just read your PM...I was thinking the problem implied an explicit solution set for x, and that's not what the problem was actually about.

-Dan

Edit: Don't worry about PMing me your solution. I'll wait along with the rest. I don't have the slightest clue of how to do this...
 
anemone said:
Find the sum of all real solutions for $x$ to the equation $\large (x^2+4x+6)^{{(x^2+4x+6)}^{(x^2+4x+6)}}=2014$.

P.S. I know this doesn't count as a challenge (no matter how you slice it) because it's quite obvious and rather a very straightforward sort of problem but I'd like to share it because I enjoyed this problem much!
I'm not sure that "it's quite obvious" – I might not have seen how to do it if I had not seen the comments about explicit solutions for $x$. But I agree that it's an enjoyable problem. Thanks for sharing it!
[sp]Let $f(y) = y^{y^y}$, for $y\geqslant1$. This is an increasing function, with $f(2) = 16 < 2014$ and $f(3) = 3^{27} > 2014$. So there is a unique value of $y$, between $2$ and $3$, with $f(y) = 2014.$

The function $y = x^2 + 4x + 6 = (x+2)^2 + 2$ represents a parabola with vertex at $(-2,2)$. It takes all values greater than $2$ exactly twice, and the sum of those two values is always $-4$. In particular, there are two values of $x$ such that $f(x^2 + 4x + 6) = 2014$, and their sum is $-4$.[/sp]
 
Opalg said:
I'm not sure that "it's quite obvious" – I might not have seen how to do it if I had not seen the comments about explicit solutions for $x$. But I agree that it's an enjoyable problem. Thanks for sharing it!
[sp]Let $f(y) = y^{y^y}$, for $y\geqslant1$. This is an increasing function, with $f(2) = 16 < 2014$ and $f(3) = 3^{27} > 2014$. So there is a unique value of $y$, between $2$ and $3$, with $f(y) = 2014.$

The function $y = x^2 + 4x + 6 = (x+2)^2 + 2$ represents a parabola with vertex at $(-2,2)$. It takes all values greater than $2$ exactly twice, and the sum of those two values is always $-4$. In particular, there are two values of $x$ such that $f(x^2 + 4x + 6) = 2014$, and their sum is $-4$.[/sp]

Hi Opalg,:)

I'm so glad that you and many others liked this problem! I solved it pretty much like the way how you solved it and I do want to explain the very reason that led me to mention how I thought this problem is not difficult:

I was enlightened by the keyword "find the sum of the real roots"! That I knew immediately this problem isn't really asking us to solve for $x$s.

Thank you Opalg for participating!

I'll reply to my many other challenges the next day because it has been a long day for me and now, my eyes are strained, I'm exhausted and I can hardly think straight too! Goodbye and good night to everyone and MHB for now!
 

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