Find the sum of all real solutions

[anemone]

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!

 

[topsquark]

(sighs) You know I just got done telling someone on another Forum that you can't solve one of these in closed form...

-Dan

 

[anemone]

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?

 

[topsquark]

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...

 

[Opalg]

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]

 

[anemone]

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:

Spoiler

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!