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Relevant wiki: Exponential Functions - Problem Solving

Let $y=x^2+4x+5$ .

$x^2+4x+(5-y)=0$

This has real solution(s) when $16-4(1)(5-y)\ge 0\\$

$\Longrightarrow y\ge 1$

${ y }^{ { y }^{ y } }=2016$

$\Longrightarrow 2<y<3$ .

This implies that ${ y }^{ { y }^{ y } }=2016$ only has one solution $2<y<3$ and it gives two real solutions of $x$ .

By Vieta's formula, the sum of the real solutions is $-\frac{4}{1}=-4$ .

(Note: In this case, we don't need to find the exact value of $y$ as long as $y\ge 1$ , because we can ​use Vieta's formula.)

Relevant wiki: Exponential Functions - Problem Solving

Suppose $x^2+4x+5=k$ satisfies the above equation. Then, we shall solve for $x$ . $x^2+4x+5=k\Rightarrow x^2+4x+(5-k)=0$ By Viete's theorem, the sum of roots is $-4/1=-4$ , regardless of the constant term. Note that there will be a real solution only when $k\geq 1$ , so we only need to observe the graph of $k^{k^k}$ when it is above 1. Noting that this graph is monotonic increasing, we conclude that there will be only one such value of $k$ that satisfies it, so the answer is $-4\times 1=-4$ .

$y^{y^y}$ is bijective from $(0,+\infty)$ to $(0,+\infty)$ so it exists one and only one positive solution for $y^{y^y}=2016$ . $x^2+4x+5$ is a quadratic polynomial which assumes only positive values and is symmetric respect to the line $x=-2$ where the vertex of its graph lies. So if $x_0$ satisfies the proposed equation, also $(-4-x_0)$ does, these are the unique two solutions and their sum is $-4$ .

Let x2+4x+5=y so y^y^y=2016. And y^y^y^y^y^y^........infinite =ywhich is written as y^y^(y^y^y^y^y^y.......)=y And also. y=2016 And solve the equation and u got sum of real solutions is -4... Thankuu...for watching my solution....!!!