Another 5000 Follower Problem

Expanding all of that is definitely not going to be fun. So instead, let's generalize the problem and let $k=5000$ . Our polynomial turns into $(x^2+kx-1)^2+(2x+k)^2$

This is much easier to expand: $\begin{aligned} (x^2+kx-1)^2+(2x+k)^2&= (x^4+2kx^3+(k^2-2)x^2-2kx+1)+(4x^2+4kx+k^2)\\ &\text{Adding the first two terms of expression 2 in:}\\ &=(x^4+2kx^3+(k^2+2)x^2+2kx+1)+(k^2)\\ &\text{We recognize the similar factorization:}\\ &=(x^2+kx+1)^2+k^2 \end{aligned}$

Substituting $k=5000$ back in, we get that $P(x)=(x^2+5000x+1)^2+25000000$

And now, by the trivial inequality, $\text{min}(P(x))=25000000$ with equality when $x^2+5000x+1=0$

We confirm that this quadratic has real solutions by checking the discriminant. Now all we need to do is find the sum of these real solutions, which by Vieta's is simply $\boxed{-5000}$ and we're done.

As an alternative solution, though it uses calc, here's how I did it.

Using the chain rule, $(f(g(x))'=f'(g(x))\cdot g'(x)$ where $f(x)=x^2$ and $g(x)=\begin{cases} x^2+5000x-1 \\ 2x+5000 \end{cases}$

$(x^2+5000x-1)^2-(2x+5000)^2~~\frac{d}{dx}$

$(x+2500)(4x^2+20000x-4)+8(x+25000)$

$4(x+2500)(x^2+5000+4)$

If we set this equal to 0 to find extreme, we need the smallest and largest roots, not the intermediate root since the graph will look like a " $W$ .

By the quadratic, $x=-2500\pm \sqrt{1250\cdot5000-4}$

Thus the smallest and greatest roots are the two which aren't -2500.

Thus by Vietas, our answer is $\boxed{-5000}$

To prove these minima have the same value, Let $x=a-2500$ this means that we are doing a horizontal shift of the graph 2500 to the right.

$((a-2500)^2+5000(a-2500)-1)^2+4((a-2500)+2500)^2$

$(a^2-5000a+5000a+2500^2-2500\cdot5000-1)^2+4a^2$

$(a^2-2500^2-1)^2+4a^2$

Obviously, this graph is symmetric about the y axis since if we plug in a=-a, we get the same result.

$P(x)\quad =\quad { ({ x }^{ 2 }+5000x-1) }^{ 2 }\quad +{ (2x+5000) }^{ 2 }\\ { P }'(x)\quad =\quad 2({ x }^{ 2 }+5000x-1)*(2x+5000)+4*(2x+5000)\quad =\quad 2({ x }^{ 2 }+5000x+1)*(2x+5000)\\ P'(x)\quad =0\quad \Rightarrow \quad Either\quad (2x+5000)=0\quad or\quad ({ x }^{ 2 }+5000x+1)=0\\ P"(x)\quad =\quad 2*{ (2x+5000) }^{ 2 }+4*({ x }^{ 2 }+5000x+1)\\ P"(x)\quad for\quad (2x+5000)=0\quad is\quad <0\quad So,\quad its\quad point\quad of\quad maxima.\\ P"(x)\quad for\quad ({ x }^{ 2 }+5000x+1)\quad is\quad of\quad form\quad 2*{ (2x+5000) }^{ 2 }\quad which\quad is\quad >0\quad \\ Hence\quad sum\quad of\quad x\quad for\quad which\quad P(x)\quad =\quad Min(P(x))\quad is\quad given\quad by\quad coefficient\quad of\quad x\quad in\quad ({ x }^{ 2 }+5000x+1)\quad i.e.\quad -5000.$

Use substitution to simplify the polynomial. Let $n:=2500$ as a short hand and complete the square in the first term:

$\begin{aligned} P(x)&=((x+n)^2-n^2-1)^2+ (2x+2n)^2\underset{y:=x+n}{=}(y^2-n^2-1)^2+4y^2=y^4-2(n^2+1)y^2+(n^2+1)^2 + 4y^2\\\\ &=y^4-2(n^2-1)y^2+(n^2+1)^2=(y^2-n^2+1)^2+(n^2+1)^2-(n^2-1)^2\geq(n^2+1)^2-(n^2-1)^2 \end{aligned}$

In the last step, we complete the square again. The lower bound really is a minimum, because the first quadratic may vanish:

$\begin{aligned} (y^2-n^2+1)^2&=0&\Leftrightarrow&&y^2&=n^2-1&\Leftrightarrow&&y_{1,2}&=\pm\sqrt{n^2-1},&x_1+x_2&\underset{y=x+n}{=}-2n=\fbox{-5000} \end{aligned}$

Take the derivative of original function: $\frac{dy}{dx}$ = $2(x^2+5000x-1)(2x+5000)+4(2x+5000)$

Because we know that the leading coefficient of the original function is positive, the minimum occurs when $\frac{dy}{dx}$ = 0 Thus, $2(x^2+5000x-1)(2x+5000)+4(2x+5000)=0$

We can easily see that $x=-2500$ is not a valid solution. It centers all the value in that quartic equation, which merely makes the P(x) extremely great for our concern of the minimum

# i.e P(-2500) is much greater than, say, P(0)

Therefore, divided by $2x+5000$ , we have:

$2(x^2+5000x-1)+4=0$

$x^2+5000x + 1=0$

By Vieta's formula, the sum is simply -5000