USA Mathematical Olympiad 1989 Problem (Third Method)

USA Mathematical Olympiad 1989 Problem (Third Method)

Which is larger, the real root of $x + x^2 + ... + x^8 = 8 - 10x^9$, or the real root of $x + x^2 + ... + x^{10} = 8 - 10x^{11}$?

My solution:

First, notice that

$1-x^9=(1-x)(x+x^2+\cdots+x^8)$

$1-x^9=(1-x)(8-10x^9)$

$1-x^9=8-10x^9-8x+10x^{10}$

$10x^{10}-9x^9-8x+7=0$   

Whereas

$1-x^{11}=(1-x)(x+x^2+\cdots+x^{10})$

$1-x^{11}=(1-x)(8-10x^{11})$

$1-x^{11}=8-10x^{11}-8x+10x^{12}$

$10x^{12}-9x^{11}-8x+7=0$

So I let

$f(x)=10x^{10}-9x^9-8x+7$ and $g(x)=10x^{12}-9x^{11}-8x+7$.

Descartes's Rule says $f(x)$ has at most two positive real roots and obviously $x=1$ is one of the root of $f(x)$ and the Intermediate Value theorem tells us the other root lies between $(0, 0.95)$ since $f(0)\cdot f(0.95)=7(-0.28)\lt 0$.

Similarly, Descartes's Rule says $g(x)$ has two positive real roots and obviously $x=1$ is one of the root of $g(x)$ and the Intermediate Value theorem tells us the other root lies between $(0, 0.95)$ since $g(0)\cdot g(0.95)=7(-0.316)\lt 0$.

Observe also that

$\begin{align*}g(x)&=10x^{12}-9x^{11}-8x+7\\&=x^2(10x^{10}-9x^9-8x+7)-8x+7+8x^3-7x^2\\&=x^2f(x)+(8x-7)(x^2-1)\end{align*}$

If $a$ is a root of the function of $f$, that means $f(a)=0$, then we have:

$\begin{align*}g(a)&=a^2f(a)+(8a-7)(a^2-1)\\&=0+(8a-7)(a^2-1)\\&=(8a-7)(a^2-1)\end{align*}$

And here is a rough sketch of the graph $g(a)=(8a-7)(a^2-1)$

So, if $\dfrac{7}{8}\lt a\lt 1$, then $g(a)\lt 0$ and it follows that the real root of $x + x^2 + ... + x^8 = 8 - 10x^9$ is greater than the real root of $x + x^2 + ... + x^{10} = 8 - 10x^{11}$.

Whereas if $0\lt a\lt \dfrac{7}{8}$, then $g(a)\gt 0$, that will suggest the real root of $x + x^2 + ... + x^8 = 8 - 10x^9$ is less than the real root of $x + x^2 + ... + x^{10} = 8 - 10x^{11}$.

Now, our objective is to find out whether the root of the function of $f$ is greater than or less than $\dfrac{7}{8}$ so we can determine if the real root of $f(x)$ or $g(x)$ is greater.

$f(\dfrac{7}{8})=10(\dfrac{7}{8})^{10}-9(\dfrac{7}{8})^9-8(\dfrac{7}{8})+7=-0.075$

$f(\dfrac{6}{8})=10(\dfrac{6}{8})^{10}-9(\dfrac{6}{8})^9-8(\dfrac{6}{8})+7=0.8873$

Hence, we can say, based on the Intermediate Value theorem that the other positive real root of $f$, i.e. $a$, lies between $\dfrac{6}{8}$ and $\dfrac{7}{8}$, i.e. $a$ is less than $\dfrac{7}{8}$.

Therefore, we have proved that the real root of $x + x^2 + ... + x^8 = 8 - 10x^9$ is less than the real root of $x + x^2 + ... + x^{10} = 8 - 10x^{11}$.