First, notice that
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[TR]
[TD]$1-x^9=(1-x)(x+x^2+\cdots+x^8)$
$1-x^9=(1-x)(8-10x^9)$
$1-x^9=8-10x^9-8x+10x^{10}$
$10x^{10}-9x^9-8x+7=0$[/TD]
[TD]$1-x^{11}=(1-x)(x+x^2+\cdots+x^{10})$
$1-x^{11}=(1-x)(8-10x^{11})$
$1-x^{11}=8-10x^{11}-8x+10x^{12}$
$10x^{12}-9x^{11}-8x+7=0$[/TD]
[/TR]
[TR]
[TD]So I let
$f(x)=10x^{10}-9x^9-8x+7$[/TD]
[TD]So I let
$g(x)=10x^{12}-9x^{11}-8x+7$[/TD]
[/TR]
[TR]
[TD]Descartes's Rule says $f(x)$ has two positive real roots and obviously $x=1$ is one of the root of $f(x)$ and the other root lies between $(0, 1)$.[/TD]
[TD]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 other root lies between $(0, 1)$.[/TD]
[/TR]
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Observe also that
$g(x)=10x^{12}-9x^{11}-8x+7=x^2(10x^{10}-9x^9-8x+7)-8x+7+8x^3-7x^2=x^2f(x)+(8x-7)(x^2-1)$
If $a$ is a root of the function of $f$, then
$g(a)=a^2f(a)+(8a-7)(a^2-1)$
$g(a)=0+(8a-7)(a^2-1)$
$g(a)=(8a-7)(a^2-1)$
And here is a rough sketch of the graph $y=(8a-7)(a^2-1)$
View attachment 1544
So, if $\dfrac{7}{8}<a<1$, then $g(a)<0$ whereas if $a<\dfrac{7}{8}$, then $g(a)>0$.
Now, our objective is to find out whether the root of the function of $f$ is greater than or less than $\dfrac{7}{8}$.
$f(\dfrac{7}{8})=10(\dfrac{7}{8})^{10}-9(\dfrac{7}{8})^9-8(\dfrac{7}{8})+7=-0.075$
$f(\dfrac{6}{8})=10(\dfrac{6}{8})^{10}-9(\dfrac{6}{8})^9-8(\dfrac{6}{8})+7=0.8873$
Hence, we can say that the other positive real root of $f$, i.e. $a$, lies between $\dfrac{6}{8}$ and $\dfrac{6}{8}$, i.e. $\dfrac{6}{8}<a<\dfrac{7}{8}$, or a is less than $\dfrac{7}{8}$ and this tells us $g(a)>0$.
The graph below shows that the real root of $g(x)$ is larger than the root of $f(x)$.
View attachment 1545