Roots of $x^{101}-100x^{100}+100=0$

Question

I do not know how to prove that $x^{101}-100x^{100}+100=0$ has exactly two positive roots.

Some can give me hint for solving this please. Thanks for your time.

Answer 1

Hint: $$f'(x)=101x^{100}-100^2x^{99}=x^{99}\left(101x-100^2\right)$$ has two solutions.

$x_{\max}=0, x_{\min}=\frac{100^2}{101}$ Then $$x_1<x_{\max}=0<x_2<x_{\min}<x_3$$ Then $x_2,x_2>0$

Answer 2

Note that the polynomial is positive at $0$, negative at (what?) and positive for large $x$, so there area least two positive roots. Now take the derivative. Where is it zero?

Answer 3

You should simply study the variations of the function $f$ defined by $f(x) = x^{101}-100x^{100}+100$ on $\mathbb R$.

Answer 4

Hint: Between any two roots of $f$ there is a root of $f'$

Answer 5

Descartes' rule of signs indicates that $P(x)=x^{101}-100x^{100}+100$ has either zero or two positive roots.

But $P(0)>0$ and $P(2)<0$ so $P(x)$ has at least one positive root, hence it has exactly two positive roots.