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

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.
