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I am trying to stablish the derivation of Brouwer Fixed Point Theorem. However, I don't understand why "the roots differ in sign or one of the roots is zero", as affirmed in the second page of http://nptel.ac.in/courses/111101002/downloads/lecture10.pdf .

I understood the geometry of the problem and I am convinced that the affirmation ir true because we could use the same ray and get a new intersection between the circle and the line, in the opposite direction. But I need to understand it from an analytic point of view!

Thanks a lot!

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That's because the product of the roots of a quadratic equation is the constant term divided by the leading coefficient. In your case, the result of this division is $-\frac{1-|f(x)|^2}{|f(x)-x|^2}\leqslant0$, since $|f(x)|^2\leqslant1$.

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  • $\begingroup$ Thanks José Carlos Santos! Muito obrigado (só para usar nosso idioma!) $\endgroup$ – Cleto Pereira Jul 15 '17 at 21:21

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