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Let $x^2-ax+b$ be a polynomial with real coefficients having two nonzero roots. Given that $|b+1|<a$, and one of the roots have modulus $<1$, show that the other root has modulus $>1$.

We can assume the two roots are $x_1,x_2$. We have the information that $|x_1x_2+1|<x_1+x_2$, and say $|x_1|<1$. How to conclude that $|x_2|>1$?

[Source: Russian competition problem]

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Let $f(x) = x^2 - ax + b$. We have $f(1) = b + 1 - a \leq |b + 1| - a < 0$. Since $f(x)$ has positive leading coefficient, it is clear then that it has a root $> 1$.

There seem to be a number of superfluous hypotheses here.

OTHER METHOD

To continue with your own suggestion, I would write $x_1 x_2 + 1 \leq |x_1 x_2 + 1| < x_1 + x_2$, hence $(1-x_1)(1-x_2) = 1 + x_1 x_2 - x_1 - x_2 < 0$. Since $1 - x_1 > 0$, it follows that $1 - x_2 < 0$, hence $x_2 > 1$.

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