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Prove: we always have at least one x>0 is a $x^3+bx^2+cx-d^2=0$ 's root (b, c, d are real numbers and $d≠0$)

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    discriminant http://en.wikipedia.org/wiki/Cubic_function#Roots_of_a_cubic_function2012-11-10
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    I think you need $d \neq 0$ for strict inequality. What happens at $x=0$ and $x \to \infty$?2012-11-10
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    Yes $d≠0$, sorry2012-11-10
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    Use Descartes' rule of signs.2012-11-10

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