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 $d≠0$)
Prove: we always have at least x>0 is a $x^3+bx^2+cx-d^2=0$ 's root
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$\begingroup$
functions
functional-equations
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0discriminant http://en.wikipedia.org/wiki/Cubic_function#Roots_of_a_cubic_function – 2012-11-10
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1I 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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0Yes $d≠0$, sorry – 2012-11-10
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0Use Descartes' rule of signs. – 2012-11-10