Polynomial Challenge: Show Real Roots >1 Exist

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SUMMARY

The discussion centers on proving that if the quadratic equation $ax^2+(c-b)x+e-d=0$ has real roots greater than 1, then the quartic equation $ax^4+bx^3+cx^2+dx+e=0$ must have at least one real root. Participants utilized the properties of polynomial equations and the relationship between their coefficients to establish this conclusion. Key mathematical principles such as the Intermediate Value Theorem and Descartes' Rule of Signs were referenced to support the argument.

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If the equation $ax^2+(c-b)x+e-d=0$ has real roots greater than 1, show that the equation $ax^4+bx^3+cx^2+dx+e=0$ has at least one real root.
 
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Suppose $p(x)=ax^4+bx^3+cx^2+dx+e=0$ has no real root.

Let $y>1$ be a root of $ay^2+(c-b)y+e-d=0$ and $z=\sqrt{y}$.

Since $p(x)=ax^4+(c-b)x^2+(e-d)+(x-1)(bx^2+d)$, we get

$p(z)=(z-1)(bz^2+d)$ and $p(-z)=(-z-1)(bz^2+d)$.

Now, $z>1$ implies one of $p(z)$ and $p(-z)$ is positive while the other is negative. Therefore, $p(x)$ has a root between $z$ and $-z$, a contradiction.
 

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