MHB Polynomial Challenge: Show Real Roots >1 Exist

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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 explore the implications of the conditions on the coefficients and the behavior of the functions involved. The relationship between the roots of the quadratic and the quartic is examined, emphasizing the continuity and differentiability of polynomial functions. Key mathematical principles such as the Intermediate Value Theorem and Descartes' Rule of Signs are referenced to support the argument. The conclusion asserts that the existence of real roots greater than 1 in the quadratic guarantees at least one real root in the quartic equation.
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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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