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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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