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[SOLVED] Polynomial challenge

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anemone

MHB POTW Director
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Feb 14, 2012
3,596
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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anemone

MHB POTW Director
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Feb 14, 2012
3,596
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.