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- #1

- Feb 14, 2012

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- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,596

- Thread starter
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- #2

- Feb 14, 2012

- 3,596

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.