I'm trying to show that the following function has no zeros $$ 60x^4-44x^3-25x^2-44x+60=0. $$ I already tried using Eisenstein's criterium, but since the first and the last coefficient are both $60$, there is no prime which holds.
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1See http://math.stackexchange.com/questions/480102/quadratic-substitution-question – lab bhattacharjee Oct 06 '14 at 13:31
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The roots are $(11\pm A(2\sqrt{574}+iB\sqrt{1183-A\sqrt{574}}))/60$, with $A,B=\pm 1$, none of which is real. – WillO Oct 06 '14 at 14:39
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In this answer I showed that the polynomial
$$ p(x) = x^4 + ax^3 + bx^2 + ax + 1 $$
has no real zeros if $|a| \leq 4$ and $b > 2 |a| - 2$.
We have
$$ 60x^4-44x^3-25x^2-44x+60 = 60 \left(x^4 - \frac{44}{60}x^3 - \frac{25}{60}x^2 - \frac{44}{60}x+1\right), $$
so in this case $a = -44/60$ and $b = -25/60$. Indeed, $|a| \leq 4$ and
$$ b = -\frac{25}{60} > -\frac{32}{60} = 2|a|-2, $$
so the given polynomial has no real zeros.
Antonio Vargas
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Following the hint in the comments, we divide by $x^2$ and rearrange to get $$60\left(x^2+\frac{1}{x^2}\right)-44\left(x+\frac{1}{x}\right)-25=0$$ Set $y=x+\frac{1}{x}$ and note that $y^2=x^2+\frac{1}{x^2}+2$ to get $$60(y^2-2)-44y-25=0$$ which rearranges to the quadratic $$60y^2-44y-145=0$$ This has roots $y\approx -1.23, 1.96$. [Alternate answer: $g(y)=60y^2-44y-145$ is an upward-pointing parabola, and $g(2)>0$ and $g(-2)>0$, so any roots are in $(-2,2)$.]
However $|x+\frac{1}{x}|\ge 2$ for all real $x$, so there are no real solutions.
Cute proof of the last fact: for $x>0$, the arithmetic-geometric mean inequality gives $x+\frac{1}{x}\ge 2\sqrt{x\cdot \frac{1}{x}}=2$.
vadim123
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