<s>[INST] If $ax^2 + (b+c)x + d+ e=0$ has solutions on $[1,\infty)$, so that $ax^4+bx^3+cx^2+dx+e=0$ has solutions on $[1,\infty)$ 
Let $ax^2 + (b+c)x + d+ e=0, a, b, c, d, e \in \mathbb R$ has solutions on the interval $[1, +\infty)$. Prove that the polynomial $f(x) = ax^4 + bx^3 + cx^2+ dx + e =0$ has solutions on $[1, +\infty)$.

I used the intermediate value theorem, but not found $x_0, x_1 \in [1, +\infty) $ such that $f(x_0)f(x_1) < 0.$
Thank for helps.
 [/INST] Nobody can prove that, since it is false. Suppose that $a=b=1$, that $c=-5$, that $d=4$, and that $e=0$. Then\begin{align}ax^2+(b+c)x+d+e=0&\iff x^2-4x+4=0\\&\iff(x-2)^2=0\\&\iff x=2.\end{align}So, yes, $ax^2+(b+c)x+d+e=0$ has solutions on $[1,+\infty)$ (actually, all of its solutions belong to that interval). However, \begin{align}ax^4+bx^3+cx^2+dx+e=0&\iff x^4+x^3-5x^2+4x=0\\&\iff x=0\vee x^3+x^2-5x+4=0\end{align}and the equation $x^3+x^2-5x+4=0$ has one and only one real root, which is approximately $-3.06$. Therefore, the equation $ax^4+bx^3+cx^2+dx+e=0$ has no real root on $[1,\infty)$.
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