Polynomial $ax^2 + (b+c)x + (d+e)$ Let $a, b, c, d$ be real number such that polynomial $ax^2 + (b+c)x + (d+e)$ has real roots greater than $1$. Prove that polynomial $ax^4+bx^3+cx^2+dx+e$ has at least one real root. 
Is my work correct ?
Let $r$ be real root of $ax^2+(c+b)x+(e+d)$, so $ar^2+cr+e=(br+d)(-1)$.
Let $P(x) = ax^4+bx^3+cx^2+dx+e$
so $P(\sqrt{r}) = ar^2+cr+e + br\sqrt{r}+d\sqrt{r}= (br+d)(\sqrt{r}-1)$
$P(-\sqrt{r}) = ar^2+cr+e - br\sqrt{r}-d\sqrt{r}= (br+d)(-\sqrt{r}-1)$
Since $\sqrt{r}>1$, so $P(\sqrt{r})>0>P(-\sqrt{r})$
By Intermediate value theorem, $P(x) = ax^4+bx^3+cx^2+dx+e$ has at least one real root.

Assume the roots are $r_1,r_2$. Then:
$$a(x-r_1)^2(x-r_2)^2=ax^2+(-ar_1-ar_2)x+ar_1r_2=0.$$
Hence the second equation:
$$f(x)=ax^4-ar_1x^3-ar_2x^2+(ar_1r_2-e)x+e=0.$$
Note:
$$f(r_1)=-er_1+e$$
$$f(0)=e$$
Now IVT is applicable.
