Given $7$. degree polynomial $p(x)$, prove that all roots of $x^{10}-10x^9+39x^8=p(x)$ are in $(-\frac{5}{2},\frac{9}{2})$.

Question

Suppose $p(x)$ is a polynomial of the degree at most $7$ and assume that the following equation has $10$ real roots $$x^{10}-10x^9+39x^8=p(x)$$ Prove that all roots are in $(-\frac{5}{2},\frac{9}{2})$.

My try: If $x_0,x_1,...,x_9$ are all roots then by Vieta's formula's we have: $$\sum_{i=0}^9x_i = 10$$ and $$\sum_{0\le i<j\le 9}x_ix_j = 39$$ Then $$\sum_{i=0}^9x_i^2 = 10^2- 2\cdot 39 = 22$$ I suppose we may assume there is a root not in $(-\frac{5}{2},\frac{9}{2})$, say $x_0$. But what now?

Answer 1

For all $i$ by C-S we obtain:

$$22-x_i^2=\frac{1}{9}\sum_{k\neq i}1^2\sum_{k\neq i}x_k^2\geq\frac{1}{9}\left(\sum_{k\neq i}x_k\right)^2=\frac{1}{9}(10-x_i)^2,$$ which gives $$5x_i^2-10x_i-49\leq0,$$ which implies $$x_i\in\left(-2.5,4.5\right).$$ Done!

Answer 2

This is just rewriting Michael's solution. So if we mark $$a= \sum_{i=1}^9x_i = 10-x_0$$ and $$b= \sum_{i=1}^9x_i^2 = 22-x_0^2$$ By Cauchy inequality we know that $$ 9b\geq a^2$$ so we get $$9(22-x_0^2) \geq (10-x_0)^2$$ and know we have $$5x_0^2 -10x_0 -49 \leq 0$$ and thus the solution. :)