Q:

Please help me solve this probem on polynomials

Let 
$$f(x)=a_0x^n+a_1x^{n-1}+a_2 x^{n-2}+\cdots+a_n $$
be a polynomial of degree $n$ with real coefficients and $a_1^2<a_2$.
  Show that not all roots of $f(x)$ can be real. 

Please help me solve this problem in a simple way as I am only a high school student. 
(image of problem)

A:

Here's a proof when $a_0=1$.
If all the roots are real, call them $b_1,b_2,\dots,b_n$ (not necessarily distinct.)   Then $$
f(x)=(x-b_1)(x-b_2)\cdots(x-b_n)$$
We have $$
a_1=-\sum_{i=1}^nb_i\\
a_2=\sum_{i<j}b_ib_j$$
So the condition is $$
\left(\sum_{i=1}^nb_i\right)^2<\sum_{i<j}b_ib_j\\
\sum_{i=1}^nb_i^2+2\sum_{i<j}b_ib_j<\sum_{i<j}b_ib_j
$$
$$
\sum_{i=1}^nb_i^2+\sum_{i<j}b_ib_j<0\tag1
$$
Since all the $b_i$ are real, we see from $(1)$ that$$
\sum_{i<j}b_ib_j<0\tag2$$ Adding $(1)$ and $(2)$ gives $$
\sum_{i=1}^nb_i^2+2\sum_{i<j}b_ib_j<0\\
\left(\sum_{i=1}^nb_i\right)^2<0$$
contradiction. 
EDIT
The statement is not true for general $a_0$.  Consider $$\frac{x^2}{100}-\frac x{50}+\frac1{100}=\frac{(x-1)^2}{100}$$
We have $a_1=-\frac1{50},\ a_2=\frac1{100}$ so that the condition holds and all the roots are real.