Real roots

Let $x_{1},x_{2} \cdots x_{5}$ be the real roots of the polynomial $P(x)$

Then $x_{1} + x_{2} + \cdots + x_{5} = S = -a$

Let $\sum x_{1} \cdot x_{2} = T$

Then you can directly prove by applying AM -GM twice

SOLUTION 2:

Lemma : For any equation $f(x)$ of degree $n$ to have all it's roots real , it must be true that all the roots of $f^{r}(x)$ must also have it's all roots real , where $1 \leq r \leq n$. BUT THE CONVERSE IS NOT TRUE.

Proof : If you see this graphically . we observe that if all the roots of $f(x)$ of degree $n$ are real , then it must be true that the graph of the polynomial must take turns $n-1$ times or we can say that $f^{'}(x)$ must be $0$ for $n-1$ values including multiplicity of roots.

So proceeding this way leads us to the desired result.

So, if the equation in the question has all it's roots real , then it's third derivative must also have all it's roots real i.e. $60x^2 +24ax + 6b = 0$ has all it's roots real.

$\Rightarrow D \geq 0 \Rightarrow \boxed{2a^2 \geq 5b}$.

But the conditions provided in the question are contrary to what is an essential for all the roots to be real.Hence , all the roots of the equation can't be real.

@Rahil Sehgal Here is an inequality problem..

SOLUTION 1:

Suppose that the roots of the equation are $x_1,x_2,x_3,x_4,x_5$ , all $\in \mathbb{R}$

By AM-GM Inequality , we have $x_i^2+x_j^2 \geq 2x_ix_j$ .

Writing similarly and adding for all pairs of $(i,j)$ such that $1 \leq i < j \leq 5$.

We get $4\displaystyle \sum_{i=1}^{5} x_{i}^2 \geq 2\displaystyle \sum_{1 \leq i < j \leq 5} x_ix_j$

Adding $8\displaystyle \sum_{1 \leq i < j \leq 5} x_ix_j$ both sides we get

$4\left(\displaystyle \sum_{i=1}^{5} x_i\right)^2 \geq 10\displaystyle \sum_{1 \leq i < j \leq 5} x_ix_j$

Using Vieta's Relations : $\displaystyle \sum_{i=1}^{5} x_i = a , \displaystyle \sum_{1 \leq i < j \leq 5} x_ix_j = b$

Therefore , we get $\boxed{2a^2 \geq 5b}$.

But the condition given in the question is just the contrary to what is an essential for all roots to be real..Hence the equation can't have all it's roots real under the given condition of $2a^2 < 5b$

@Md Zuhair @Aditya Narayan Sharma @Anirudh Sreekumar@Brian Charlesworth@Pi Han Goh Please post your solutions!

It will help everyone know alternate solutions to the problem,

@Tapas Mazumdar@Kushal Bose@Akshat Sharda Post your solutions guys.