Polynomial Doubt Question :
$x^4 + px^3 + qx^2 + px + 1 =0 $ has real roots. Then what is the minimum value of $ p^2 +q^2 $ .
How I started ? 
I started by dividing the whole equation by $x^2$ then we get 
$ (x + \frac{1}{x} ) ^2 + p (x + \frac{1}{x} ) + q - 2 = 0 $
Then put  $(x + \frac{1}{x} ) = t$. Then discriminant should be greater than equal to zero. But now the problem arises that $t$ does not belong to $(-2,2)$ , so taking care of that part leads to solving inequality which I am unable to do .
Have I started the right way? 
One more thing to notice is that the sum of roots of the equation is equal to the sum of reciprocal of the roots . 
How to proceed further ?
 A: I don't know how much this is going to help, but if you use this you can find the roots
\begin{eqnarray}
x_1&=& -\frac{1}{4} \sqrt{p^2-4 q+8}-\frac{1}{2}
   \sqrt{\frac{p^2}{2}-\frac{-p^3+4 p q-8 p}{2 \sqrt{p^2-4
   q+8}}-q-2}-\frac{p}{4},\\
x_2&=& -\frac{1}{4} \sqrt{p^2-4
   q+8}+\frac{1}{2} \sqrt{\frac{p^2}{2}-\frac{-p^3+4 p q-8 p}{2
   \sqrt{p^2-4 q+8}}-q-2}-\frac{p}{4},\\
x_3&=& \frac{1}{4}
   \sqrt{p^2-4 q+8}-\frac{1}{2} \sqrt{\frac{p^2}{2}+\frac{-p^3+4 p q-8
   p}{2 \sqrt{p^2-4 q+8}}-q-2}-\frac{p}{4}, \\
x_4&=&
   \frac{1}{4} \sqrt{p^2-4 q+8}+\frac{1}{2}
   \sqrt{\frac{p^2}{2}+\frac{-p^3+4 p q-8 p}{2 \sqrt{p^2-4
   q+8}}-q-2}-\frac{p}{4}
\end{eqnarray}
And from here the constraint
$$
p^2 - 4q+8 > 0 \tag{1}
$$
A: Not a full solution, but could be a direction to go with:
$$t = x + \frac{1}{x} \in (-\infty, -2] \cup [2, +\infty) \text{ for } x \in \mathbb R$$
We want $t^2 + pt + q -2 = 0$ to have solution for $t$ as above：
$$p^2-4(q-2) \ge 0 \text{ and }$$
$$\frac{-p+\sqrt{p^2-4(q-2)}}{2} \ge 2 \text{ or } \frac{-p-\sqrt{p^2-4(q-2)}}{2} \le -2$$
Then we could get relations about $p$ and $q$, and get the min of $p^2 + q^2$.
