Pairs of Quadratic equations If each pair of the three equations
 $$x^2 + P_1x + q_1 = 0$$
$$x^2 + P_2x + q_2 = 0$$
$$x^2 + P_3x + q_3 = 0$$ 
have a common root, then prove that 
$$P_1^2 + P_2^2 + P_3^2 + 4(q_1 + q_2 + q_3) = 2 ( P_1.P_2 + P_2.P_3 + P_3.P_1)$$
 A: This statement is false.  Rewrite the first two equations in the form:
$\left[\begin{array}{rcl}x^2&x&1\end{array}\right]\left[\begin{array}{c}1\\P_1\\q_1\end{array}\right] = \left[\begin{array}{rcl}x^2&x&1\end{array}\right]\left[\begin{array}{c}1\\P_2\\q_2\end{array}\right]
%\left[\begin{array}{c}0\\0\\0\end{array}\right]$.
To solve this system of equations, we move $x$'s to one side and cancelling $x^2$'s, to get $x (P_1-P_2) = q_2-q_1$.  So for this solution to exist, it is necessary and sufficient that $P_1\neq P_2$ or $q_2 = q_1$.
This does not tell you anything about the relationship between the $P_i$ and $q_i$.  In particular, if we let all $q_i=0$, then all three pairs will have the solution $x=0$; however, we can choose $P_1=1=P_3$ and $P_2=-1$ to get 1+1+1+4(0) = 2(-1+1+-1) = -2, contradiction.
A: There are, seemingly, two cases:
Either there is a root which is common to the three equations (case 1) or not (case 2).
Let us treat case (2) first. Let $x_1$, $x_2$, $x_3$ be the common roots between equ. (2) and (3), (1) and (3) and (1) and (2) resp.
The only possibility is that equ. (1) has roots $x_2$ and $x_3$, equ. (2) has roots $x_3$ and $x_1$, equ. (3) has roots $x_1$ and $x_2$.
Using the classical formulas for the sum and the product of roots in a second degree equation:
$$P_1 = -x_2 - x_3 \ ; \ P_2 = -x_1 - x_3 \ ; \ P_3 = -x_2 - x_1$$
$$q_1 = x_2 x_3 \ ; \  q_2=x_1 x_3 \ ; \  q_3 = x_1 x_2$$
Plugging these expressions into 
$$P_1^2 + P_2^2 + P_3^2 + 4(q_1 + q_2 + q_3) - 2 ( P_1.P_2 + P_2.P_3 + P_3.P_1) \ \ (*) $$
yields 0.
Case (1). Let $x_0$ be the common root to the three equations. Let $x_1$, $x_2$, $x_3$ be the other roots (of the first, second and third equ. resp.). We have, in this case :
$$P_1 = -x_0 - x_1 \ ; \ P_2 = -x_0 - x_2 \ ; \ P_3 = -x_0 - x_3$$
$$q_1 = x_0 x_1 \ ; \ q_2 = x_0 x_2 \ ; \ q_3 = x_0 x_3 $$
BUT, in this case, plugging these expressions into (*) do not give 0 !
So, the text of the problem should be more explicit by excluding this case.
A final remark: the case of double roots has not been treated. It can be considered as a limit case to which continuity arguments can be applied. 
