Nature of the common root Let $p,q,r$ be positive real numbers not all equal ,such that two of the equations $$px^2+2qx+r=0$$ $$qx^2+2rx+p=0$$$$rx^2+2px+q=0$$ have a common root, say $a$,then the question is to comment about the nature of the common root $a$,i.e. whether the root is real and positive or real and negative or imaginary.
My attempt at the solution 
I found out the discriminant of the given equations and also I tried to get the value of $a$ from first two equations by cross multiplication. However I couldn't conclude anything.
Please help me in this regard.Thanks.
 A: Let's say $a$ is the common root of the first two equations. By multiplying the first equation with $q$ and the second with $p$ we get:
$$2(q^2-pr)a=p^2-rq$$
This equation cannot be $0=0$ for some positive $p$ and $q$, otherwise we would have $q^2-pr=p^2-rq=0$ so $r=\frac{p^2}{q}=\frac{q^2}{p}$, so $p=q=r$, contradiction with the hypothesis that $p,q,r$ are not all equal
So the equation must have at most one solution, which implies $a=\frac{p^2-rq}{2(q^2-pr)}$ is real.
$a$ cannot be $\ge 0$ as it would imply $pa^2+2qa+r>0$ by the positivity of $p,q,r$
So $a$ has to be negative.
A: Since you are just cyclically permuting the variables after renaming you can assume you have a shared root for the first two equations. Lets assume that $p,q\neq 0$ so that we are talking about a system of quadratic equations. If $p=0$ or $q=0$, then the linear polynomial is a factor of the quadratic polynomial so that case is easy. Next let's divide by the lead terms, to get that $\alpha_1$ is a root of both
$$x^2+2\frac{q}{p}x+\frac{r}{q}, $$ and
$$x^2+2\frac{r}{q}x+\frac{p}{q}.$$
Suppose that other root of the first equation is $\alpha_2$, and the other root of the second equation is $\alpha_3$.  This tells us that $\alpha_1\alpha_2=\frac{r}{q}$, $\alpha_1\alpha_3=\frac{p}{q}$, $\alpha_1+\alpha_2=-2\frac{q}{p}$ and $\alpha_1+\alpha_3=-2\frac{r}{q}$.
Solving the second two equations for $\alpha_2$ and $\alpha_3$ respectively, and plugging into the first two we get,
$$\alpha_1(-\alpha_1-2\frac{q}{p})=\frac{r}{q}$$ and $$\alpha_1(-\alpha_1-2\frac{r}{q})=\frac{p}{q}.$$
Taking the difference of the two equations you get an equation that is linear in $\alpha_1$, that allows you to solve for it in terms of $r,p,q$,
In the case such an $\alpha_1$ exists.
