Let $p$ and $q$ be real numbers such that $p\ne q$. If the equations $x^2+px+q=0$ and $x^2+qx+p=0$ have a common root, then the value of $p+q$ is equal to:

Solution

1. Let $\alpha$ be the common root such that ${\alpha }^2+p\alpha +q=0$ and ${\alpha }^2+q\alpha +p=0$.
2. Subtracting the two equations yields $(p-q)\alpha +(q-p)=0$, which simplifies to $(p-q)(\alpha -1)=0$.
3. Since $p\ne q$, we must have $\alpha -1=0$, meaning the common root is $\alpha =1$.
4. Substituting $\alpha =1$ into the first equation gives $1^2+p(1)+q=0$, which simplifies to $p+q=-1$.
5. Hence the answer is (A).