If `x^2+p x+q=0a n dx^2+q x+p=0,(p!=q)` have a common roots, show that `p+q=0` . Also, show that their other roots are the roots of the equation `x^2+x+p q=0.`

Explanation

To solve the problem, we need to show that if the two quadratic equations \(x^2 + px + q = 0\) and \(x^2 + qx + p = 0\) have a common root, then \(p + q = 0\). We also need to show that the other roots of these equations are the roots of the equation \(x^2 + x + pq = 0\).

### Step 1: Let the common root be \(r\).
Assuming \(r\) is the common root, we can substitute \(r\) into both equations:

1. From the first equation:
\[
r^2 + pr + q = 0 \quad \text{(1)}
...