2230.62 – Two Quadratics

Well,

$$\begin{align*}0 &= x^2 + a x + b \\&= (x - p)(x - q) \\&= x^2 -(p + q) x + pq\end{align*}$$

Therefore, $a = -p - q$ and $b = pq$. Similarly, $p = -a - b$ and $q = ab$. Therefore,

$$\begin{align*}a+b+p+q &= (-p - q) + pq + p + q \\&= pq\end{align*}$$

Using similar reasoning, $a+b+p+q = ab$, therefore,

$$pq = ab = a(pq)$$

Now, if $pq \ne 0$ then $pq$ cancels in the preceding equation and we learn that $a = 1.$ Parallel reasoning says that $p = 1$, and from this follows that $q = b = -2$. So the answer to the problem is $-2$, and the given equations are the single equation $x^2 + x - 2 = 0$.

However, if $pq \ne 0$ then it is easy to show that $a = b = p = q = 0$ so both given equations are $x^2 = 0$ and the answer is $0$.