## 2230.62 – Two Quadratics

Suppose the quadratic equation $x^2 + a x + b = 0$ has roots $p$ and $q$ AND the quadratic equation $x^2 + p x + q = 0$ has roots $a$ and $b$. Then what is the value of $a+b+p+q$?

## Solution

Well,

$0 = x^2 + a x + b = (x - p)(x - q) = x^2 -(p + q) x + pq.$

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

$a+b+p+q = (-p - q) + pq + p + q = pq.$

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 $0$.