2230.62 – Two Quadratics

Suppose the quadratic equation x2+ax+b=0x^2 + a x + b = 0 has roots pp and qq AND the quadratic equation x2+px+q=0x^2 + p x + q = 0 has roots aa and bb. Then what is the value of a+b+p+qa+b+p+q?

Solution

Well,

0=x2+ax+b=(x−p)(x−q)=x2−(p+q)x+pq.0 = x^2 + a x + b = (x - p)(x - q) = x^2 -(p + q) x + pq.

Therefore, a=−p−qa = -p - q and b=pqb = pq. Similarly, p=−a−bp = -a - b and q=abq = ab. Therefore,

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

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

pq=ab=a(pq). pq = ab = a(pq).

Now if pq≠0pq \ne 0 then pqpq cancels in the preceding equation and we learn that a=1.a = 1. Parallel reasoning says that p=1,p = 1, and from this follows that q=b=−2q = b = -2. So the answer to the problem is −2-2, and the given equations are the single equation x2+x−2=0 x^2 + x - 2 = 0.

However, if pq≠0pq \ne 0 then it is easy to show that a=b=p=q=0a = b = p = q = 0 so both given equations are x2=0x^2 = 0 and the answer is 00.