2230.62 – Two Quadratics


Tags:Problem Set 8Algebra/TrigonometryAlgebra

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

Therefore, a=pqa = -p - q and b=pqb = pq. Similarly, p=abp = -a - b and q=abq = ab. Therefore,

a+b+p+q=(pq)+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 pq0pq \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+x2=0 x^2 + x - 2 = 0.

However, if pq0pq \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.