If p,q,r and s are real numbers such that pr=2(q+s), then show that at least one of the equations x2+px+q=0 and x2+rx+s=0 has real roots.
We have x2+px+q=0....(i)
x2+rx+s=0....(ii)
Let D1 and D2 be the discriminants of equations (i) and (ii). Then
D1=b2−4ac=p2−4q
similarly,
D2=r2−4s
⇒D1+D2=p2−4q+r2−4s=(p2+r2)−4(q+s)
⇒D1+D2=p2+r2−4(pr2)[∵pr=2(q+s)∴q+s=pr2]
⇒D1+D2=p2+r2−2pr=(p−r)2≥0
At least one of D1 and D2 is greater than or equal to zero
At least one of the two equations has real roots.