If b1b2=2(c1+c2), then at least one of the equations x2+b1x+c1=0 and x2+b2x+c2=0 has

Let D1 and D2 be discriminants of x2+b1x+c1=0and x2+b2x+c2=0 respectively. ThenD1+D2=b12-4c1+b22-4c2=(b12+b22)-4(c1+c2) b12+b22-2b1b2,     (∵b1b2=2(c1+c2)) =(b1-b2)2≥0 ⇒D1≥0 or D2≥0 or D1 and D2 both are positive.