If `P(x)=ax^2+bx+c`, `Q(x)=-ax^2+dx+c` where `ac!=0` then `P(x).Q(x)=0` has

If P(x)=ax^(2)+bx+cQ(x)=-ax^(2)+dx+c where ac!=0 then P(x)*Q(x)=0 has

If p(x) = ax^2 + bx + c and Q(x) = -ax^2 + dx +c where ac ne 0 then p(x). Q(x) = 0 has at least …………. Real roots

P(X) = ax^2+bx+c and Q(x)=-ax^2+dx+c=0 where ac!=0 then the equation P(x).Q(x)=0 has a) four real roots b) exactly 2 real roots c) either 2 or 4 roots d) at most 2 real roots

If P(x) = ax^(2) + bx + c and Q(x) = - ax^(2) + dx + c, where ac ne 0 , then P(x) . Q(x) = 0 has at least :

If (x)=ax^(2)+bx+c&Q(x)=-ax^(2)+dx+c,ac!=0P(x)=ax^(2)+bx+c&Q(x)=-ax^(2)+dx+c,ac!=0 then the equation P(x)*Q(x)=0 has (a) Exactly two real roots (b) Atleast two real roots (c)Exactly four real roots (d) No real roots

If f(x)=ax^(2)+bx+c,g(x)=-ax^(2)+bx+c, where ac !=0 then prove that f(x)g(x)=0 has at least two real roots.