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

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 (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.

If (x)=ax^(2)+bx+c amd Q(x)=-ax^(2)+dx+c,ac!=0P(x)=ax^(2)+bx+camdQ(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 ne 0 " then " f(x).g(x)=0 has

Let f(x)=ax^(2)+bx+c , g(x)=ax^(2)+px+q , where a , b , c , q , p in R and b ne p . If their discriminants are equal and f(x)=g(x) has a root alpha , then