`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 (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 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

If P(x) = ax^2+bx+c , and Q(x) = -ax^2+dx+c, acne 0, then prove that P(x).Q(x) =0 has at least two real roots.

If P(x) = ax^(2) + bx + c , and Q(x) = -ax^(2) + dx + c, ax ne 0 then prove that P(x), Q(x) =0 has atleast two 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 a ,b ,c ,d in R , then the equation (x^2+a x-3b)(x^2-c x+b)(x^2-dx+2b)=0 has a) 6 real roots b) at least 2 real roots c) 4 real roots d) none of these

If a+b+c=0, then the equation 3ax^(2)+2bx+c=0 has (i) imaginary roots (ii) real and equal roots (ii) real and unequal roots (iv) rational roots

The quadratic equation ax^(2)+bx+c=0 has real roots if:

The quadratic equation ax^(2)+bx+c=0 has real roots if: