The equation `(x/(x+1))^2+(x/(x-1))^2=a(a-1)` has a. Four real roots if `a >2` b.Four real roots if `a<-1` c Two real roots if `1

If a,b,c,d in R, then the equation (x^(2)+ax-3b)(x^(2)-cx+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 the equation ax^(2)+bx+c=x has no real roots,then the equation a(ax^(2)+bx+c)^(2)+b(ax^(2)+bx+c)+c=x will have a.four real roots b.no real root c.at least two least roots d.none of these

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

(x^(2)+1)^(2)-x^(2)=0 has (i) four real roots (ii) two real roots (iii) no real roots (iv) one real root

If f(x)=x has non real roots, then the equation f(f(x))=x (A) has all real and unequal roots (B) has some real and non real roots (C) has all real and equal roots (D) has all non 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

The equation x^(5)+x^(4)+1=0 has (1) One real and four imaginary roots (2) Five real roots (3) Three real and two imaginary roots (4) Three real roots in which two are same and two imaginary roots

[x^2] = x + 1 how many real roots

Show that (x^(2)+1)^(2)-x^(2)=0 has no real roots.