[" Let "f(x)=ax^(2)+bx+c,a,b in Ra!=0],[" satisfying "f(1)+f(2)=0" .Then the equation "f(x)],[=0" has "]

Let f(x)=a x^(2)+b x+c , a, b, in R, a neq 0 satisfying f(1)+f(2)=0 . Then the equation f(x)=0 has

Let f(x)=ax^(2)+bx+c AA a,b,c in R,a!=0 satisfying f(1)+f(2)=0. Then,the quadratic equation f(x)=0 must have :

Let f(x)=ax^2+bx+c,b,cinR,aphi0 , satisfying f(1)+f(2)

Let f(x)=3ax^(2)-4bx+c(a,b,c in R.a!=0) where a,b,c are in A.P.Then the equation f(x)=0 has

Let f(x)=ax^(2)+bx+c, where a,b,c,in R and a!=0, If f(2)=3f(1) and 3 is a roots of the equation f(x)=0, then the other roots of f(x)=0 is

If f(x) is continuous in [0,2] and f(0)=f(2). Then the equation f(x)=f(x+1) has

If f(x) is continuous in [0,2] and f(0)=f(2). Then the equation f(x)=f(x+1) has

Let f (x) =ax ^(2) +bx + c,a ne 0, such the f (-1-x)=f (-1+ x) AA x in R. Also given that f (x) =0 has no real roots and 4a + b gt 0. Let p =b-4a, q=2a +b, then pq is:

Let f (x) =ax ^(2) +bx + c,a ne 0, such the f (-1-x)=f (-1+ x) AA x in R. Also given that f (x) =0 has no real roots and 4a + b gt 0. Let p =b-4a, q=2a +b, then pq is: