If coefficients of the equation ax^{2} + bx + c = 0, a ¹ 0 are real and roots of the equation are non-real complex and a + c < b, then
(A) 4a + c > 2b (B) 4a + c < 2b (C) 4a + c = 2b (D) none of these
Please give the solution of this question.
Solution": Let assume x=-1
a-b+c but
a+c
hence f(x)<0 for all real values of x
therefore
putting x=-2
we get f(X)=4a+c-2b<0
or 4a+c<2b