Quadratic EquationHard
Question
It both roots of the equation ax2 + x + c - a = 0 are imaginary and c > -1, then :-
Options
A.4c + 2 > 3a
B.4c + 2 < 3a
C.c < a
D.None
Solution
f(x) = ax2 + x + c - a = 0
f(l) = a + 1 + c - a = c + 1 > 0 ∵ c > - 1
1 
> 0
+ c - a > 0
4c - 3a + 2 > 0
4c + 2 > 3a
f(l) = a + 1 + c - a = c + 1 > 0 ∵ c > - 1
1 > 0 + c - a > 04c - 3a + 2 > 0
4c + 2 > 3a
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