If a, b ε R, a ≠ 0 and the quadratic equation ax2−bx+2 =0 has imaginary roots, then a + b + 2 is
Positive
Observe that for given expression f(x) = ax2−bx+2
f(−1) = a+b+2
f(x) = 0 has imaginary roots
So, b2−4ac < 0
b2 < 4ac
b2 is always > 0 and for 4ac to be greater than b2
4ac > 0 ⇒ac > 0
⇒ a & c should be of same sign
Given c = 2(> 0) hence a > 0
As a > 0 & D < 0, graph looks as below
Hence, for ∀ x ε R, f(x) is positive
Hence f(-1) is also positive