2015/074) If $ax^2+bx+c=0$ has complex roots and $a + c \lt b$, then prove $4a+c\lt 2b$

because $f(x) = ax^2+bx+c=0$ has complex roots so this is positive for all x or -ve for all x

$f(-1) = a – b + c$
from given condition $a + c – b \lt 0$ so $f(-1) \lt 0$
so $f(-2) = 4a – 2b + c \lt 0$ or $4a + c \lt 2b$
proved