2015/096) $ax^2 + bx + c = 0$ has imaginary roots and $a + c \lt b$ then prove that $4a + c \lt 2b$

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

$f(-1)= a - b + c < 0$ as$ a + c < b$

so

$f(-2) = 4a - 2b +c < 0$ or $4a + c < 2b$