Question:

Step 1: Since $f(x) = ax^2 + 2bx - 3c$ has no real roots, the discriminant is negative: $(2b)^2 - 4a(-3c) < 0 \Rightarrow 4b^2 + 12ac < 0$. Step 2: From $4b^2 + 12ac < 0$: since $b^2 \geq 0$, we need $12ac < 0$, so $ac < 0$. Step 3: Since $f(x)$ has no real roots and is a quadratic, it is either always positive or always negative. Evaluate at $x=0$: $f(0) = -3c$. The parabola opens upward ($a > 0$) or downward ($a < 0$). Step 4: Given $\frac{3c}{4} < a + b$, evaluate $f(1) = a + 2b - 3c > 0$ (if $a > 0$, $f$ is always positive). Also $f(-3/2) = a\cdot\frac{9}{4} - 3b - 3c$. Since $ac < 0$ and we need to determine sign of $a$: if $a < 0$, $f$ is always negative, so $f(0) = -3c < 0 \Rightarrow c > 0$, but then $ac < 0$ ✓. Check $\frac{3c}{4} < a+b$: with $a<0, c>0$, this gives a negative LHS... Actually $\frac{3c}{4}>0$ and $a+b$ could be negative, contradiction. So $a > 0$. Step 5: Since $a > 0$ and $ac < 0$, we get $c < 0$. Options (a) and (b) are correct. Step 6: Since $c < 0$, $\frac{3c}{4} < 0 \leq a + |b|$, so $a + |b| > \frac{3c}{4}$. Option (c) is correct. Option (d) $b < 0$ is not necessarily true.