Question

If a + b + c >$\frac{9c}{4}$and equation ax² + 2bx – 5c = 0 has non-real complex roots, then-

• A. a > 0, c > 0
• B. a > 0, c < 0
• C. a < 0, c < 0
• D. a < 0, c > 0

Answer 1

Answer: (B)

a > 0, c < 0

Answer 2

4a + 4b + 4c – 9c > 0

4a + 4b – 5c > 0

f(x) = ax² + 2bx – 5c = 0, D < 0

f(2) = 4a + 4b – 5c > 0

a > 0

$$$\Rightarrow c < 0$Ž $\begin{matrix} \begin{aligned} & a > 0 \\ & c < 0 \end{aligned} \end{matrix}$