If $a, b, c$ arc positive and $a = 2 b + 3 c,$ then roots of the equation $a x^{2} + b x + c = 0$ are real for

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$|\frac{c}{a} - 11| \geq 4 \sqrt{7}$

$|\frac{b}{c} + 4| \geq 2 \sqrt{7}$

$|\frac{a}{c} - 11| \geq 4 \sqrt{7}$

$|\frac{c}{b} - 4| \geq 2 \sqrt{7}$

answer is A.

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Detailed Solution

For real roots, we must have

$b^{2} - 4 a c \geq 0$

$\begin{array}{l} \Rightarrow {(\frac{a - 3 c}{2})}^{2} - 4 a c \geq 0 \\ \Rightarrow a^{2} + 9 c^{2} - 6 a c - 16 a c \geq 0 \\ \Rightarrow {(\frac{a}{c})}^{2} - 22 (\frac{a}{c}) + 9 \geq 0 \Rightarrow {(\frac{a}{c} - 11)}^{2} \geq (4 \sqrt{7})^{2} \\ \Rightarrow |\frac{a}{c} - 11| \geq 4 \sqrt{7} \end{array}$

Again,

$b^{2} - 4 a c \geq 0$ and $a = 2 b + 3 c$

$\begin{array}{l} \Rightarrow b^{2} - 4 c (2 b + 3 c) \geq 0 \\ \Rightarrow b^{2} - 8 b c - 12 c^{2} \geq 0 \\ \Rightarrow {(\frac{b}{c})}^{2} - 8 (\frac{b}{c}) - 12 \geq 0 \\ \Rightarrow {(\frac{b}{c} - 4)}^{2} \geq (2 \sqrt{7})^{2} \Rightarrow |\frac{b}{c} - 4| \geq 2 \sqrt{7} \end{array}$