Correct Answer - Option 4 :
\(\left|\dfrac{c}{a}-7\right|<4\sqrt{3}\)
Concept:
For any quadratic equation ax2 + bx + c = 0, where a, b, c are real numbers, there exists-
Formula used:
If a, b, c are in AP, then,
2b = a + c
⇒ b = (a + c)/2 ----(1)
Calculation:
For imaginary roots, we have,
b2 - 4ac < 0
\(⇒ \left ( \frac{a + c}{2} \right )^{2}-4ac < 0\) [using (1)]
⇒ a2 + c2 - 14ac < 0
On dividing the equation by a2, we get,
\(\Rightarrow(\frac{c}{a})^{2} + 1 + \frac{14c}{a}<0\)
\(\Rightarrow \left ( \frac{c}{a}-7 \right )^{2}<48\)
\(\Rightarrow \left|\dfrac{c}{a}-7\right|<4\sqrt{3}\)
Hence, the roots of the quadratic equation ax2 + bx + c = 0 are imaginary for \(\left|\dfrac{c}{a}-7\right|<4\sqrt{3}\)