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If a, b, c ∈ ℝ+ and are in arithmetic progression, then the roots of the quadratic equation ax2 + bx + c = 0 are imaginary for:
1. \(\left|\dfrac{c}{a}-7\right|<3\sqrt{2}\)
2. \(\left|\dfrac{c}{a}-7\right|\ge4\sqrt{2}\)
3. \(\left|\dfrac{c}{a}-7\right|>3\sqrt{2}\)
4. \(\left|\dfrac{c}{a}-7\right|<4\sqrt{3}\)

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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-

  • Real roots when b2 - 4ac ≥ 0
  • Imaginary roots when b2 - 4ac < 0

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 \(\left|\dfrac{c}{a}-7\right|<4\sqrt{3}\)

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