JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(a, b \in \mathbb{C}\). Let \(\alpha, \beta\) be the roots of the equation \(x^2 + ax + b = 0\). If \(\beta - \alpha = \sqrt{11}\) and \(\beta^2 - \alpha^2 = 3i\sqrt{11}\), then \((\beta^3 - \alpha^3)^2\) is equal to:
- A \(160\)
- B \(176\)
- C \(194\)
- D \(187\)
Answer & Solution
Correct Answer
(B) \(176\)
Step-by-step Solution
Detailed explanation
Given \(\beta - \alpha = \sqrt{11}\) and \(\beta^2 - \alpha^2 = 3i\sqrt{11}\) \(\beta + \alpha = \dfrac{\beta^2 - \alpha^2}{\beta - \alpha} = \dfrac{3i\sqrt{11}}{\sqrt{11}} = 3i\) Using the identity \((\beta - \alpha)^2 = (\beta + \alpha)^2 - 4\alpha\beta\)…
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