Let α,β be the roots of the quadratic equation x2+√6x+3=0. Then \(\frac{α^23+β^23 + α ^14 + β ^14 }{α ^15 + β ^15 + α 160 + β ^{10}}\) is equal to
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When dealing with complex roots of quadratic equations, converting to polar form can be beneficial for simplification. Review trigonometric identities for eval uating cosine and sine of various angles. Pay careful attention to signs and pow ers when simplifying.
- 9
- 12
- 81
- 729
The Correct Option is C
Solution and Explanation
Given: \[ \alpha, \beta = \frac{-\sqrt{6} \pm \sqrt{6 - 12}}{2}. \]
We can rewrite this as: \[ \alpha, \beta = \sqrt{3} e^{\pm 3\pi i / 4}. \]
Required Expression: \[ \left(\sqrt{3}\right)^{23} 2\cos\left(69\pi / 4\right) = \frac{-\sqrt{6} \pm \sqrt{6} i}{2} + \left(\sqrt{3}\right)^{14} 2\cos\left(42\pi / 4\right). \]
Simplifying further: \[ \left(\sqrt{3}\right)^{15} 2\cos\left(45\pi / 4\right). \]
Additionally, we know: \[ \left(\sqrt{3}\right)^{10} 2\cos\left(30\pi / 4\right) \sqrt{3}^{8} = 81. \]
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