JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(\alpha, \beta\) be two roots of the equation \(x^{2}+(20)^{\frac{1}{4}} x+(5)^{\frac{1}{2}}=0\). Then \(\alpha^{8}+\beta^{8}\) is equal to:
- A \(10\)
- B \(50\)
- C \(160\)
- D \(100\)
Answer & Solution
Correct Answer
(B) \(50\)
Step-by-step Solution
Detailed explanation
\(\left(x^{2}+\sqrt{5}\right)^{2}=\sqrt{20} x^{2}\) \(x^{4}=-5 \Rightarrow x^{8}=25\) \(\alpha^{8}+\beta^{8}=50\)
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