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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:

  1. A \(10\)
  2. B \(50\)
  3. C \(160\)
  4. D \(100\)
Verified Solution

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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