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JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations

यदि \(\alpha\) तथा \(\beta\), समीकरण \(x ^{2}-64 x +256=0\) के दो मूल हैं, तो \(\left(\frac{\alpha^{3}}{\beta^{5}}\right)^{\frac{1}{8}}+\left(\frac{\beta^{3}}{\alpha^{5}}\right)^{\frac{1}{8}}\) का मान है 

  1. A \(1\)
  2. B \(3\)
  3. C \(4\)
  4. D \(2\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(2\)

Step-by-step Solution

Detailed explanation

\(x^{2}-64 x+256=0\) \(\frac{\alpha+\beta}{(25)^{5 / 8}}=64, \alpha \beta=256\) \(\left(\frac{\alpha^{3}}{\beta^{1 / 8}}+\left(\frac{\beta^{3}}{\alpha^{5}}\right)^{1 / 8}\right.\) \(=\frac{\alpha^{3 / 8}}{\beta^{5 / 8}}+\frac{\beta^{3 / 8}}{\alpha^{5 / 8}}\)…
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