JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
If \(\alpha\) and \(\beta\) be two roots of the equation \(x^{2}-64 x+256=0\) Then the value of \(\left(\frac{\alpha^{3}}{\beta^{5}}\right)^{\frac{1}{8}}+\left(\frac{\beta^{3}}{\alpha^{5}}\right)^{\frac{1}{8}}\) is
- A \(1\)
- B \(3\)
- C \(4\)
- D \(2\)
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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