4-2.Quadratic Equations and Inequations
easy

मान $\alpha, \beta$ समीकरण $x ^{2}+(20)^{1 / 4} x +(5)^{1 / 2}=0$ के दो मूल हैं। तो $\alpha^{8}+\beta^{8}$ बराबर है

A

$10$

B

$50$

C

$160$

D

$100$

(JEE MAIN-2021)

Solution

$\left(x^{2}+\sqrt{5}\right)^{2}=\sqrt{20} x^{2}$

$x^{4}=-5 \Rightarrow x^{8}=25$

$\alpha^{8}+\beta^{8}=50$

Standard 11
Mathematics

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