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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