Let α, β be two roots of equation x^{2}+(20)^{frac{1}{4}} x+(5)^{frac{1}{2}}=0 . Then α^{8}+β^{8

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4-2.Quadratic Equations and Inequations

easy

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:

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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