Question:
Let $\alpha, \beta$ be two roots of the equation
$x^{2}+(20)^{1 / 4} x+(5)^{1 / 2}=0 .$ Then $\alpha^{8}+\beta^{8}$ is equal to
Correct Option: , 3
Solution:
$\left(x^{2}+\sqrt{5}\right)^{2}=\sqrt{20} x^{2}$
$x^{4}=-5 \Rightarrow x^{8}=25$
$\alpha^{8}+\beta^{8}=50$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.