JEE Main 2021 (Online) 27th July Morning ShiftQuestion 1
Let $\alpha,\beta$ be two roots of the equation $x^2+(20)^{\frac14}x+(5)^{\frac12}=0$. Then $\alpha^8+\beta^8$ is equal to :
Quadratic Equation · Powers of Roots (αⁿ ± βⁿ Recurrence)
Solution
Concept. Build $\alpha^8+\beta^8$ by squaring step by step. Formulas used. (1) $\alpha^{2k}+\beta^{2k}=\left(\alpha^k+\beta^k\right)^2-2(\alpha\beta)^k$Step 1 - Basics. $\alpha+\beta=-20^{1/4}$, $\alpha\beta=\sqrt5$. Step 2 - Use (1) three times. $$\alpha^2+\beta^2=\sqrt{20}-2\sqrt5=0,$$ $$\alpha^4+\beta^4=0-2\cdot5=-10,$$ $$\alpha^8+\beta^8=100-2\cdot25=50.$$ Answer: (C)