JEE PYQ: Quadratic Equation Question 23
Question 23 - 2020 (06 Sep Shift 1)
If $\alpha$ and $\beta$ be two roots of the equation $x^2 - 64x + 256 = 0$. Then the value of $\left(\frac{\alpha^3}{\beta^5}\right)^{1/8} + \left(\frac{\beta^3}{\alpha^5}\right)^{1/8}$ is:
(1) 2
(2) 3
(3) 1
(4) 4
Show Answer
Answer: (1)
Solution
$\therefore \alpha + \beta = 64$, $\alpha\beta = 256$
$\frac{\alpha^{3/8}}{\beta^{5/8}} + \frac{\beta^{3/8}}{\alpha^{5/8}} = \frac{\alpha + \beta}{(\alpha\beta)^{5/8}} = \frac{64}{(256)^{5/8}} = \frac{64}{32} = 2$
BETA
