JEE PYQ: Quadratic Equation Question 37
Question 37 - 2019 (09 Apr Shift 2)
If $m$ is chosen in the quadratic equation $(m^2 + 1)x^2 - 3x + (m^2 + 1)^2 = 0$ such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is:
(1) $10\sqrt{5}$
(2) $8\sqrt{3}$
(3) $8\sqrt{5}$
(4) $4\sqrt{3}$
Show Answer
Answer: (3)
Solution
Sum of roots $= \frac{3}{m^2 + 1}$
$\therefore$ sum of roots is greatest $\therefore m = 0$
Hence equation becomes $x^2 - 3x + 1 = 0$
Now, $\alpha + \beta = 3$, $\alpha\beta = 1 \Rightarrow |\alpha - \beta| = \sqrt{5}$
$|\alpha^3 - \beta^3| = |(\alpha - \beta)(\alpha^2 + \beta^2 + \alpha\beta)| = \sqrt{5}(9 - 1) = 8\sqrt{5}$
BETA
