QJEEM22Ma9e23: JEE Main 2022 Mathematics - Class 11 JEEM22Ma9e23 Solution: The minimum value of the sum of the squares of the roots of is:

$\alpha + \beta = a - 3,\,\alpha \beta = 1 - 2a$

$\Rightarrow {\alpha ^2} + {\beta ^2} = {(a - 3)^2} - 2(1 - 2a)$

$= {a^2} - 6a + 9 - 2 + 4a$

$= {a^2} - 2a + 7$

$= {(a - 1)^2} + 6$

So, ${\alpha ^2} + {\beta ^2} \ge 6$