CUET PG 2024 — Mathematics PYQ

$x^2+4x+{\alpha }^2-3\alpha =0$  

$\text{Roots real}$  

$D\ge 0$  

$D=\sqrt{(4{)}^2-4(1\times ({\alpha }^2-3\alpha ))}\ge 0$  

$16-4({\alpha }^2-3\alpha )\ge 0$  

$4-({\alpha }^2-3\alpha )\ge 0$  

$-({\alpha }^2-3\alpha )\ge -4$  

$-\alpha (\alpha -3)\ge -4$  

$\alpha =+4,-1$  

$\alpha \in [-1,4]$