JEE Main 2025 — 7 April, Evening Shift — Question 29

$y=\frac{5-x}{x^{2}-3 x+2}$

$x^{2} y-3 x y+2 y=5-x$

$x^{2} y+x(1-3 y)+2 y-5=0$

For $x$ to be real $D>0$

$(1-3 y)^{2}-4 y(2 y-5)>0$

$1+9 y^{2}-6 y-8 y^{2}+20 y>0$

$y^{2}+14 y+1>0$

$y \in(-\infty, \alpha) \cup(\beta, \infty)$

$\alpha+\beta=-14$

$\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$

$=194$