JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Quadratic Equations and Inequalities

The condition requires the map $r\mapsto r^2-2$ to send the set of roots to itself. For a repeated root $\alpha$, ${\alpha }^2-2=\alpha$ gives $\alpha =2$ or $\alpha =-1$, yielding $(a,b)=(-4,4)$ and $(2,1)$. For distinct roots $\alpha ,\beta$: if both are fixed, $\{\alpha ,\beta \}=\{2,-1\}$ gives $(a,b)=(-1,-2)$; if they are interchanged, ${\alpha }^2-2=\beta$ and ${\beta }^2-2=\alpha$ give $\alpha +\beta =-1$ and $\alpha \beta =-1$, so $(a,b)=(1,-1)$; if both map to one common fixed root, $\{2,-2\}$ gives $(0,-4)$ and $\{1,-1\}$ gives $(0,-1)$. A complex-conjugate pair yields no solution. There are six such pairs $(a,b)$.