If $\mathrm{\alpha },\mathrm{\beta }$are the roots of ${\mathrm{x}}^2+(\mathrm{a}-2)\mathrm{x}-(\mathrm{a}+1)=0$ where 'a' is a variable then the minimum value of ${\mathrm{\alpha }}^2+{\mathrm{\beta }}^2$ is

$$\begin{gathered} \mathrm{\alpha }+\mathrm{\beta }=\mathrm{a}-2 \\ \mathrm{\alpha \beta }=-\mathrm{a}-1 \\ \mathrm{S}={\mathrm{\alpha }}^2+{\mathrm{\beta }}^2 \\ = {\left(\mathrm{\alpha }+\mathrm{\beta }\right)}^2-2\mathrm{\alpha \beta } \\ ={\left(\mathrm{a}-2\right)}^2+2\left(\mathrm{a}+1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{S}={\mathrm{a}}^2+4-4\mathrm{a}+2\mathrm{a}+2 \\ ={\mathrm{a}}^2-2\mathrm{a}+6 \\ ={\left(\mathrm{a}-1\right)}^2+5 \end{gathered}$$

$$\begin{gathered} {\left(\mathrm{a}-1\right)}^2\ge 0 \\ \Rightarrow {\left(\mathrm{a}-1\right)}^2+5\ge 5 \\ \Rightarrow \mathrm{S}\ge 5 \\ \therefore \min \mathrm{value} \mathrm{of} \mathrm{S} \mathrm{is} 5 \end{gathered}$$