The sum of the squares of the roots of $ |x-2|^2 + |x-2| - 2 = 0 $ and the squares of the roots of $ x^2 - 2|x-3| - 5 = 0 $, is

<p>$$\begin{aligned} & |\mathrm{x}-2|^2+2|\mathrm{x}-2|-|\mathrm{x}-2|-2=0 \\ & \Rightarrow(|\mathrm{x}-2|+2)(|\mathrm{x}-2|-1)=0 \\ & \Rightarrow|\mathrm{x}-2|=1 \\ & \Rightarrow \mathrm{x}=2 \pm 1=3,1 \\ & \Rightarrow \text { sum of square of roots }=9+1=10 \\ & \mathrm{x}^2-2|\mathrm{x}-3|-5=0 \\ & \text { Case-I } \mathrm{x}-3 \geq 0 \\ & \Rightarrow \mathrm{x}^2-2 \mathrm{x}+1=0 \\ & \Rightarrow(\mathrm{x}-1)^2=0 \\ & \Rightarrow \mathrm{x}=1 \\ & \text { But } \mathrm{x} \geq 3 \\ & \Rightarrow \mathrm{x} \in \phi \\ & \text { Case-II } \mathrm{x}-3<0 \\ & \mathrm{x}^2+2 \mathrm{x}-11=0, \mathrm{D}>0 \Rightarrow \text { Real & distinct roots } \\ & \mathrm{f}(\mathrm{x})=\mathrm{x}^2+2 \mathrm{x}-11 \\ & \mathrm{f}(3)>0, \frac{-\mathrm{p}}{2 \mathrm{a}}=-1<3 \\ & \Rightarrow \text { both roots }<3, \text { both roots acceptable } \\ & \text { Sum of square of roots }=(\alpha+\beta)^2-2 \alpha \beta \\ & =4+22=26 \\ & \Rightarrow \text { Final sum }=10+26=36 \end{aligned}$$</p>