Mathematics - Quadratic Equation Question with Solution | TestHub

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