The sum, of the squares of all the roots of the equation x² + |2x – 3| – 4 = 0, is

$\begin{array}{l}{\mathrm{x}}^2+|2\mathrm{x}-3|-4=0\\ \text{ Case }\mathrm{I}:\mathrm{x}\ge \frac{3}{2}\\ {\mathrm{x}}^2+2\mathrm{x}-3-4=0\end{array}$

$\begin{array}{l}\begin{array}{c}{x}^2+2x-7=0\\ x==2\sqrt{2}-1\end{array}\\ \text{ Case II : }x<\frac{3}{2}\\ x^2+3-2x-4=0\\ x^2-2x-1=0\\ x=1-\sqrt{2}\\ \text{ Sum of squares }=(2\sqrt{2}-1{)}^2+(1-\sqrt{2}{)}^2\\ =8-4\sqrt{2}+1+1-2\sqrt{2}+2\\ =6(2-\sqrt{2})\end{array}$