QJEEM25Ma5c77: JEE Main 2025 Mathematics - Class 11 JEEM25Ma5c77 Solution: The sum, of the squares of all the roots of the equation , is

$\begin{aligned}& \mathrm{x}^2+|2 \mathrm{x}-3|-4=0 \\& \text { Case } \mathrm{I}: \mathrm{x} \geq \frac{3}{2} \\& \qquad \begin{array}{l}\mathrm{x}^2+2 \mathrm{x}-3-4=0 \\\mathrm{x}^2+2 \mathrm{x}-7=0 \\\mathrm{x}=2 \sqrt{2}-1\end{array}\end{aligned}$

$\begin{aligned}& \text { Case } \mathrm{II}: x<\frac{3}{2} \\& \begin{aligned}& x^2+3-2 x-4=0 \\& x^2-2 x-1=0 \\& x=1-\sqrt{2}\end{aligned}\end{aligned}$

$\begin{aligned}\text { Sum of squares } & =(2 \sqrt{2}-1)^2+(1-\sqrt{2})^2 \\& =8-4 \sqrt{2}+1+1-2 \sqrt{2}+2 \\& =6(2-\sqrt{2}) \quad \therefore(3)\end{aligned}$