Linear Inequalities

$$\begin{gathered} x^2-|2x-3|-4=0 \\ |2x-3|=\left\{\begin{array}{l}2x-3, x\ge \frac{3}{2}\\ -2x+3, x<\frac{3}{2}\end{array}\right. \\ Case I: when x\ge \frac{3}{2} \\ {x}^2-2x-1=0 \\ x= \frac{2\pm \sqrt{8}}{2}=1\pm \sqrt{2} \\ but 1-\sqrt{2} is discarded \sin ce x should be greater than 3/2. \\ So, x = 1+\sqrt{2} \\ Now Similarly from case II, we get x = -1-2\sqrt{2} \\ Now Sum of square of roots = {x_1}^2+{x_2}^2 \\ ={(1+\sqrt{2})}^2+{(-1-2\sqrt{2})}^2 \\ =1+2+2\sqrt{2}+1+8+4\sqrt{2} \\ =12+6\sqrt{2} \end{gathered}$$

$$\begin{gathered} x^2-|2x-3|-4=0 \\ |2x-3|=\left\{\begin{array}{l}2x-3, x\ge \frac{3}{2}\\ -2x+3, x<\frac{3}{2}\end{array}\right. \\ Case I: when x\ge \frac{3}{2} \\ {x}^2-2x-1=0 \\ x= \frac{2\pm \sqrt{8}}{2}=1\pm \sqrt{2} \\ but 1-\sqrt{2} is discarded \sin ce x should be greater than 3/2. \\ So, x = 1+\sqrt{2} \\ Now Similarly from case II, we get x = -1-2\sqrt{2} \\ Now Sum of square of roots = {x_1}^2+{x_2}^2 \\ ={(1+\sqrt{2})}^2+{(-1-2\sqrt{2})}^2 \\ =1+2+2\sqrt{2}+1+8+4\sqrt{2} \\ =12+6\sqrt{2} \end{gathered}$$