Mathematics - Quadratic Equation Question with Solution | TestHub

$\begin{array}{rl}& \text{ For }x\ge \frac{3}{2}\\ & x^2+2x-3-4=0\\ & x^2+2x-7=0\\ & x=\frac{-2\pm \sqrt{4+28}}{2}=-1\pm 2\sqrt{2}\end{array}$ Only $2\sqrt{2}-1$ is acceptable root For $x<\frac{3}{2}$ $x^2-2x+3-4=0$ $\begin{array}{rl}& x^2-2x-1=0\\ & x=\frac{2\pm \sqrt{4+4}}{2}=1\pm \sqrt{2}\end{array}$ Only $1-\sqrt{2}$ is acceptable Sum of the square $=(1-\sqrt{2}{)}^2+(2\sqrt{2}-1{)}^2$ $=6(2-\sqrt{2})$