Mathematics - Quadratic Equation Question with Solution | TestHub

$$\begin{array}{rl}& x^2+|2x-3|-4=0\\ & |2x-3|=\{\begin{array}{lll}(2x-3)& \text{ if }& \mathrm{x}>\frac{3}{2}\\ -(2x-3)& \text{ if }& \mathrm{x}<\frac{3}{2}\end{array}\end{array}$$ $$\begin{array}{rl}& \text{ for }x>\frac{3}{2},x^2+2x-3-4=0\\ & x^2+2x-7=0\\ & x=\frac{-2\pm \sqrt{4+28}}{2}\\ & =\frac{-2\pm 4\sqrt{2}}{2}=-1\pm 2\sqrt{2}\\ & \text{ Here }x=2\sqrt{2}-1\\ & \left\{2\sqrt{2}-1<\frac{3}{2}\right\}\\ & \text{ for }x<\frac{3}{2}\\ & x^2-2x+3-4=0\\ & \Rightarrow x^2-2x-1=0\\ & \Rightarrow x=\frac{2\pm \sqrt{4+4}}{2}=\frac{2\pm 2\sqrt{2}}{2}=1\pm \sqrt{2}\\ & \text{ Here }x=1-\sqrt{2}\left\{(1-\sqrt{2})<\frac{3}{2}\right\}\end{array}$$ Sum of roots: $(2\sqrt{2}-1)+(1-\sqrt{2})=\sqrt{2}$