A) \[2\]
B) \[-2\]
C) \[\sqrt{2}\]
D) \[-\sqrt{2}\]
Correct Answer: C
Solution :
\[{{x}^{2}}+|2x-3|-4=0\] \[|2x-3|=\left\{ \begin{matrix} (2x-3)\,\,\,\text{if} & x>\frac{3}{2} \\ -(2x-3)\,\,\text{if} & x<\frac{3}{2} \\ \end{matrix} \right.\] for\[x>\frac{3}{2},{{x}^{2}}+2x-3-4=0\] \[{{x}^{2}}+2x-7=0\] \[x=\frac{-2\pm \sqrt{4+28}}{2}=\frac{-2\pm 4\sqrt{2}}{2}=-1\pm 2\sqrt{2}\] Here \[x=2\sqrt{2}-1\] \[\left\{ 2\sqrt{2}-1<\frac{3}{2} \right\}\] for\[x<\frac{3}{2}\] \[{{x}^{2}}-2x+3-4=0\] \[\Rightarrow \]\[{{x}^{2}}-2x-1=0\] \[\Rightarrow \]\[x=\frac{2\pm \sqrt{4+4}}{2}=\frac{2\pm 2\sqrt{2}}{2}=1\pm \sqrt{2}\] Here \[x=1-\sqrt{2}\left\{ (1-\sqrt{2})<\frac{3}{2} \right\}\] Sum of roots : \[(2\sqrt{2}-1)+(1-\sqrt{2})=\sqrt{2}\]
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