A) 2
B) 4
C) 1
D) 3
Correct Answer: B
Solution :
\[{{x}^{2}}-3\left| x \right|+2=0\] lf \[x>0\], then \[\left| x \right|=x\] \[\therefore \] \[{{x}^{2}}-2x-x+2=0\] \[\Rightarrow \] \[{{x}^{2}}-2x-x+2=0\] \[\Rightarrow \] \[x(x-2)-1\,(x-2)=0\] \[\Rightarrow \] \[(x-1)\,(x-2)=0\] \[\Rightarrow \] \[x=1,2\] lf\[x<0\], then \[\left| x \right|=-x\] \[\therefore \] \[{{x}^{2}}+3x+2=0\] \[\Rightarrow \] \[{{x}^{2}}+2x+x+2=0\] \[\Rightarrow \] \[x(x+2)+1(x+2)=0\] \[\Rightarrow \] \[(x+1)\,(x+2)=0\] \[\Rightarrow \] \[x=-1,-2\] Hence, four solutions are possible.
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