Given equations is \( \Large x^{2}-3|x|+2=0 \). If x > 0 then lxl = x

=> \( \Large x^{2}-3x+2=0 \)

=> \( \Large x^{2}-2x-x+2=0 \)

=> \( \Large x \left(x-2\right)-1 \left(x-2\right)=0 \)

=> \( \Large \left(x-1\right) \left(x-2\right)=0 \)

x = 1,2

If x < 0, then \( \Large |x|=-x \)

=> \( \Large x^{2}+3x+2=0 \)

=> \( \Large x^{2}+2x+x+2=0 \)

=> \( \Large x \left(x+2\right)+1 \left(x+2\right)=0 \)

=> \( \Large \left(x+1\right) \left(x+2\right)=0 \)

=> x = -1, -2

Hence four solutions are possible.