A) 2 |
B) 4 |
C) 1 |
D) 3 |
B) 4 |
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.