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| As we have |x| in the equation so we will have two cases | |
| -Case 1: x ≥ 0 => |x| = x => \(x^2 − 3x + 2 = 0\) => \(x^2 − 2x - x + 2 = 0\) => x * ( x-2) - 1*(x - 2) = 0 => (x - 2) * (x - 1) = 0 => x = 1 or 2 But condition was x ≥ 0 => x = 1 or 2 are both SOLUTIONS | -Case 2: x < 0 => |x| = -x => \(x^2 + 3x + 2 = 0\) => \(x^2 + 2x + x + 2 = 0\) => x * ( x+2) + 1*(x + 2) = 0 => (x + 2) * (x + 1) = 0 => x = -1 or -2 But condition was x < 0 => x = -1, -2 are both SOLUTIONS |
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