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| As we have |x| in the equation so we will have two cases | |
| -Case 1: x ≥ 0 => |x| = x => \(x^2 – 5x + 6 = 0\) => \(x^2 -2x -3x + 6 = 0\) => x*(x - 2) - 3*(x - 2) = 0 => (x - 2) * (x - 3) = 0 => x = 2, 3 But condition was x ≥ 0 and both 2 and 3 are ≥ 0 => x = 2, 3 are SOLUTIONS | -Case 2: x ≤ 0 => |x| = -x => \( x^2 + 5x + 6 = 0 \) => \(x^2 + 2x + 3x + 6 = 0\) => x*(x + 2) + 3*(x + 2) = 0 => (x + 2) * (x + 3) = 0 => x = -2, -3 But condition was x ≤ 0 and both -2 and -3 are ≤ 0 => x = -2, -3 are SOLUTIONS |
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