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| Last visit was: 23 Apr 2026, 11:13 |
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| As we have |x| in the equation so we will have two cases | |
| -Case 1: x ≥ 0 => |x| = x => \(x^2 – 11x - 60 = 0\) => \(x^2 + 4x - 15x - 60 = 0\) => x*(x + 4) -15*(x + 4) = 0 => (x + 4) * (x - 15) = 0 => x = -4, 15 But condition was x ≥ 0 => x = 15 is a SOLUTION | -Case 2: x < 0 => |x| = -x => \(x^2 + 11x - 60 = 0\) => \(x^2 + 15x - 4x - 60 = 0\) => x*(x + 15) -4*(x + 15) = 0 => (x + 15) * (x - 4) = 0 => x = 4, -15 But condition was x < 0 => x = -15 is a SOLUTION |
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