解:∵ Δ = b² - 4ac = 16 - 12 = 4 > 0 ∴ 方程有两个不等实根 x = [-b ± √(b²-4ac)] / 2a = [4 ± √4] / 2 = 2 ± 1 ∴ x₁=3, x₂=1
Show that the equation x² - 4x + 3 = 0 has two distinct real roots. Solution: Discriminant Δ = b² - 4ac = (-4)² - 4×1×3 = 16 - 12 = 4 Since Δ > 0, the equation has two distinct real roots. Using the quadratic formula: x = [4 ± √4] / 2 x = 2 ± 1 Therefore, x = 3 or x = 1.
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