代数技巧 - 第九部分
一元二次方程的求解
📐 什么是一元二次方程?
一元二次方程是以下形式的方程:
\(ax^2 + bx + c = 0\)
其中 \(a \neq 0\)
例子:
- \(x^2 - 5x + 6 = 0\) (a=1, b=-5, c=6)
- \(2x^2 + 3x - 2 = 0\) (a=2, b=3, c=-2)
- \(x^2 - 9 = 0\) (a=1, b=0, c=-9)
📊 求根公式
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
💡 两个解:
\(x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}\)
\(x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}\)
Δ 判别式
\(\Delta = b^2 - 4ac\)
💡 判别式决定方程解的个数:
| \(\Delta > 0\) | 两个不同的解 |
| \(\Delta = 0\) | 一个解(重根) |
| \(\Delta < 0\) | 无解(在实数范围内) |
✏️ 例题 - 公式法应用
例 1: 解方程 \(x^2 - 5x + 6 = 0\)
\(a = 1, b = -5, c = 6\)
\(\Delta = (-5)^2 - 4(1)(6) = 25 - 24 = 1\)
\(x = \frac{-(-5) \pm \sqrt{1}}{2(1)} = \frac{5 \pm 1}{2}\)
\(x_1 = \frac{5 + 1}{2} = 3\)
\(x_2 = \frac{5 - 1}{2} = 2\)
答:\(x = 2\) 或 \(x = 3\)
例 2: 解方程 \(x^2 - 4x + 4 = 0\)
\(a = 1, b = -4, c = 4\)
\(\Delta = 16 - 16 = 0\)
\(x = \frac{4}{2} = 2\)
答:\(x = 2\)(重根)
🔧 其他求解方法
1. 因式分解法:
当方程可以分解为乘积时使用。
\(x^2 - 5x + 6 = 0\)
\((x - 2)(x - 3) = 0\)
\(x = 2\) 或 \(x = 3\)
2. 开平方法(当 b=0 时):
\(x^2 - 9 = 0\)
\(x^2 = 9\)
\(x = \pm 3\)
3. 提公因式法(当 c=0 时):
\(x^2 - 5x = 0\)
\(x(x - 5) = 0\)
\(x = 0\) 或 \(x = 5\)
📐 韦达定理(根与系数的关系)
设 \(x_1, x_2\) 是方程 \(ax^2 + bx + c = 0\) 的两个根:
\(x_1 + x_2 = -\frac{b}{a}\) (两根之和)
\(x_1 \cdot x_2 = \frac{c}{a}\) (两根之积)
例子: \(x^2 - 5x + 6 = 0\)
和:\(2 + 3 = 5 = -\frac{-5}{1}\) ✓
积:\(2 \cdot 3 = 6 = \frac{6}{1}\) ✓
💡 考试提示
先确定 a、b、c再开始计算!
单独计算 Δ
用代入法检验答案!
📝 总结
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(\Delta > 0\) → 2 个解 | \(\Delta = 0\) → 1 个解 | \(\Delta < 0\) → 无解