📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are one of the most important topics in the IGCSE Edexcel Mathematics syllabus. In this article, we will explore the standard form, solving methods, the discriminant, and real-world applications, with step-by-step examples designed for exam success.
二次方程是 IGCSE Edexcel 数学大纲中最重要的内容之一。本文将系统讲解二次方程的标准形式、求解方法、判别式以及实际应用，并通过逐步例题帮助你获得考试高分。
1. Standard Form of a Quadratic Equation | 二次方程的标准形式
c are constants, and a ≠ 0. The highest power of x is 2, which gives the equation its name "quadratic".
二次方程是可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数，且 a ≠ 0。未知数 x 的最高次数为 2，因此称为“二次”方程。
- a is the coefficient of x², b is the coefficient of x, and c is the constant term.
- a 是 x² 的系数，b 是 x 的系数，c 是常数项。
- If a = 0, the equation becomes linear, not quadratic.
- 如果 a = 0，方程变为一次方程，而不是二次方程。
ax² + bx + c = 0, a ≠ 0
2. Solving by Factorisation | 因式分解法
Factorisation is often the fastest method when the quadratic can be written as a product of two linear factors. You look for two numbers that multiply to give ac and add to give b, then split the middle term or use direct factor pairs.
当二次式可以写成两个一次因式的乘积时，因式分解法通常是最快的方法。你需要找到两个数，它们的乘积等于 ac，和等于 b，然后拆项或直接使用因式对。
ultiply to 12 and add to 7 are 3 and 4, so (x + 3)(x + 4) = 0. Therefore x = -3 or x = -4.
例：解方程 x² + 7x + 12 = 0。乘积为 12 且和为 7 的两个数是 3 和 4，所以 (x + 3)(x + 4) = 0。因此 x = -3 或 x = -4。
- Write the equation in standard form ax² + bx + c = 0. | 将方程写成标准形式 ax² + bx + c = 0。
- Factorise the left-hand side. | 对左边进行因式分解。
- Set each factor equal to zero and solve. | 令每个因式等于零并求解。
3. Solving by Completing the Square | 配方法
method works for all quadratics and is especially useful when factorisation is difficult.
配方法将二次式改写为 (x + p)² = q 的形式。这种方法对所有二次方程都适用，尤其在因式分解困难时特别有用。
6 is 3, so (x + 3)² - 9 + 1 = 0, which gives (x + 3)² = 8. Taking square roots: x + 3 = ±√8, so x = -3 ± 2√2.
例：解方程 x² + 6x + 1 = 0。6 的一半是 3，所以 (x + 3)² - 9 + 1 = 0，即 (x + 3)² = 8。开平方根：x + 3 = ±√8，所以 x = -3 ± 2√2。
x² + 6x + 1 = 0 ⇒ (x + 3)² = 8
4. The Quadratic Formula | 求根公式
ny quadratic equation directly. For ax² + bx + c = 0, the solutions are given by x = (-b ± √b² - 4ac) / (2a).
求根公式可以直接求解任何二次方程。对于 ax² + bx + c = 0，解为 x = (-b ± √b² - 4ac) / (2a)。
x = (−b ± √b² − 4ac) / 2a
Solve 2x² - 4x - 3 = 0. Here a = 2, b = -4, c = -3. Substitute into the formula: x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2.
例：解方程 2x² - 4x - 3 = 0。这里 a = 2，b = -4，c = -3。代入公式：x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2。
- Always simplify the surd if possible. | 尽量化简根式。
- Check your answers by substituting back. | 代回原方程检验答案。
5. The Discriminant | 判别式
The expression b² - 4ac is called the discriminant. It tells us how many real roots a quadratic equation has.
表达式 b² - 4ac 称为判别式。它告诉我们二次方程有多少个实数根。
| Discriminant Δ | Value of Δ | Number of Real Roots | 判别式 Δ | Δ 的值 | 实数根个数 |
|---|---|---|---|---|---|
| Δ > 0 | Positive | Two distinct real roots | Δ > 0 | 正数 | 两个不同的实数根 |
| Δ = 0 | Zero | One repeated real root | Δ = 0 | 零 | 一个重根 |
| Δ < 0 | Negative | No real roots | Δ < 0 | 负数 | 没有实数根 |
For example, x² + 2x + 5 = 0 has Δ = 4 - 20 = -16, so it has no real roots.
例如，x² + 2x + 5 = 0 的判别式 Δ = 4 - 20 = -16，因此没有实数根。
6. Solving Quadratic Inequalities | 解二次不等式
t solve the corresponding equation x² - x - 6 = 0 to get critical points x = -2 and x = 3.
解二次不等式（如 x² - x - 6 < 0）时，首先解对应的方程 x² - x - 6 = 0，得到临界点 x = -2 和 x = 3。
he coefficient of x² is positive, the parabola opens upward. The inequality x² - x - 6 < 0 is satisfied between the roots, so -2 < x < 3.
画出 y = x² - x - 6 的草图。由于 x² 的系数为正，抛物线开口向上。不等式 x² - x - 6 < 0 在两根之间成立，因此 -2 < x < 3。
x² - x - 6 < 0 ⇒ -2 < x < 3
- If the inequality is > 0, the solution is x < -2 or x > 3. | 如果不等式为 > 0，解为 x < -2 或 x > 3。
- Always consider whether endpoints are included (≤ or ≥). | 始终注意端点是否包含（≤ 或 ≥）。
7. Roots and Coefficients | 根与系数的关系
= 0 with roots α and β, the sum of the roots is α + β = -b/a, and the product is αβ = c/a.
对于二次方程 ax² + bx + c = 0，若其根为 α 和 β，则两根之和 α + β = -b/a，两根之积 αβ = c/a。
α + β = −b/a, αβ = c/a
5/2 and αβ = -3/2. You can use these to form new equations or to check your factorisation.
例：若 2x² + 5x - 3 = 0，则 α + β = -5/2，αβ = -3/2。你可以利用这些关系构造新方程或验证因式分解。
8. Solving Equations Reducible to Quadratics | 可化为二次方程的方程
mmon type involves substitution, such as letting u = x² for the equation x⁴ - 5x² + 4 = 0.
有些方程乍看不是二次方程，但可以通过变换化为二次方程。常见类型涉及换元，例如令 u = x² 来解方程 x⁴ - 5x² + 4 = 0。
4 = 0, so (u - 1)(u - 4) = 0. Thus u = 1 or u = 4. Since u = x², we get x = ±1 or x = ±2.
例：解方程 x⁴ - 5x² + 4 = 0。令 u = x²，则 u² - 5u + 4 = 0，所以 (u - 1)(u - 4) = 0。因此 u = 1 或 u = 4。由于 u = x²，得到 x = ±1 或 x = ±2。
Also, equations with fractions can be multiplied by a common denominator to obtain a quadratic equation.
此外，含有分式的方程可以乘以公分母，从而化为二次方程。
9. Word Problems with Quadratics | 二次方程应用题
Many exam problems describe a real-life situation that leads to a quadratic equation. Read the problem carefully, define a variable, form an equation, then solve and interpret the answer.
许多考试题目描述一个现实情境，最终会得到二次方程。仔细阅读题目，定义变量，建立方程，求解并解释答案。
Example: A rectangle has length (x + 3) cm and width (x - 1) cm. Its area is 36 cm². Find x.
例：一个长方形的长为 (x + 3) cm，宽为 (x - 1) cm，面积为 36 cm²。求 x。
(x + 3)(x − 1) = 36 ⇒ x² + 2x − 3 = 36 ⇒ x² + 2x − 39 = 0
Using the quadratic formula: x = (-2 ± √4 + 156) / 2 = (-2 ± √160) / 2 = -1 ± 2√10. Since x must be positive, x = -1 + 2√10 ≈ 5.32. The answer must make sense in the context.
使用求根公式：x = (-2 ± √4 + 156) / 2 = (-2 ± √160) / 2 = -1 ± 2√10。由于 x 必须为正，x = -1 + 2√10 ≈ 5.32。答案必须符合实际意义。
10. Sketching Quadratic Graphs | 二次函数图像
a < 0). The x-intercepts are the real roots of ax² + bx + c = 0, and the y-intercept is c.
二次函数 y = ax² + bx + c 的图像是抛物线。a 的正负决定开口方向：a > 0 时开口向上，a < 0 时开口向下。与 x 轴的交点就是方程 ax² + bx + c = 0 的实数根，与 y 轴的交点是 c。
ing the formula x = -b/(2a). The axis of symmetry is the vertical line through the vertex.
顶点（转折点）可以通过配方法求得，也可以用公式 x = -b/(2a)。对称轴是穿过顶点的竖直直线。
- When sketching, label the roots, the y-intercept and the vertex. | 画草图时，标注根、y 轴截距和顶点。
- If there are no real roots, the graph does not cross the x-axis. | 如果没有实数根，图像不与 x 轴相交。
11. Common Exam Mistakes | 常见考试错误
Many students lose marks in quadratic equation questions due to avoidable mistakes. Here are the most frequent ones.
许多学生在二次方程相关题目中因为可避免的错误而失分。以下是最常见的几点。
- range the equation into the form ax² + bx + c = 0 before factorising or using the formula.
- 忘记在因式分解或使用公式前将方程整理成 ax² + bx + c = 0 的形式。
- Misapplying the quadratic formula by using the sign of b incorrectly.
- 错误地使用求根公式，尤其是 b 的符号。
- Not simplifying surds such as √50 to 5√2.
- 没有化简根式，例如 √50 应化为 5√2。
- When solving inequalities, forgetting to reverse the sign when multiplying by a negative number.
- 解不等式时，两边乘以负数忘记改变不等号方向。
12. Practice Questions | 练习题目
Try these questions to test your understanding. Answers are provided below, but make sure you attempt them first.
尝试以下题目来检验你的理解。答案附在后面，但请务必先自己做一遍。
- Solve x² - 5x + 6 = 0. | 解方程 x² - 5x + 6 = 0。
- Solve 3x² + 2x - 1 = 0 using the quadratic formula. | 用求根公式解 3x² + 2x - 1 = 0。
- Find the discriminant of 2x² - 4x + 1 = 0 and state the number of real roots. | 求 2x² - 4x + 1 = 0 的判别式，并说明实数根的个数。
- Solve the inequality x² - 4x < 12. | 解不等式 x² - 4x < 12。
Answers: 1) x = 2 or x = 3 2) x = (−2 ± √16)/6 = 1/3 or −1 3) Δ = 8, two distinct real roots 4) −2 < x < 6
答案：1) x = 2 或 x = 3 2) x = (−2 ± √16)/6 = 1/3 或 −1 3) Δ = 8，两个不同的实数根 4) −2 < x < 6
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