Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations are equations of the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. Solving them is a core skill in IGCSE Mathematics and appears in both Paper 1 and Paper 2. This article explains the key methods, the discriminant, and common exam tips.

二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。解二次方程是 IGCSE 数学的核心技能，在 Paper 1 和 Paper 2 中都会出现。本文介绍主要解法、判别式以及常见考试技巧。

1. Standard Form and Real Roots | 标准形式与实数根

A quadratic equation must be rearranged into the standard form ax² + bx + c = 0 before solving. The solutions are called roots, and they represent the x-values where the parabola crosses the x-axis.

解二次方程前必须先整理成标准形式 ax² + bx + c = 0。解称为“根”，它代表抛物线与 x 轴交点的 x 坐标。

If the equation is given as (x − 3)(x + 2) = 0, it is already factorised.
If it is given as x² = 5x − 6, rearrange to x² − 5x + 6 = 0.

For example, solve x² − 5x + 6 = 0. Then factorise to (x − 2)(x − 3) = 0, so x = 2 or x = 3.

例如，解 x² − 5x + 6 = 0，因式分解得 (x − 2)(x − 3) = 0，所以 x = 2 或 x = 3。

2. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic has simple integer factors. The principle is that if the product of two expressions is zero, then at least one of them must be zero.

当二次式有明显整数因式时，因式分解法最快。原理是：两个表达式的乘积为零，则至少一个表达式为零。

Write the equation in the form ax² + bx + c = 0.
Find two numbers that multiply to ac and add to b.
Split the middle term and factorise by grouping.

Example: Solve 2x² + 7x + 3 = 0. Multiply a and c: 2 × 3 = 6. Find two numbers whose product is 6 and sum is 7: they are 6 and 1. Split: 2x² + 6x + x + 3 = 0. Then 2x(x + 3) + 1(x + 3) = 0 → (2x + 1)(x + 3) = 0 → x = −½ or x = −3.

例：解 2x² + 7x + 3 = 0。a × c = 2 × 3 = 6。找两个数积为 6、和为 7：6 和 1。拆项：2x² + 6x + x + 3 = 0。然后 2x(x + 3) + 1(x + 3) = 0 → (2x + 1)(x + 3) = 0 → x = −½ 或 x = −3。

3. The Quadratic Formula | 二次求根公式

For any quadratic equation ax² + bx + c = 0, the roots can be found using the formula:

对于任意二次方程 ax² + bx + c = 0，可用求根公式求出根：

x = (−b ± √(b² − 4ac)) ÷ 2a

This method works for all quadratics, even when factorisation is difficult or impossible. You should memorise it and be able to substitute values correctly.

这种方法适用于所有二次方程，即使难以因式分解或无法因式分解。你需要牢记公式并正确代入数值。

Example: Solve 3x² − 4x − 2 = 0 using the formula. Here a = 3, b = −4, c = −2. Substitute: x = (4 ± √(16 + 24)) ÷ 6 = (4 ± √40) ÷ 6 = (4 ± 2√10) ÷ 6 = (2 ± √10) ÷ 3. So x ≈ 1.72 or x ≈ −0.39.

例：用公式解 3x² − 4x − 2 = 0。a = 3，b = −4，c = −2。代入：x = (4 ± √(16 + 24)) ÷ 6 = (4 ± √40) ÷ 6 = (4 ± 2√10) ÷ 6 = (2 ± √10) ÷ 3。所以 x ≈ 1.72 或 x ≈ −0.39。

4. Completing the Square | 配方法

Completing the square rewrites a quadratic as a perfect square plus a constant. The general form is:

配方法将二次式改写为完全平方加一个常数。一般形式为：

x² + bx + (b/2)² − (b/2)² + c = 0

This method is useful when solving equations or finding the turning point of a parabola.

该方法在解方程或求抛物线顶点时很有用。

Example: Solve x² + 6x + 2 = 0 by completing the square. First, x² + 6x + 9 − 9 + 2 = 0 → (x + 3)² − 7 = 0. Then (x + 3)² = 7 → x + 3 = ±√7 → x = −3 ± √7.

例：用配方法解 x² + 6x + 2 = 0。首先 x² + 6x + 9 − 9 + 2 = 0 → (x + 3)² − 7 = 0。然后 (x + 3)² = 7 → x + 3 = ±√7 → x = −3 ± √7。

5. The Discriminant | 判别式 Δ (Delta)

The expression b² − 4ac is called the discriminant. It tells you the nature of the roots without solving the equation.

表达式 b² − 4ac 称为判别式。它可以在不解方程的情况下判断根的性质。

Discriminant Δ = b² − 4ac	Nature of roots	根的性质
Δ > 0	Two distinct real roots	两个不相等实数根
Δ = 0	Two equal real roots (one repeated root)	两个相等实数根（重根）
Δ < 0	No real roots (complex roots)	无实数根（复数根）

Example: For 2x² + 3x + 5 = 0, Δ = 9 − 40 = −31 < 0, so there are no real roots.

例：对于 2x² + 3x + 5 = 0，Δ = 9 − 40 = −31 < 0，所以没有实数根。

6. Factorising Special Quadratics | 特殊二次式的因式分解

Some quadratics can be factorised using special identities. The difference of two squares and perfect squares appear often in IGCSE papers.

有些二次式可以用特殊恒等式因式分解。平方差和完全平方式在 IGCSE 考试中经常出现。

Difference of squares: x² − a² = (x − a)(x + a)
Perfect square: x² + 2abx + a²b² = (x + ab)²

Example: Factorise 4x² − 9. This is (2x)² − 3² → (2x − 3)(2x + 3).

例：分解 4x² − 9。即 (2x)² − 3² → (2x − 3)(2x + 3)。

7. Equations with Fractions or in Disguise | 分式方程与“伪装”的二次方程

Many IGCSE questions present equations that are not obviously quadratic. You must multiply through by the denominator or substitute a new variable to reveal a quadratic form.

许多 IGCSE 题目给出的方程并不是明显的二次方程。需要通过乘以分母或换元来揭示二次形式。

Example 1: Solve x + 1/x = 3. Multiply by x (x ≠ 0): x² + 1 = 3x → x² − 3x + 1 = 0. Use the formula to get x = (3 ± √5) ÷ 2.

例 1：解 x + 1/x = 3。两边乘以 x（x ≠ 0）：x² + 1 = 3x → x² − 3x + 1 = 0。用求根公式得 x = (3 ± √5) ÷ 2。

Example 2: Solve x⁴ − 5x² + 4 = 0. Let y = x², then y² − 5y + 4 = 0 → (y − 1)(y − 4) = 0 → y = 1 or y = 4 → x = ±1 or x = ±2.

例 2：解 x⁴ − 5x² + 4 = 0。令 y = x²，则 y² − 5y + 4 = 0 → (y − 1)(y − 4) = 0 → y = 1 或 y = 4 → x = ±1 或 x = ±2。

8. Word Problems and Graphs | 应用题与图像

Quadratic equations often model real-life situations, such as projectile motion, areas, and profit calculations. When solving word problems, define a variable carefully and check that your answer makes sense in the context.

二次方程常用来建模现实情境，如抛体运动、面积和利润计算。解应用题时要谨慎设变量，并检查答案在情境中是否合理。

Example: The area of a rectangle is 40 cm² and its length is 3 cm longer than its width. Find the width. Let width = x, length = x + 3. Then x(x + 3) = 40 → x² + 3x − 40 = 0 → (x + 8)(x − 5) = 0 → x = 5 (since width cannot be negative).

例：矩形面积为 40 cm²，长比宽长 3 cm。求宽。设宽 = x，长 = x + 3。则 x(x + 3) = 40 → x² + 3x − 40 = 0 → (x + 8)(x − 5) = 0 → x = 5（因为宽不能为负）。

Graphically, the roots of a quadratic equation correspond to the x-intercepts of the parabola y = ax² + bx + c. If the quadratic has no real roots, the graph does not cross the x-axis.

从图像上看，二次方程的根对应抛物线 y = ax² + bx + c 与 x 轴的交点。如果二次方程无实数根，则图像不与 x 轴相交。

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Solving Quadratic Equations | 解一元二次方程