Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations appear in almost every IGCSE Mathematics paper. This revision guide covers the four essential methods of solution, the discriminant, graph sketching, and common examination pitfalls — all explained step by step for both Paper 2 and Paper 4.

二次方程几乎出现在每一份 IGCSE 数学试卷中。本篇复习指南涵盖四种必备解法、判别式、图像草图以及常见考试陷阱——无论是 Paper 2 还是 Paper 4，均分步详解。

1. What Is a Quadratic Equation? | 什么是二次方程？

A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the variable is 2, which is why it is called ‘quadratic’ (from the Latin quadratus, meaning square).

二次方程是指任何可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数，且 a ≠ 0。变量的最高次数为 2，因此称为“二次”（源自拉丁语 quadratus，意为“平方”）。

For example, 2x² − 5x + 3 = 0 is a quadratic equation, while x³ + x² = 0 is not, because its highest power is 3. A quadratic equation always has at most two solutions; these solutions are also called roots or solutions.

例如，2x² − 5x + 3 = 0 是二次方程，而 x³ + x² = 0 不是，因为它的最高次数是 3。二次方程至多有两个解；这些解也称为根或解。

ax² + bx + c = 0 , a ≠ 0

2. Expanding and Factorising | 展开与因式分解

Before solving, you must be fluent in two algebraic skills. Expanding means removing brackets, e.g. (x + 3)(x − 2) = x² + x − 6. Factorising is the reverse process: x² + x − 6 = (x + 3)(x − 2). In IGCSE exams, factorising a quadratic with integer coefficients is a routine first step.

在求解之前，你必须熟练掌握两项代数技能。展开意味着去掉括号，例如 (x + 3)(x − 2) = x² + x − 6。因式分解是逆过程：x² + x − 6 = (x + 3)(x − 2)。在 IGCSE 考试中，对整数系数的二次式进行因式分解是常规第一步。

To factorise x² + bx + c, find two numbers whose product is c and whose sum is b. For example, factorise x² − 7x + 12: we need two numbers with product 12 and sum −7; these are −3 and −4. Hence x² − 7x + 12 = (x − 3)(x − 4).

要对 x² + bx + c 进行因式分解，需要找到两个数，使其乘积为 c，和为 b。例如，分解 x² − 7x + 12：我们需要乘积为 12、和为 −7 的两个数，即 −3 和 −4。因此 x² − 7x + 12 = (x − 3)(x − 4)。

When the coefficient of x² is not 1 (i.e. a ≠ 1), use the method of factorising by grouping. For 2x² + 7x + 3, multiply a and c: 2 × 3 = 6. Find two numbers with product 6 and sum 7: these are 1 and 6. Then split the middle term:

当 x² 的系数不为 1（即 a ≠ 1）时，使用分组分解法。对于 2x² + 7x + 3，先计算 a 与 c 的乘积：2 × 3 = 6。找到乘积为 6、和为 7 的两个数：即 1 和 6。然后拆分中间项：

2x² + 7x + 3 = 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)

3. Solving by Factorisation | 用因式分解求解

The factorisation method relies on the zero-product property: if A × B = 0, then either A = 0 or B = 0. This gives a fast and elegant way to solve many quadratic equations, and examiners expect it whenever factorisation is possible.

因式分解法依赖于零积性质：若 A × B = 0，则 A = 0 或 B = 0。这为解许多二次方程提供了一种快速而优雅的方法，只要可以因式分解，考官就期望使用此法。

Example: Solve x² − 5x + 6 = 0. First factorise: (x − 2)(x − 3) = 0. Then apply the zero-product rule:

例：解 x² − 5x + 6 = 0。首先因式分解：(x − 2)(x − 3) = 0。然后应用零积法则：

x − 2 = 0 → x = 2 ; x − 3 = 0 → x = 3

So the solutions are x = 2 and x = 3. Always check your answers by substituting back into the original equation: 2² − 5(2) + 6 = 4 − 10 + 6 = 0 ✓, and 3² − 5(3) + 6 = 9 − 15 + 6 = 0 ✓.

因此解为 x = 2 和 x = 3。务必通过代回原方程来检验：2² − 5(2) + 6 = 4 − 10 + 6 = 0 ✓，3² − 5(3) + 6 = 9 − 15 + 6 = 0 ✓。

Example with a ≠ 1: Solve 3x² − 5x − 2 = 0. Multiply 3 × (−2) = −6. Two numbers with product −6 and sum −5 are −6 and 1. Split the middle term: 3x² − 6x + x − 2 = 0. Group: 3x(x − 2) + 1(x − 2) = 0. Hence (3x + 1)(x − 2) = 0, so x = −1⁄3 or x = 2.

a ≠ 1 的示例：解 3x² − 5x − 2 = 0。计算 3 × (−2) = −6。乘积为 −6、和为 −5 的两个数是 −6 和 1。拆分中间项：3x² − 6x + x − 2 = 0。分组：3x(x − 2) + 1(x − 2) = 0。因此 (3x + 1)(x − 2) = 0，所以 x = −1⁄3 或 x = 2。

4. The Quadratic Formula | 二次公式

When factorisation is not straightforward, the quadratic formula is a universal method. For ax² + bx + c = 0, the solutions are given by:

当因式分解不直接时，二次公式是一种通用方法。对于 ax² + bx + c = 0，解由下式给出：

x = (−b ± √(b² − 4ac)) ÷ (2a)

This formula is provided on the IGCSE formula sheet, but you must know exactly how to substitute and simplify. Be especially careful with negative values of b.

该公式在 IGCSE 公式表中提供，但你必须会准确代入并化简。尤其要小心 b 为负数的情况。

Example: Solve 2x² + 3x − 5 = 0 using the formula. Here a = 2, b = 3, c = −5. Substitute:

例：用公式解 2x² + 3x − 5 = 0。这里 a = 2，b = 3，c = −5。代入：

x = (−3 ± √(3² − 4(2)(−5))) ÷ (2(2)) = (−3 ± √(9 + 40)) ÷ 4 = (−3 ± √49) ÷ 4 = (−3 ± 7) ÷ 4

This gives two solutions: x = (−3 + 7) ÷ 4 = 4 ÷ 4 = 1, and x = (−3 − 7) ÷ 4 = −10 ÷ 4 = −2.5. Both check out when substituted back. Note that the formula always works, even when factorisation would be quicker.

这给出两个解：x = (−3 + 7) ÷ 4 = 4 ÷ 4 = 1，以及 x = (−3 − 7) ÷ 4 = −10 ÷ 4 = −2.5。代回检验均成立。注意，这个公式总是有效，即使因式分解可能更快。

5. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. This form reveals the turning point of the graph and is also used to solve equations. The key identity is (x + p)² = x² + 2px + p², so the coefficient of x tells us what p is.

配方法将二次式改写为 (x + p)² + q 的形式。这种形式能揭示图像的顶点，也用于解方程。关键恒等式为 (x + p)² = x² + 2px + p²，因此 x 的系数告诉我们 p 的值。

Example: Complete the square for x² + 6x + 2. Take half of 6, which is 3, and write (x + 3)² = x² + 6x + 9. To keep the expression equal to the original, subtract 9 and add the constant 2:

例：对 x² + 6x + 2 配方。取 6 的一半，即 3，写出 (x + 3)² = x² + 6x + 9。为使表达式与原式相等，减去 9 再加上常数 2：

x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7

To solve (x + 3)² − 7 = 0, isolate the square: (x + 3)² = 7. Take square roots of both sides: x + 3 = ±√7, so x = −3 ± √7. This method is especially useful when the solutions are irrational, as in this case.

要解 (x + 3)² − 7 = 0，先隔离平方项：(x + 3)² = 7。两边开平方：x + 3 = ±√7，所以 x = −3 ± √7。当解为无理数时，此方法特别有用，本例即是。

When a ≠ 1, first factor a out of the x² and x terms. For example, 2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3.

当 a ≠ 1 时，先把 a 从 x² 项和 x 项中提出。例如，2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3。

6. The Discriminant | 判别式

The expression b² − 4ac inside the quadratic formula is called the discriminant, often written as Δ. It tells us how many roots a quadratic equation has, without actually solving it — a very common IGCSE question type.

二次公式中的表达式 b² − 4ac 称为判别式，通常记作 Δ。它能告诉我们二次方程有多少个根，而无需实际求解——这是 IGCSE 中非常常见的题型。

Discriminant value	Nature of roots	判别式取值	根的性质
b² − 4ac > 0	Two distinct real roots	b² − 4ac > 0	两个不同的实数根
b² − 4ac = 0	One repeated root (two equal roots)	b² − 4ac = 0	一个重根（两个相等根）
b² − 4ac < 0	No real roots	b² − 4ac < 0	无实数根

Example: Find the value of k such that x² + 4x + k = 0 has exactly one root. Set the discriminant equal to zero: b² − 4ac = 4² − 4(1)(k) = 16 − 4k = 0, so k = 4. The equation x² + 4x + 4 = (x + 2)² = 0 indeed has only the root x = −2.

例：求使 x² + 4x + k = 0 恰好有一个根时的 k 值。令判别式等于零：b² − 4ac = 4² − 4(1)(k) = 16 − 4k = 0，所以 k = 4。方程 x² + 4x + 4 = (x + 2)² = 0 确实只有根 x = −2。

7. Graphs of Quadratic Functions | 二次函数图像

The graph of y = ax² + bx + c is a curve called a parabola. If a > 0, the parabola opens upwards (a ‘smile’); if a < 0, it opens downwards (a 'frown'). The curve is symmetric about a vertical line through its vertex (turning point).

y = ax² + bx + c 的图像是一条称为抛物线的曲线。若 a > 0，抛物线开口向上（“微笑”形）；若 a < 0，开口向下（“皱眉”形）。曲线关于通过其顶点（转折点）的竖直线对称。

To sketch a quadratic graph, you need four key features: the y-intercept, the x-intercepts (roots), the axis of symmetry, and the vertex. The y-intercept is simply c. The x-intercepts are found by solving ax² + bx + c = 0. The axis of symmetry is x = −b ÷ (2a), and the vertex lies on this line.

要画二次函数草图，需要四个关键特征：y 截距、x 截距（根）、对称轴和顶点。y 截距就是 c。x 截距通过解 ax² + bx + c = 0 求得。对称轴为 x = −b ÷ (2a)，顶点位于该直线上。

Example: Sketch y = x² − 2x − 3. The y-intercept is −3. Factorise the quadratic: (x − 3)(x + 1) = 0, so roots are x = 3 and x = −1. The axis of symmetry is x = −(−2) ÷ (2 × 1) = 1. Substitute x = 1 into the equation to find the vertex: y = 1 − 2 − 3 = −4. So the vertex is (1, −4).

例：画出 y = x² − 2x − 3 的草图。y 截距为 −3。因式分解：(x − 3)(x + 1) = 0，所以根为 x = 3 和 x = −1。对称轴为 x = −(−2) ÷ (2 × 1) = 1。将 x = 1 代入方程求顶点：y = 1 − 2 − 3 = −4。所以顶点为 (1, −4)。

8. Applications and Word Problems | 实际应用与应用题

IGCSE papers frequently test quadratics through word problems. Common contexts include finding unknown lengths in rectangles, projectile motion, and number puzzles. The key skill is translating the English description into an algebraic equation.

IGCSE 试卷常通过应用题考查二次方程。常见情境包括求矩形未知边长、抛体运动以及数字谜题。关键技能是把文字描述转化为代数方程。

Example: A rectangle has length 4 cm longer than its width. Its area is 96 cm². Find its dimensions. Let the width be x, so the length is x + 4. The area gives x(x + 4) = 96, i.e. x² + 4x − 96 = 0. Factorise: (x + 12)(x − 8) = 0. Since a length cannot be negative, x = −12 is discarded, so the width is 8 cm and the length is 12 cm.

例：一个矩形的长比宽长 4 cm，面积为 96 cm²。求其尺寸。设宽为 x，则长为 x + 4。面积给出 x(x + 4) = 96，即 x² + 4x − 96 = 0。因式分解：(x + 12)(x − 8) = 0。因为长度不能为负，舍去 x = −12，所以宽为 8 cm，长为 12 cm。

Always check whether the solutions make sense in context. Reject any negative or fractional answers that are physically impossible, and clearly state the units in your final answer.

始终检查解在情境中是否有意义。舍弃任何在物理上不可能的负数或分数答案，并在最终答案中明确写出单位。

9. Common Mistakes and Exam Tips | 常见错误与考试技巧

One of the most frequent errors is forgetting to rearrange the equation into the standard form ax² + bx + c = 0 before solving. For example, to solve x² = 3x, students often divide both sides by x and lose the root x = 0. The correct method is to rearrange: x² − 3x = 0, then factorise: x(x − 3) = 0, giving x = 0 or x = 3.

最常见的错误之一是忘记先将方程整理成标准形式 ax² + bx + c = 0 再求解。例如，解 x² = 3x 时，学生常两边同除以 x，从而丢失根 x = 0。正确做法是移项：x² − 3x = 0，再因式分解：x(x − 3) = 0，得到 x = 0 或 x = 3。

Always set the equation equal to zero before factorising or using the formula. 求解前务必使方程等于零。
Check the sign of b when substituting into the quadratic formula — (−b) with a negative b becomes positive. 代入二次公式时注意 b 的符号——若 b 为负，则 −b 为正。
In word problems, reject solutions that do not make sense in the real-world context. 在应用题中，舍弃在现实情境中不合理的解。
Show all working; even if your final answer is wrong, method marks can be awarded. 写出完整过程；即使最终答案错误，步骤分仍可获得。
Use a calculator for arithmetic, but not as a substitute for algebraic understanding. 用计算器做算术，但不能替代代数理解。

10. Summary of Key Formulas | 关键公式总结

Having all the essential results in one place helps with revision. Mastery of these four tools — factorisation, the formula, completing the square, and the discriminant — will let you solve any quadratic problem the IGCSE syllabus can offer.

将所有关键结论集中在一处有助于复习。熟练掌握这四种工具——因式分解、二次公式、配方法和判别式——能让你解决 IGCSE 考纲中任何二次方程问题。

Standard form: ax² + bx + c = 0
Quadratic formula: x = (−b ± √(b² − 4ac)) ÷ (2a)
Discriminant: Δ = b² − 4ac
Completed square: (x + p)² + q
Axis of symmetry: x = −b ÷ (2a)

With consistent practice and careful attention to algebraic detail, solving quadratic equations will become one of the most reliable scoring areas in your IGCSE Mathematics examinations.

通过持续练习和对代数细节的细心关注，解二次方程将成为你 IGCSE 数学考试中最稳定的得分点之一。

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Solving Quadratic Equations | 解二次方程