Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most frequently tested topics in IGCSE Mathematics. Whether you are studying the Extended or Core syllabus, understanding how to solve equations of the form ax² + bx + c = 0 is essential for success in Paper 2 and Paper 4. This revision guide covers every standard method — factorisation, completing the square, the quadratic formula, and graphical interpretation — with clear worked examples and exam-style advice.

二次方程是 IGCSE 数学中考查频率最高的知识点之一。无论你学习的是 Extended（扩展）还是 Core（核心）大纲，掌握如何解形如 ax² + bx + c = 0 的方程，都是 Paper 2 和 Paper 4 取得高分的关键。本复习指南涵盖所有标准方法——因式分解法、配方法、二次求根公式以及图像解法——并配有清晰的例题和考试技巧。

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

A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the unknown variable is 2. The general form is written as ax² + bx + c = 0, where a, b and c are real numbers, and a must not be zero. If a = 0, the equation becomes linear rather than quadratic.

二次方程是次数为 2 的多项式方程，即未知变量的最高次幂是 2。其一般形式写作 ax² + bx + c = 0，其中 a、b、c 为实数，且 a 不能为零。如果 a = 0，方程就变成一次（线性）方程，而不是二次方程。

The values of x that satisfy the equation are called roots or solutions. A quadratic equation always has exactly two roots in the complex number system, but in real numbers, it may have two distinct real roots, one repeated real root, or no real roots at all. This depends on the value of the discriminant, which we will explore later.

满足方程的 x 值称为根或解。在复数范围内，二次方程始终正好有两个根；但在实数范围内，它可能有两个不同实根、一个重根，或者没有实根。这取决于判别式的值，我们将在后文详细探讨。

2. Standard Form and Key Terminology | 标准形式与关键术语

Before solving, every quadratic equation should be rearranged into its standard form: ax² + bx + c = 0. The term ax² is called the quadratic term, bx is the linear term, and c is the constant term. The coefficient a determines the width and direction of the parabola, while b and c affect its position.

在求解之前，每个二次方程都应化为标准形式：ax² + bx + c = 0。其中 ax² 称为二次项，bx 称为一次项，c 称为常数项。系数 a 决定抛物线的开口方向和宽窄，而 b 和 c 影响抛物线的位置。

For example, in the equation 2x² – 5x + 3 = 0, we identify a = 2, b = -5 and c = 3. Notice that b is negative, and we must carry the sign into all calculations. Many student errors arise simply from misreading negative coefficients, so always write down the values of a, b and c explicitly before applying any method.

例如，在方程 2x² – 5x + 3 = 0 中，我们识别出 a = 2，b = -5，c = 3。注意 b 是负数，计算时必须带上符号。许多学生犯错仅仅是因为误读了负系数，所以在运用任何方法之前，务必明确写出 a、b、c 的值。

3. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic expression factorises neatly into linear factors. The principle is based on the zero product property: if the product of two expressions equals zero, then at least one of them must be zero. So if we can write ax² + bx + c = 0 as (px + q)(rx + s) = 0, then the solutions are found by setting each bracket equal to zero.

当二次表达式能干净利落地分解为线性因式时，因式分解法是最快捷的方法。其原理基于零积性质：如果两个表达式的乘积为零，则至少有一个必须为零。因此，若能将 ax² + bx + c = 0 写成 (px + q)(rx + s) = 0，则只需令每个括号等于零即可求出解。

Worked Example 1: Solve x² + 5x + 6 = 0.

例题 1：解方程 x² + 5x + 6 = 0。

We look for two numbers that multiply to give 6 and add to give 5. The numbers 2 and 3 satisfy both conditions, so x² + 5x + 6 = (x + 2)(x + 3). Setting each factor to zero gives x + 2 = 0 or x + 3 = 0, hence x = -2 or x = -3.

我们寻找两个数，它们相乘得 6，相加得 5。数字 2 和 3 同时满足这两个条件，所以 x² + 5x + 6 = (x + 2)(x + 3)。令每个因式为 0，得到 x + 2 = 0 或 x + 3 = 0，故 x = -2 或 x = -3。

Worked Example 2: Solve 2x² – 7x + 3 = 0.

例题 2：解方程 2x² – 7x + 3 = 0。

Here the leading coefficient is not 1. We multiply a and c: 2 × 3 = 6. We then find two numbers that multiply to 6 and add to -7, which are -1 and -6. Split the middle term: 2x² – x – 6x + 3 = 0. Now factor by grouping: x(2x – 1) – 3(2x – 1) = 0, giving (2x – 1)(x – 3) = 0. Thus x = ½ or x = 3.

这里首项系数不是 1。我们将 a 和 c 相乘：2 × 3 = 6。然后寻找两个数相乘得 6、相加得 -7，它们是 -1 和 -6。拆分中间项：2x² – x – 6x + 3 = 0。接着分组分解：x(2x – 1) – 3(2x – 1) = 0，得到 (2x – 1)(x – 3) = 0。因此 x = ½ 或 x = 3。

If AB = 0, then A = 0 or B = 0.

Always check your factorised form by expanding the brackets — this takes only a few seconds and catches sign errors immediately.

务必通过展开括号来检查你的因式结果——这只需几秒钟，却能立刻发现符号错误。

4. Completing the Square | 配方法

Completing the square rewrites a quadratic expression in the form a(x + p)² + q. This method is particularly useful when the quadratic does not factorise easily, and it also helps us find the turning point (vertex) of the parabola and solve equations more elegantly.

配方法将二次表达式改写成 a(x + p)² + q 的形式。当二次式不易因式分解时，这种方法尤为有用，它还能帮助我们找到抛物线的顶点（转向点），并更优雅地求解方程。

Worked Example 3: Solve x² + 6x – 7 = 0 by completing the square.

例题 3：用配方法解方程 x² + 6x – 7 = 0。

Take half of the coefficient of x, which is 6 ÷ 2 = 3. Write (x + 3)² and subtract 3² = 9 to keep the expression equivalent: x² + 6x = (x + 3)² – 9. Substituting back: (x + 3)² – 9 – 7 = 0, so (x + 3)² – 16 = 0. Then (x + 3)² = 16, giving x + 3 = ±4. Hence x = 1 or x = -7.

取 x 系数的一半，即 6 ÷ 2 = 3。写出 (x + 3)²，再减去 3² = 9 以保持等值：x² + 6x = (x + 3)² – 9。代入原方程：(x + 3)² – 9 – 7 = 0，即 (x + 3)² – 16 = 0。于是 (x + 3)² = 16，得到 x + 3 = ±4。因此 x = 1 或 x = -7。

The general procedure for a quadratic of the form x² + bx + c = 0 is:

对于形如 x² + bx + c = 0 的二次方程，一般步骤如下：

Identify the coefficient of x, which is b. | 找出 x 的系数 b。
Write (x + b/2)² – (b/2)² + c = 0. | 写出 (x + b/2)² – (b/2)² + c = 0。
Simplify the constant terms and solve. | 化简常数项并求解。

When the coefficient of x² is not 1, factor out a first: for example, 2x² + 8x + 5 = 0 becomes 2[x² + 4x] + 5 = 0, then complete the square inside the brackets.

当 x² 的系数不为 1 时，先把 a 提出来：例如，2x² + 8x + 5 = 0 变为 2[x² + 4x] + 5 = 0，然后在括号内配方。

5. The Quadratic Formula | 二次求根公式

The quadratic formula is the most universal method for solving any quadratic equation. It is derived from completing the square on the general form ax² + bx + c = 0, and it always works, even when factorisation is impossible. The formula is stated below and is provided on the IGCSE formula sheet, but you must know how and when to apply it.

二次求根公式是解任意二次方程的通用方法。它通过对一般形式 ax² + bx + c = 0 配方推导而来，无论能否因式分解都始终有效。公式如下所示，IGCSE 公式表会提供它，但你必须知道如何以及何时使用它。

x = (-b ± √(b² – 4ac)) / 2a

Worked Example 4: Solve 2x² + 3x – 5 = 0 using the quadratic formula.

例题 4：用二次求根公式解方程 2x² + 3x – 5 = 0。

Identify a = 2, b = 3, c = -5. Substitute into the formula:

识别 a = 2，b = 3，c = -5。代入公式：

x = (-3 ± √(3² – 4 × 2 × (-5))) / (2 × 2)

Simplify step by step: the discriminant is 9 – 4 × 2 × (-5) = 9 + 40 = 49. Therefore x = (-3 ± √49) / 4 = (-3 ± 7) / 4. This gives x = (-3 + 7) / 4 = 1, or x = (-3 – 7) / 4 = -2.5.

逐步化简：判别式为 9 – 4 × 2 × (-5) = 9 + 40 = 49。因此 x = (-3 ± √49) / 4 = (-3 ± 7) / 4。得到 x = (-3 + 7) / 4 = 1，或 x = (-3 – 7) / 4 = -2.5。

When using the formula, calculate the discriminant first. If it is a perfect square, the roots are rational; if not, you may need to leave your answer in exact surd form or round to a specified number of decimal places.

使用公式时，先计算判别式。如果它是一个完全平方数，根为有理数；如果不是，你可能需将答案保留为精确根式形式，或四舍五入到指定的小数位数。

6. The Graphical Method | 图像法

The graph of a quadratic function y = ax² + bx + c is a smooth U-shaped curve called a parabola. When a is positive, the parabola opens upwards; when a is negative, it opens downwards. The solutions to the equation ax² + bx + c = 0 correspond to the x-coordinates of the points where the parabola crosses the x-axis, that is, where y = 0.

二次函数 y = ax² + bx + c 的图像是一条平滑的 U 形曲线，称为抛物线。当 a 为正时，抛物线开口向上；当 a 为负时，开口向下。方程 ax² + bx + c = 0 的解对应抛物线与 x 轴交点的 x 坐标，即 y = 0 的位置。

Worked Example 5: Sketch y = x² – 2x – 3 and state the roots.

例题 5：画出 y = x² – 2x – 3 的草图并写出根。

Factorising gives y = (x – 3)(x + 1), so the x-intercepts are at x = 3 and x = -1. The y-intercept is found by substituting x = 0, giving y = -3. The line of symmetry lies halfway between the roots, at x = (3 + (-1)) / 2 = 1. Substituting x = 1 gives the turning point: y = 1² – 2(1) – 3 = -4. The vertex is therefore (1, -4).

因式分解得 y = (x – 3)(x + 1)，所以 x 轴截距在 x = 3 和 x = -1 处。y 轴截距通过代入 x = 0 得到 y = -3。对称轴位于两根的中点，即 x = (3 + (-1)) / 2 = 1。代入 x = 1 得转向点：y = 1² – 2(1) – 3 = -4。因此顶点为 (1, -4)。

In non-calculator examinations, you may be asked to read approximate roots from a given graph, or to add a straight line to the graph to solve a related equation. For example, solving x² – 2x – 3 = 2 is equivalent to finding where y = x² – 2x – 3 intersects the horizontal line y = 2.

在不能用计算器的考试中，你可能会被要求从给定图像中读取近似根，或者画一条直线来解决相关方程。例如，解 x² – 2x – 3 = 2 等价于找 y = x² – 2x – 3 与水平线 y = 2 的交点。

7. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root sign, given the symbol Δ. It tells us how many real roots a quadratic equation has without solving it fully.

判别式是二次求根公式中根号内的部分，用符号 Δ 表示。它无需完整求解就能告诉我们二次方程有多少个实根。

Δ = b² – 4ac

The three cases are summarised in the table below:

三种情况总结如下表：

Discriminant Δ	Number of Real Roots	Graph Interpretation
Δ > 0	Two distinct real roots	Parabola crosses the x-axis at two points
Δ = 0	One repeated real root	Parabola touches the x-axis at the vertex
Δ < 0	No real roots	Parabola does not intersect the x-axis

A common exam question asks you to find the range of k for which a given quadratic has two distinct real roots. For example, for x² + kx + 4 = 0 to have two distinct roots, we require Δ > 0, so k² – 16 > 0, meaning k > 4 or k < -4.

一个常见考题是要求找出使给定二次方程有两个不同实根时 k 的取值范围。例如，要使 x² + kx + 4 = 0 有两个不同实根，需满足 Δ > 0，即 k² – 16 > 0，意味着 k > 4 或 k < -4。

8. Solving Quadratic Inequalities | 二次不等式

Quadratic inequalities extend the solving techniques covered above. The key steps are to rearrange the inequality to one side, factorise, find the critical values (roots), and then determine which intervals satisfy the inequality by sketching a quick parabola or testing points.

二次不等式扩展了上述求解技巧。关键步骤是：将不等式整理到一边、因式分解、找出临界值（根），然后通过画草图或测试点来判断哪些区间满足不等式。

Worked Example 6: Solve x² – x – 12 < 0.

例题 6：解不等式 x² – x – 12 < 0。

Factorise: x² – x – 12 = (x – 4)(x + 3) < 0. The critical values are x = 4 and x = -3. Since the coefficient of x² is positive, the parabola opens upwards. The expression is negative between the roots, so the solution is -3 < x < 4.

因式分解：x² – x – 12 = (x – 4)(x + 3) < 0。临界值为 x = 4 和 x = -3。由于 x² 的系数为正，抛物线开口向上。表达式在两根之间为负，因此解为 -3 < x < 4。

For an inequality like x² – x – 12 > 0, the solution would instead be the two separate intervals x < -3 or x > 4. Remember that strict inequalities use < or >, while non-strict ones use ≤ or ≥, and the critical values are included only in the latter case.

对于像 x² – x – 12 > 0 这样的不等式，解则是两个单独区间 x < -3 或 x > 4。记住严格不等式使用 < 或 >，而非严格不等式使用 ≤ 或 ≥，只有后者才包含临界值。

9. Applications and Word Problems | 实际应用与文字题

Quadratic equations appear frequently in real-world contexts, including geometry, projectile motion, and economics. In IGCSE exams, word problems often require you to translate a written situation into a quadratic equation, solve it, and then interpret the roots in context.

二次方程在现实场景中频繁出现，包括几何、抛体运动和经济领域。在 IGCSE 考试中，文字题通常要求你将文字情境转化为二次方程、求解，然后在具体情境中解释根的含义。

Worked Example 7: A rectangle has length (x + 5) cm and width (x – 2) cm. Its area is 60 cm². Find the dimensions of the rectangle.

例题 7：一个矩形的长为 (x + 5) cm，宽为 (x – 2) cm，面积为 60 cm²。求该矩形的尺寸。

Write the area equation: (x + 5)(x – 2) = 60. Expanding gives x² + 3x – 10 = 60, which rearranges to x² + 3x – 70 = 0. Factorising: (x + 10)(x – 7) = 0, so x = -10 or x = 7. Since a length cannot be negative, we discard x = -10. Therefore the length is 7 + 5 = 12 cm and the width is 7 – 2 = 5 cm.

写出面积方程：(x + 5)(x – 2) = 60。展开得 x² + 3x – 10 = 60，整理为 x² + 3x – 70 = 0。因式分解：(x + 10)(x – 7) = 0，所以 x = -10 或 x = 7。由于长度不能为负，我们舍去 x = -10。因此长为 7 + 5 = 12 cm，宽为 7 – 2 = 5 cm。

A key skill in word problems is checking which root is physically meaningful. Always relate your answers back to the question: a negative length, time, or distance should alert you to discard that solution. Also, always include the correct units in your final answer.

文字题的关键技能是判断哪个根在物理意义上合理。始终将答案与题目对照：负数长度、时间或距离应提醒你舍去该解。同时，最终答案务必包含正确的单位。

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

Even high-achieving students lose marks on quadratic equations due to avoidable errors. Below are the most common pitfalls and how to avoid them.

即使是成绩优秀的学生也会因可避免的错误在二次方程上丢分。以下是最常见的陷阱及规避方法。

Forgetting to set the equation to zero before factorising — always rearrange to ax² + bx + c = 0 first. | 忘记在因式分解前将方程化为零——务必先整理成 ax² + bx + c = 0。
Incorrect signs when substituting into the quadratic formula — write b = -5 not b = 5 when the coefficient is negative. | 将数值代入二次公式时符号错误——当系数为负时，写 b = -5 而不是 b = 5。
Dividing both sides by x to simplify — this loses the root x = 0 and is forbidden in examinations. | 两边同除以 x 来化简——这会丢失根 x = 0，在考试中是禁止的。
Discarding valid negative roots without justification in word problems. | 在文字题中无正当理由就舍弃负数根。
Not checking solutions by substitution. | 不通过代入检验解的正确性。

For calculator papers, you can use your calculator to verify roots quickly, but always show full working — a correct answer with no method earns very few marks. For non-calculator papers, practise arithmetic with fractions and surds, especially when using the quadratic formula.

在可用计算器的试卷中，你可以用计算器快速验证根，但务必展示完整步骤——只有正确答案而没有方法几乎得不到分。在不能用计算器的试卷中，要练习分数和根式的运算，尤其是在使用二次公式时。

11. Quick Revision Checklist | 快速复习清单

Use this checklist before your exam to confirm you have mastered every aspect of quadratic equations.

考试前使用此清单，确认你已掌握二次方程的所有要点。

Skill	Confidence
Rearranging to standard form ax² + bx + c = 0	☐
Factorising quadratics with a = 1 and a ≠ 1	☐
Completing the square and finding the vertex	☐
Applying the quadratic formula accurately	☐
Using the discriminant Δ = b² – 4ac	☐
Solving quadratic inequalities	☐
Interpreting graphs of quadratic functions	☐
Setting up and solving word problems	☐

If any item feels uncertain, revisit the relevant section of this guide and attempt at least ten practice questions on that skill. Consistent practice is the single most reliable predictor of success in IGCSE Mathematics.

如果任何一项让你感到不确定，请重新阅读本指南的相关章节，并针对该技能至少做十道练习题。持续练习是 IGCSE 数学取得成功最可靠的预测指标。

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Solving Quadratic Equations | 解二次方程