Solving Quadratic Equations | 一元二次方程

📚 Solving Quadratic Equations | 一元二次方程

Quadratic equations appear in almost every IGCSE Mathematics paper, from short multiple-choice questions to extended problem-solving tasks. To score well, you must be confident with factorisation, the quadratic formula, completing the square, and the interpretation of quadratic graphs. This article gives you all the tools you need.

一元二次方程几乎出现在每份 IGCSE 数学试卷中，无论是简短的选择题还是综合应用题。想拿高分，你必须熟练掌握因式分解、求根公式、配方法以及二次函数图像的分析。本文为你梳理所有必备工具。

1. What Is a Quadratic Equation? | 什么是一元二次方程？

A quadratic equation is an equation where the highest power of the variable is 2. The general form is written as:

ax² + bx + c = 0 (a ≠ 0)

Here a is the coefficient of x², b is the coefficient of x, and c is the constant term. For example, 2x² − 3x + 1 = 0 is a quadratic equation, while 3x + 2 = 0 is linear because it contains no x² term. In IGCSE questions, the equation may not look like this at first; you may need to expand brackets or multiply both sides to reveal the standard form.

一元二次方程是指变量的最高次数为 2 的方程，一般形式为：

ax² + bx + c = 0 (a ≠ 0)

其中 a 称为二次项系数，b 称为一次项系数，c 是常数项。例如 2x² − 3x + 1 = 0 是一元二次方程，而 3x + 2 = 0 是一元一次方程，因为没有 x² 项。在 IGCSE 题目中，方程未必一开始就是标准形式，可能需要先展开括号或去分母。

2. Expressing Equations in Standard Form | 化为标准形式

Many examination questions give you an equation that must be rearranged before you can solve it. For example, consider: (x + 1)² = 3x + 7. Expanding the left side gives x² + 2x + 1 = 3x + 7. Subtracting 3x and 7 from both sides gives x² − x − 6 = 0, which is now in the standard form ax² + bx + c = 0.

许多考题给出的方程需要先整理才能求解。例如：

(x + 1)² = 3x + 7

展开左边得 x² + 2x + 1 = 3x + 7，两边同时减去 3x 和 7，得到 x² − x − 6 = 0，这就是标准形式 ax² + bx + c = 0。

Another common situation is an equation involving fractions, such as 2/x + x/3 = 1 with x ≠ 0. Multiplying through by the lowest common denominator, 3x, gives 6 + x² = 3x, which rearranges to x² − 3x + 6 = 0. Always write down the rearrangement steps clearly; method marks are awarded even if your final roots are wrong.

另一种常见情况是含分母的方程，例如 2/x + x/3 = 1（其中 x ≠ 0）。两边同时乘以最简公分母 3x，得到 6 + x² = 3x，整理得 x² − 3x + 6 = 0。务必清晰写出整理步骤，因为即使最后答案有误，过程也能得到步骤分。

3. Solving by Factorisation | 因式分解法求解

Factorisation is often the quickest method for IGCSE quadratics. The principle is simple: if the product of two expressions is zero, then at least one of them must be zero. To solve x² − x − 6 = 0, you look for two numbers that multiply to −6 and add to −1. Those numbers are −3 and 2, so the equation factorises as:

(x − 3)(x + 2) = 0

Therefore x − 3 = 0 or x + 2 = 0, giving the two roots x = 3 and x = −2. Always check your factors by expanding: (x − 3)(x + 2) = x² − x − 6, so the factorisation is correct.

因式分解法是 IGCSE 解二次方程最快的方法之一。核心原理是：如果两个表达式的乘积为零，那么其中至少一个必须为零。要解 x² − x − 6 = 0，需要找到两个数，它们相乘等于 −6，相加等于 −1。这两个数是 −3 和 2，因此方程分解为：

(x − 3)(x + 2) = 0

所以 x − 3 = 0 或 x + 2 = 0，得到两个根 x = 3 和 x = −2。一定要通过展开检验因式分解是否正确：(x − 3)(x + 2) = x² − x − 6，说明分解正确。

4. Difference of Two Squares and Special Patterns | 平方差公式与特殊模式

When the quadratic has no x term, it may be a difference of two squares. The rule is: a² − b² = (a + b)(a − b). For example, x² − 25 = 0 can be written as (x + 5)(x − 5) = 0, so x = 5 or x = −5. Similarly, 9x² − 49 = 0 becomes (3x + 7)(3x − 7) = 0, giving x = 7/3 or x = −7/3.

当二次方程没有一次项时，往往是平方差形式。规则是：a² − b² = (a + b)(a − b)。例如 x² − 25 = 0 可写成 (x + 5)(x − 5) = 0，因此 x = 5 或 x = −5。同理，9x² − 49 = 0 分解为 (3x + 7)(3x − 7) = 0，得到 x = 7/3 或 x = −7/3。

Another special pattern is the perfect square: x² + 2ab + b² = (a + b)². For instance, x² + 6x + 9 = 0 is exactly (x + 3)² = 0, which has one repeated root, x = −3. Recognising these patterns saves valuable time in the exam.

另一个特殊模式是完全平方：x² + 2ab + b² = (a + b)²。例如 x² + 6x + 9 = 0 恰好是 (x + 3)² = 0，只有一个重根 x = −3。在考试中识别这些模式可以节省宝贵时间。

5. The Quadratic Formula | 求根公式

If a quadratic cannot be factorised easily, use the quadratic formula. For any quadratic in the form ax² + bx + c = 0, the solutions are:

x = (−b ± √(b² − 4ac)) / (2a)

For example, solve 2x² + 3x − 5 = 0. Here a = 2, b = 3, c = −5. Substitute these values:

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

Thus x = (−3 + 7)/4 = 1, or x = (−3 − 7)/4 = −2.5. The formula must be memorised exactly; copy it into your working carefully because one sign error can destroy the whole solution.

当二次方程难以因式分解时，可以使用求根公式。对于任何形如 ax² + bx + c = 0 的方程，解为：

x = (−b ± √(b² − 4ac)) / (2a)

例如解 2x² + 3x − 5 = 0，这里 a = 2，b = 3，c = −5。代入得：

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

所以 x = (−3 + 7)/4 = 1，或 x = (−3 − 7)/4 = −2.5。公式必须准确记忆，书写计算过程时要格外小心，一个符号错误就会毁掉整个解答。

6. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root: Δ = b² − 4ac. It tells you how many real roots a quadratic equation has, without actually solving it. If Δ > 0, there are two distinct real roots. If Δ = 0, there is exactly one repeated real root. If Δ < 0, there are no real roots.

判别式是求根公式中根号内的部分：Δ = b² − 4ac。它可以在不解方程的情况下判断一元二次方程实根的个数。若 Δ > 0，有两个不相等的实数根；若 Δ = 0，有一个二重实数根；若 Δ < 0，则没有实数根。

For instance, the equation x² + 4x + 5 = 0 has Δ = 16 − 20 = −4, which is negative, so the curve y = x² + 4x + 5 never crosses the x-axis. In contrast, the equation x² − 4x + 4 = 0 has Δ = 16 − 16 = 0, so its graph touches the x-axis at one point. Exam questions may ask you to “show that the equation has two real roots”, which is done by calculating the discriminant and stating its sign.

例如方程 x² + 4x + 5 = 0 的判别式 Δ = 16 − 20 = −4，为负数，因此曲线 y = x² + 4x + 5 与 x 轴无交点。而方程 x² − 4x + 4 = 0 的 Δ = 16 − 16 = 0，图像与 x 轴相切于一点。考试中常要求”证明方程有两个实数根”，只需计算判别式并说明其符号即可。

7. Completing the Square | 配方法

Completing the square rewrites a quadratic as a perfect square plus a constant. For an expression x² + bx, you add and subtract (b/2)². For example:

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

This form is extremely useful for finding the turning point of a parabola and for solving equations. When the coefficient of x² is not 1, factor it out first: 2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3.

配方法是将二次式改写为完全平方加常数的形式。对于形如 x² + bx 的表达式，需要加上再减去 (b/2)²。例如：

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

这种形式在求抛物线顶点以及解方程时非常有用。当 x² 的系数不是 1 时，先提取系数：2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3。

8. Solving by Completing the Square | 用配方法解方程

Completing the square also provides a systematic way to solve any quadratic. Take x² + 6x + 2 = 0. From the previous section, we know that (x + 3)² − 7 = 0. Adding 7 to both sides gives (x + 3)² = 7. Taking square roots gives x + 3 = ±√7, so x = −3 ± √7. These are the exact solutions, which is valuable when the answer has a surd.

配方法也为解任意二次方程提供了系统的途径。以 x² + 6x + 2 = 0 为例，由上一节可知 (x + 3)² − 7 = 0。两边加 7 得 (x + 3)² = 7，开平方得 x + 3 = ±√7，因此 x = −3 ± √7。这就是精确解，当答案含根号时特别有用。

If the coefficient of x² is not 1, such as 2x² + 8x + 5 = 0, first write 2(x + 2)² − 3 = 0, then divide by 2 and rearrange: (x + 2)² = 1.5, so x = −2 ± √1.5. Remember to give your answer in the form requested by the question: exact, to 3 significant figures, or to 2 decimal places.

如果 x² 的系数不是 1，例如 2x² + 8x + 5 = 0，先写成 2(x + 2)² − 3 = 0，再除以 2 并整理：(x + 2)² = 1.5，所以 x = −2 ± √1.5。注意题目要求的最终形式：精确值、3 位有效数字还是 2 位小数。

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

The graph of y = ax² + bx + c is a parabola. When a > 0, the parabola opens upwards and has a minimum point; when a < 0, it opens downwards and has a maximum point. The roots of the quadratic equation ax² + bx + c = 0 are exactly the x-coordinates where the graph crosses the x-axis.

函数 y = ax² + bx + c 的图像是一条抛物线。当 a > 0 时，抛物线开口向上，有最小值点；当 a < 0 时，开口向下，有最大值点。方程 ax² + bx + c = 0 的根，正是图像与 x 轴交点的横坐标。

The turning point lies on the axis of symmetry, which has equation x = −b/(2a). For y = 2(x + 2)² − 3, the vertex is at (−2, −3). To sketch a quadratic graph, you should find the y-intercept (set x = 0), the roots (set y = 0), and the turning point. A typical exam sketch needs all three features labelled clearly.

顶点位于对称轴上，对称轴方程为 x = −b/(2a)。对于 y = 2(x + 2)² − 3，顶点坐标为 (−2, −3)。画二次函数草图时，需要求出 y 轴截距（令 x = 0）、根（令 y = 0）和顶点。考试作图通常要求将这三点清晰地标在图上。

10. Word Problems and Exam Tips | 应用题与应试技巧

Quadratic equations are also used to model real-life situations. A typical problem: the length of a rectangle is 3 cm more than its width, and the area is 40 cm². Let the width be x; then length = x + 3, so x(x + 3) = 40. This expands to x² + 3x − 40 = 0, which factorises as (x + 8)(x − 5) = 0. Since a length cannot be negative, x = 5 and the length is 8 cm. Always reject impossible negative solutions in context questions.

二次方程也常用于建模实际生活问题。典型题目：一个长方形的长比宽多 3 cm，面积为 40 cm²。设宽为 x，则长为 x + 3，因此 x(x + 3) = 40。展开得 x² + 3x − 40 = 0，因式分解为 (x + 8)(x − 5) = 0。因为长度不能为负，所以 x = 5，长为 8 cm。在应用题中，务必舍去不符合实际的负数解。

Final tips for the exam: first rearrange every equation into the form ax² + bx + c = 0. Try factorisation first; if you cannot spot factors within a few seconds, switch to the quadratic formula or completing the square. Show all working and substitute carefully into the formula. Practise past-paper questions until each method becomes automatic, and always check your roots by substituting them back into the original equation.

考试最后建议：先把每个方程整理成 ax² + bx + c = 0 的形式。优先尝试因式分解；如果几秒内找不到因子，就改用求根公式或配方法。写出完整的解题过程，代入公式时格外仔细。多做历年真题，直到每种方法都驾轻就熟，最后把根代回原方程检验。

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Solving Quadratic Equations | 一元二次方程