Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Welcome to this focused revision guide on solving quadratic equations — one of the most frequently tested topics in IGCSE Mathematics. Whether you are preparing for Paper 2 or Paper 4, mastering quadratics is essential because they appear in algebra, geometry, graphing and problem-solving questions throughout the syllabus.

欢迎阅读本期专题复习指南。解一元二次方程是 IGCSE 数学中考查频率最高的知识点之一。无论你准备的是 Paper 2 还是 Paper 4，掌握二次方程都至关重要，因为它贯穿代数、几何、函数图像与应用题等整个考纲。

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

A quadratic equation is any equation that can be written in the general form shown below, where a, b and c are real numbers and a ≠ 0. The highest power of the variable x is 2, which is why it is called ‘quadratic’.

一元二次方程是指可以化为如下一般形式的方程，其中 a、b、c 为实数且 a ≠ 0。未知数 x 的最高次数为 2，因此称为”二次”方程。

ax² + bx + c = 0 (a ≠ 0)

For example, 3x² − 5x + 2 = 0 is a quadratic equation. However, x² + 4x − 1 = 5x + 3 is also quadratic, because after rearranging it becomes x² − x − 4 = 0. Always ensure all terms are on one side and the equation equals zero before you begin solving.

例如，3x² − 5x + 2 = 0 是一元二次方程。而 x² + 4x − 1 = 5x + 3 同样是二次方程，因为移项整理后可化为 x² − x − 4 = 0。在开始求解之前，务必把所有项移到等号一侧，使方程等于零。

The coefficient a is the quadratic coefficient, b is the linear coefficient, and c is the constant term.

系数 a 称为二次项系数，b 称为一次项系数，c 称为常数项。
If a = 0, the equation becomes linear, so the condition a ≠ 0 is essential.

若 a = 0，方程就退化为一次方程，因此 a ≠ 0 是必要条件。

2. Method 1: Solving by Factorisation | 方法一：因式分解法

Factorisation is often the quickest method when the equation has simple integer roots. We write the quadratic as a product of two linear brackets, then set each bracket equal to zero.

当方程具有简单的整数根时，因式分解通常是最快的方法。我们将二次式写成两个一次括号相乘的形式，然后令每个括号等于零。

Step-by-step example: Solve x² − 5x + 6 = 0.

分步示例：解方程 x² − 5x + 6 = 0。

Find two numbers that multiply to give 6 and add to give −5. Those numbers are −2 and −3.

寻找两个数，使其乘积为 6、和为 −5。这两个数是 −2 和 −3。
Write the factorised form: (x − 2)(x − 3) = 0.

写出因式分解形式：(x − 2)(x − 3) = 0。
Set each bracket to zero: x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.

令每个括号为零：x − 2 = 0 或 x − 3 = 0，解得 x = 2 或 x = 3。

Be alert for special cases. The difference of two squares, x² − 9 = 0, factorises as (x − 3)(x + 3) = 0. A perfect square, x² − 6x + 9 = 0, factorises as (x − 3)² = 0 and has one repeated root x = 3.

注意特殊情形。平方差 x² − 9 = 0 可分解为 (x − 3)(x + 3) = 0。完全平方式 x² − 6x + 9 = 0 可分解为 (x − 3)² = 0，此时有一个重根 x = 3。

If the coefficient of x² is not 1, for example 2x² + 5x + 3 = 0, look for a pair of brackets such as (2x + 3)(x + 1) = 0. Expanding carefully will help you verify your factorisation instantly.

如果 x² 的系数不为 1，例如 2x² + 5x + 3 = 0，需要寻找形如 (2x + 3)(x + 1) = 0 的括号组合。通过展开可以立即验证你的因式分解是否正确。

3. Method 2: Completing the Square | 方法二：配方法

Completing the square rewrites the quadratic in the form a(x + p)² + q. This method is especially useful for solving equations that do not factorise and for finding the turning point of a quadratic graph.

配方法将二次式改写为 a(x + p)² + q 的形式。这种方法特别适用于无法因式分解的方程，以及求二次函数图像的顶点坐标。

Worked example: Solve x² + 6x + 4 = 0 by completing the square.

示例：用配方法解方程 x² + 6x + 4 = 0。

Halve the coefficient of x, which is 6, to get 3. Write (x + 3)².

将 x 的系数 6 取半得 3，写出 (x + 3)²。
Expand (x + 3)² = x² + 6x + 9, which is 5 more than the original expression. Hence subtract 9.

展开 (x + 3)² = x² + 6x + 9，比原表达式多出 5，因此要减去 9。
The equation becomes (x + 3)² − 9 + 4 = 0, so (x + 3)² − 5 = 0.

方程化为 (x + 3)² − 9 + 4 = 0，即 (x + 3)² − 5 = 0。
Then (x + 3)² = 5. Take square roots: x + 3 = ±√5, so x = −3 ± √5.

于是 (x + 3)² = 5。两边开平方：x + 3 = ±√5，因此 x = −3 ± √5。

x = −3 + √5 or x = −3 − √5

In general, for a quadratic of the form x² + bx + c = 0, completing the square gives the expression shown below. You should memorise the pattern because it also connects directly to the quadratic formula.

一般地，对于 x² + bx + c = 0 形式的方程，配方法给出如下结果。建议牢记这一模式，因为它与求根公式直接相关。

(x + b/2)² = (b/2)² − c

4. Method 3: The Quadratic Formula | 方法三：求根公式

The quadratic formula is the most powerful method because it works for every quadratic equation, whether the roots are rational, irrational or even non-real. You are given this formula in the IGCSE formula sheet, but you must know how to apply it correctly.

求根公式是最强大的方法，因为它适用于所有一元二次方程，无论根是有理数、无理数还是没有实数解。IGCSE 公式表中会给出这个公式，但你必须正确运用它。

x = (−b ± √(b² − 4ac)) / 2a

Worked example: Solve 2x² − 4x − 3 = 0, giving your answers correct to 3 significant figures.

示例：解方程 2x² − 4x − 3 = 0，答案精确到 3 位有效数字。

Identify a = 2, b = −4, c = −3.

确定 a = 2，b = −4，c = −3。
Substitute into the formula: x = (4 ± √(16 − 4 × 2 × (−3))) / (2 × 2).

代入公式：x = (4 ± √(16 − 4 × 2 × (−3))) / (2 × 2)。
Simplify: x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4.

化简：x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4。
Therefore x = (4 + √40) / 4 ≈ 2.58 or x = (4 − √40) / 4 ≈ −0.581.

因此 x = (4 + √40) / 4 ≈ 2.58 或 x = (4 − √40) / 4 ≈ −0.581。

When using a calculator, be careful to use brackets around the numerator and the denominator. A common mistake is to compute −b ± √(b² − 4ac) correctly but then forget to divide the entire numerator by 2a.

使用计算器时，注意分子和分母都要加括号。常见错误是正确算出了 −b ± √(b² − 4ac)，但忘记将整个分子除以 2a。

5. The Discriminant | 判别式

The expression inside the square root, b² − 4ac, is called the discriminant and is usually denoted by Δ. It tells us the number and type of roots without solving the equation fully.

根号内的表达式 b² − 4ac 称为判别式，通常记作 Δ。它可以在不完全解方程的情况下告诉我们根的个数与类型。

Δ = b² − 4ac

Value of Δ	Type of Roots	Graph Meaning
Δ > 0	Two distinct real roots	Graph crosses the x-axis at two points
Δ = 0	One repeated real root	Graph touches the x-axis at one point
Δ < 0	No real roots (two complex roots)	Graph does not intersect the x-axis

For example, 2x² − 4x − 3 = 0 has Δ = (−4)² − 4 × 2 × (−3) = 40 > 0, so it has two distinct real roots. In contrast, x² + 2x + 5 = 0 has Δ = 4 − 20 = −16 < 0, meaning no real roots exist.

例如，2x² − 4x − 3 = 0 的判别式 Δ = (−4)² − 4 × 2 × (−3) = 40 > 0，因此有两个不相等的实数根。而 x² + 2x + 5 = 0 的判别式 Δ = 4 − 20 = −16 < 0，说明没有实数根。

You may also be asked a reverse question: ‘Find the value of k such that the equation kx² + 4x + 1 = 0 has a repeated root.’ Set Δ = 0 and solve for k: 16 − 4k = 0, so k = 4.

还可能遇到逆向问题：”求 k 的值，使方程 kx² + 4x + 1 = 0 有重根。”令 Δ = 0 并解出 k：16 − 4k = 0，因此 k = 4。

6. Solving Quadratic Word Problems | 二次方程应用题

In IGCSE examinations, quadratics are often embedded in real-world contexts such as geometry, area, projectile motion or number problems. The key is to translate the words into an equation and then solve it.

在 IGCSE 考试中，二次方程常被嵌入实际情境中，如几何、面积、抛体运动或数字问题。关键是将文字转化为方程并求解。

Example: A rectangle has length (x + 3) cm and width x cm. Its area is 40 cm². Find the value of x.

示例：一个长方形的长为 (x + 3) cm，宽为 x cm，面积为 40 cm²。求 x 的值。

Write the area equation: x(x + 3) = 40, which expands to x² + 3x = 40.

列出面积方程：x(x + 3) = 40，展开得 x² + 3x = 40。
Rearrange to zero: x² + 3x − 40 = 0.

移项化为零：x² + 3x − 40 = 0。
Factorise: (x + 8)(x − 5) = 0, so x = −8 or x = 5.

因式分解：(x + 8)(x − 5) = 0，所以 x = −8 或 x = 5。
Since a length cannot be negative, reject x = −8. The width is 5 cm and the length is 8 cm.

由于长度不能为负数，舍去 x = −8。宽为 5 cm，长为 8 cm。

Always check whether your solutions make sense in the original context. If a question asks for a distance, time or number of items, a negative answer is usually invalid and must be discarded with a clear line of reasoning.

始终检查解在原始情境中是否合理。如果题目求的是距离、时间或物品数量，负解通常无效，必须写明理由后舍去。

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

The graph of a quadratic function is a parabola. The x-coordinates of the points where it crosses the x-axis are exactly the roots of the equation ax² + bx + c = 0. Therefore, solving quadratics is intimately linked to graphing.

二次函数的图像是一条抛物线。图像与 x 轴交点的横坐标正是方程 ax² + bx + c = 0 的根。因此，解二次方程与画图密不可分。

If Δ > 0, the parabola crosses the x-axis twice. If Δ = 0, it touches the axis at exactly one point. If Δ < 0, it never touches the x-axis at all.

若 Δ > 0，抛物线与 x 轴有两个交点；若 Δ = 0，它与 x 轴相切于一点；若 Δ < 0，它与 x 轴无交点。

The axis of symmetry and the turning point can be found using the completed square form. For y = a(x + p)² + q, the turning point is (−p, q) and the axis is x = −p. Alternatively, the axis of symmetry is x = −b/(2a).

抛物线的对称轴和顶点可通过配方法形式求得。对于 y = a(x + p)² + q，顶点为 (−p, q)，对称轴为 x = −p。另一种方法是对称轴为 x = −b/(2a)。

Axis of symmetry: x = −b / (2a)

For example, y = x² + 6x + 4 = (x + 3)² − 5 has its vertex at (−3, −5). Since the coefficient of x² is positive, the parabola opens upward and the vertex is a minimum point.

例如，y = x² + 6x + 4 = (x + 3)² − 5 的顶点为 (−3, −5)。由于 x² 的系数为正，抛物线开口向上，顶点为最小值点。

8. Common Mistakes and How to Avoid Them | 常见错误与规避

After years of marking IGCSE papers, examiners consistently report the same pitfalls. Knowing these errors in advance can earn you easy marks in the examination.

经过多年阅卷统计，考官反复报告相同的易错点。提前了解这些错误，能帮你在考试中轻松拿分。

Forgetting to rearrange to zero: You must solve ax² + bx + c = 0. Factorising x² + 3x = 40 directly gives incorrect results. Always subtract 40 first.

忘记移项归零：必须求解 ax² + bx + c = 0。直接对 x² + 3x = 40 因式分解会出错，务必先移项。
Sign errors with negative coefficients: In 2x² − 4x − 3 = 0, note that b = −4 and c = −3. When substituting into the formula, write −(−4) as +4 carefully.

负系数符号错误：在 2x² − 4x − 3 = 0 中，b = −4，c = −3。代入公式时，注意 −(−4) 要写成 +4。
Dropping ± when taking square roots: If (x + 3)² = 5, then x + 3 = ±√5. Write both solutions, not just the positive one.

开平方漏写 ±：若 (x + 3)² = 5，则 x + 3 = ±√5。必须写出两个解，不能只写正的。
Premature rounding: Do not round intermediate values when using the quadratic formula. Keep √40 as √40 until the final step, then round your final answer to the required accuracy.

过早四舍五入：使用求根公式时，不要在中间步骤四舍五入。保留 √40 的形式到最后一步，再按要求的精度取近似值。

9. Exam Tips and Quick Summary | 考试技巧与快速总结

In the exam, always read the question carefully to decide which method is most efficient. If the question says ‘show your working’, factorisation or completing the square is often preferable because the quadratic formula may look like a black-box computation.

考试时，仔细审题并选择最有效的方法。如果题目要求”写出过程”，因式分解或配方法往往更合适，因为求根公式看起来像直接套用计算器。

Check your answers by substitution: Plug each solution back into the original equation. A quick mental check can catch sign errors.

代入检验：将每个解代回原方程，快速心算可以揪出符号错误。
Use your calculator wisely: The IGCSE calculator can solve quadratics directly, but marks are awarded for written working. Use the calculator to verify, not to replace, your algebra.

善用计算器：IGCSE 计算器可以直接解二次方程，但过程分是写在卷面上的。用计算器验证而不是替代你的代数步骤。
Match the required form: If the question asks for answers in exact form, leave them as fractions or surds such as x = −3 ± √5. If it asks for 2 decimal places, only then round.

匹配答案形式：如果题目要求精确值，保留分数或根式，如 x = −3 ± √5；如果要求保留 2 位小数，才进行近似。
Memorise the three key tools: factorisation for nice integer roots, completing the square for the turning point, and the quadratic formula as the universal fallback.

牢记三大工具：整数根用因式分解，求顶点用配方法，求根公式作为万能备用方案。

To finish, here is a quick flow chart in words: write the equation in standard form, check the discriminant Δ, then choose your method. With regular practice, solving quadratic equations will become one of your most reliable strengths in the IGCSE Mathematics examination.

最后，用一句话总结思路：先将方程化为标准形式，再看判别式 Δ，然后选择方法。通过持续练习，解一元二次方程将成为你在 IGCSE 数学考试中最稳定的得分点之一。

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Solving Quadratic Equations | 解一元二次方程