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, whether in Paper 1 (non-calculator) or Paper 2 (calculator). They test your algebraic manipulation, factorisation skills, and ability to interpret results in real-world contexts. This article covers every method you need, from factorisation to the quadratic formula, completing the square, and the discriminant.

二次方程几乎出现在每一份 IGCSE 数学试卷中，无论是 Paper 1（不可用计算器）还是 Paper 2（可用计算器）。它考查代数变形、因式分解能力，以及在实际情境中解读结果的能力。本文将涵盖所有必备方法：因式分解法、公式法、配方法和判别式。

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

A quadratic equation is any equation that can be rearranged into the standard form:

ax² + bx + c = 0, where a ≠ 0

Here a, b and c are constants. The highest power of the unknown x is 2, which is why the equation is called “quadratic” (from the Latin word quadratus, meaning square). The condition a ≠ 0 is essential because if a = 0, the equation becomes linear, not quadratic.

其中 a、b、c 为常数。未知数 x 的最高次数是 2，因此称为二次方程（quadratic 源自拉丁语 quadratus，意为“平方”）。条件 a ≠ 0 至关重要，因为若 a = 0，方程就变成一次方程而非二次方程了。

For example, 2x² + 3x − 5 = 0 and x² − 4 = 0 are quadratic equations. However, x³ + x² − 1 = 0 is a cubic equation, not quadratic. You must always check the highest exponent before deciding which solving method to use.

例如，2x² + 3x − 5 = 0 和 x² − 4 = 0 都是二次方程。而 x³ + x² − 1 = 0 是三次方程，不是二次方程。在决定使用哪种解法之前，必须先确认最高指数。

2. Solving by Factorisation | 因式分解法

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

因式分解法在二次方程具有简单整数因子时是最快的解法。其原理是：若两个表达式的乘积为零，则其中至少有一个必为零。这称为“零积性质”。

To solve x² + 5x + 6 = 0, look for two numbers that multiply to 6 and add to 5. These numbers are 2 and 3 because 2 × 3 = 6 and 2 + 3 = 5. Therefore:

x² + 5x + 6 = (x + 2)(x + 3) = 0

So x + 2 = 0 or x + 3 = 0, giving x = −2 or x = −3. Always check by substituting both values back into the original equation.

于是 x + 2 = 0 或 x + 3 = 0，得到 x = −2 或 x = −3。务必把两个解代回原方程检验。

When the coefficient of x² is not 1, use the “multiply ac, then split the middle term” method. For 2x² + 7x + 3 = 0, multiply a × c = 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7: these are 6 and 1. Split 7x into 6x + 1x, then factor by grouping:

当 x² 的系数不为 1 时，使用“先乘 ac，再拆中间项”的方法。对于 2x² + 7x + 3 = 0，先算 a × c = 2 × 3 = 6。找两个数相乘为 6、相加为 7：即 6 和 1。将 7x 拆成 6x + 1x，再分组因式分解：

2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0

Thus x = −½ or x = −3.

因此 x = −½ 或 x = −3。

3. Special Cases: Difference of Two Squares and Common Factors | 特殊情况：平方差与公因式

The difference of two squares is a special factorisation pattern you must recognise instantly:

a² − b² = (a + b)(a − b)

For example, x² − 25 = 0 becomes (x + 5)(x − 5) = 0, so x = 5 or x = −5. This pattern also works with fractions and decimals, such as 4x² − 9 = 0 → (2x + 3)(2x − 3) = 0.

例如，x² − 25 = 0 可化为 (x + 5)(x − 5) = 0，所以 x = 5 或 x = −5。这个模式也适用于分数和小数，如 4x² − 9 = 0 → (2x + 3)(2x − 3) = 0。

Always look for a common factor first. In 3x² − 12x = 0, factor out 3x: 3x(x − 4) = 0, giving x = 0 or x = 4. Forgetting x = 0 is a very common IGCSE mistake.

始终先寻找公因式。在 3x² − 12x = 0 中，提出 3x：3x(x − 4) = 0，得到 x = 0 或 x = 4。漏掉 x = 0 是 IGCSE 考试中非常常见的错误。

4. The Quadratic Formula | 公式法

When factorisation is difficult or impossible, use the quadratic formula. This formula works for every quadratic equation, provided a ≠ 0.

当因式分解困难或不可行时，使用公式法。只要 a ≠ 0，该公式对所有二次方程都适用。

x = (−b ± √(b² − 4ac)) / (2a)

To use it, write the equation in the form ax² + bx + c = 0, identify a, b and c, then substitute. For 2x² + 3x − 2 = 0, we have a = 2, b = 3, c = −2. Substituting:

使用时，先把方程写成 ax² + bx + c = 0 的形式，确定 a、b、c，再代入。对于 2x² + 3x − 2 = 0，有 a = 2，b = 3，c = −2。代入得：

x = (−3 ± √(3² − 4 × 2 × (−2))) / (2 × 2) = (−3 ± √25) / 4

Since √25 = 5, we get x = (−3 + 5)/4 = ½ or x = (−3 − 5)/4 = −2.

因为 √25 = 5，得到 x = (−3 + 5)/4 = ½ 或 x = (−3 − 5)/4 = −2。

In the IGCSE exams, the quadratic formula is given on the formula sheet, but you still need to know how to substitute correctly and simplify surds such as √72 = 6√2.

在 IGCSE 考试中，公式表会给出二次公式，但你仍需掌握正确的代入方法以及化简根式，例如 √72 = 6√2。

5. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. This is useful for solving equations, finding turning points, and deriving the quadratic formula itself.

配方法将二次式改写为 (x + p)² + q 的形式。它可用于解方程、求顶点坐标，以及推导二次公式本身。

To complete the square for x² + 6x + 5 = 0, take half of 6 (which is 3), square it (giving 9), and write:

对于 x² + 6x + 5 = 0，取 6 的一半（即 3），平方得到 9，然后写成：

(x + 3)² − 9 + 5 = 0

Simplifying gives (x + 3)² = 4. Taking the square root of both sides:

化简得 (x + 3)² = 4。对两边开平方：

x + 3 = ±2, so x = −1 or x = −5

If the coefficient of x² is not 1, divide the whole equation by that coefficient first. For example, 2x² − 8x + 3 = 0 becomes x² − 4x + 1.5 = 0 before completing the square.

若 x² 的系数不为 1，先将整个方程除以该系数。例如，2x² − 8x + 3 = 0 需先化为 x² − 4x + 1.5 = 0 再进行配方。

6. The Discriminant | 判别式

The expression b² − 4ac inside the quadratic formula is called the discriminant, often denoted by Δ. It tells us how many real solutions a quadratic equation has, without actually solving it.

二次公式中 b² − 4ac 这一部分称为判别式，通常用 Δ 表示。它告诉我们在不解方程的情况下，方程有几个实数解。

Δ = b² − 4ac	Number of Real Solutions	Graph Interpretation
Δ > 0	Two distinct real roots	Curve crosses the x-axis twice
Δ = 0	One repeated root	Curve touches the x-axis once
Δ < 0	No real roots	Curve never meets the x-axis

For the equation 3x² − 6x + 2 = 0, Δ = (−6)² − 4 × 3 × 2 = 36 − 24 = 12. Since 12 > 0, the equation has two distinct real roots.

对于方程 3x² − 6x + 2 = 0，Δ = (−6)² − 4 × 3 × 2 = 36 − 24 = 12。因为 12 > 0，方程有两个不同的实数根。

Exam questions often say “find the range of values of k for which x² + kx + 9 = 0 has two distinct real roots.” You set b² − 4ac > 0, substitute b = k, and solve k² − 36 > 0, giving k < −6 or k > 6.

考试中常出现“求 k 的取值范围，使 x² + kx + 9 = 0 有两个不同的实数根”。你需令 b² − 4ac > 0，代入 b = k，解 k² − 36 > 0，得 k < −6 或 k > 6。

7. Solving Quadratic Equations by Graphs | 用图像解二次方程

The real solutions of ax² + bx + c = 0 are the x-coordinates of the points where the curve y = ax² + bx + c crosses the x-axis. Reading these points from a graph gives approximate solutions.

ax² + bx + c = 0 的实数解是曲线 y = ax² + bx + c 与 x 轴交点的 x 坐标。从图像上读出这些点即可得到近似解。

For example, if a curve crosses the x-axis at x = 1.4 and x = 3.6, then the equation has approximate solutions 1.4 and 3.6. If the curve sits entirely above the x-axis, there are no real solutions — the discriminant is negative.

例如，若曲线与 x 轴交于 x = 1.4 和 x = 3.6，则方程有近似解 1.4 和 3.6。若曲线完全位于 x 轴上方，则无实数解——判别式为负。

You may also be asked to solve a quadratic by drawing a straight line on the graph. To solve x² − 2x − 3 = 0 using the graph of y = x² − 2x − 5, add 2 to both sides so x² − 2x − 3 = 0 becomes x² − 2x − 5 = −2. Draw the line y = −2 and read the intersection points.

你也可能被要求通过在图像上画一条直线来解二次方程。若要用 y = x² − 2x − 5 的图像解 x² − 2x − 3 = 0，先在两边加 2，使 x² − 2x − 3 = 0 变为 x² − 2x − 5 = −2。画出直线 y = −2，读出交点即可。

8. Word Problems with Quadratic Equations | 二次方程应用题

IGCSE frequently tests quadratic equations through word problems. The most common type is the rectangle area problem. For instance: “A rectangle has length 5 cm more than its width and an area of 36 cm². Find its width.”

IGCSE 经常通过应用题考查二次方程，最常见的类型是矩形面积问题。例如：“一个矩形的长比宽多 5 cm，面积为 36 cm²。求它的宽。”

Let the width be x. Then the length is x + 5, and the area gives x(x + 5) = 36. Expanding:

设宽为 x，则长为 x + 5，面积方程为 x(x + 5) = 36。展开得：

x² + 5x = 36 → x² + 5x − 36 = 0 → (x + 9)(x − 4) = 0

So x = −9 or x = 4. Since a length cannot be negative, the width is 4 cm and the length is 9 cm. Always reject negative solutions in measurement contexts.

所以 x = −9 或 x = 4。由于长度不能为负，宽度为 4 cm，长度为 9 cm。在涉及测量的情境中，必须舍弃负数解。

Other common contexts include projectile motion (height = ut − ½gt²), number problems, and compound shapes where the total area leads to a quadratic equation.

其他常见情境包括抛体运动（高度 = ut − ½gt²）、数字问题，以及由复合图形总面积引出的二次方程。

9. Common Mistakes and Exam Tips | 常见错误与应试技巧

Based on examiner reports, these are the most frequent errors in quadratic equation questions on IGCSE papers.

根据考官报告，以下是 IGCSE 试卷中二次方程题最常见的错误。

Forgetting to set the equation to zero before factorising. Always rearrange to ax² + bx + c = 0 first. 忘记将方程整理为零再因式分解。必须先把方程化为 ax² + bx + c = 0。
Losing one solution when taking square roots. Remember x² = 9 has two solutions: x = 3 and x = −3. 开平方时丢失一个解。记住 x² = 9 有两个解：x = 3 和 x = −3。
Misidentifying a and c when b or c is negative. Write down a = 2, b = −3, c = −5 explicitly before substituting into the formula. 当 b 或 c 为负数时搞错 a 和 c。代入公式前明确写下 a = 2，b = −3，c = −5。
Arithmetic errors in the discriminant. Use brackets carefully: b² − 4ac means b² − (4ac), not (b² − 4a) × c. 判别式的运算错误。注意括号：b² − 4ac 表示 b² − (4ac)，而不是 (b² − 4a) × c。
Giving approximate answers when exact values are required. Leave answers as fractions or surds such as (1 + √5)/2. 在要求精确值时给出近似答案。应保留分数或根式，如 (1 + √5)/2。

10. Practice Questions | 练习巩固

Try these five questions on your own, then check your answers against the solutions below.

请独立完成以下五道练习，再对照下方的答案检验。

Q1. Solve x² − 7x + 12 = 0 by factorisation.

Q2. Solve 2x² + 5x − 3 = 0 using the quadratic formula.

Q3. Solve x² + 10x + 9 = 0 by completing the square.

Q4. Find the value of the discriminant of 4x² − 4x + 1 = 0 and state the number of real roots.

Q5. A right-angled triangle has legs x and x + 7, and hypotenuse 13. Find x.

Answers:

Q1: (x − 3)(x − 4) = 0 → x = 3 or x = 4.

Q2: x = (−5 ± √(25 + 24))/4 = (−5 ± 7)/4 → x = ½ or x = −3.

Q3: (x + 5)² − 25 + 9 = 0 → (x + 5)² = 16 → x + 5 = ±4 → x = −1 or x = −9.

Q4: Δ = (−4)² − 4 × 4 × 1 = 16 − 16 = 0 → one repeated real root.

Q5: x² + (x + 7)² = 13² → 2x² + 14x + 49 = 169 → x² + 7x − 60 = 0 → (x + 12)(x − 5) = 0 → x = 5 (reject x = −12).

11. Summary | 总结

There are four key methods for solving quadratic equations: factorisation (fastest when possible), the quadratic formula (always works), completing the square (also gives the turning point), and graphical solution (approximate but visual). The discriminant b² − 4ac determines how many real roots exist.

解二次方程有四种核心方法：因式分解法（可行时最快）、公式法（总能使用）、配方法（同时给出顶点坐标）、图像法（近似但直观）。判别式 b² − 4ac 决定实数根的个数。

In the exam, always check whether the equation is already in standard form, whether a common factor can be extracted, and whether your final answers satisfy the original equation. With systematic practice, quadratic equations become one of the most reliable marks on your paper.

考试时，务必检查方程是否已化为标准形式、是否有公因式可提取，并检验最终答案是否满足原方程。通过系统练习，二次方程将成为你试卷上最有把握的得分点之一。

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Solving Quadratic Equations | 解二次方程