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 as a direct algebra question, a graph-sketching task, or a word problem. This revision guide covers the four essential methods of solution and the theory you need to secure full marks.

二次方程几乎出现在每一份 IGCSE 数学试卷中，无论是直接的代数题、函数作图题，还是应用题。本复习指南涵盖四种核心解法以及拿满分所需的理论基础。

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 standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the unknown is 2, which is why it is called “quadratic”, from the Latin quadratus meaning “square”.

二次方程是任何可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 是常数，且 a ≠ 0。未知数的最高次数为 2，因此称为”二次”（quadratic），源自拉丁语 quadratus，意为”平方”。

For example, 2x² − 5x + 3 = 0 is quadratic, while x³ − 2x + 1 = 0 is not. Note that the coefficient a may be negative, and b or c may be zero; for instance, 4x² − 9 = 0 is still a quadratic equation.

例如，2x² − 5x + 3 = 0 是二次方程，而 x³ − 2x + 1 = 0 不是。注意系数 a 可以为负，b 或 c 也可以为零；例如 4x² − 9 = 0 依然是二次方程。

The solutions of a quadratic equation are called its roots. A quadratic equation always has at most two real roots, and some equations have none at all in the real number system. Understanding the conditions for the number of roots requires the discriminant, discussed in Section 6.

二次方程的解称为根。一个二次方程至多有两个实数根，有些方程在实数范围内没有根。要判断根的个数，需要用到第 6 节介绍的判别式。

2. Expanding and Factorising Quadratics | 展开与因式分解二次式

Before solving, you must be fluent in expanding and factorising. Expanding means removing brackets: (x + p)(x + q) = x² + (p + q)x + pq. For example, (x + 3)(x − 5) = x² − 2x − 15.

解题之前，你必须熟练掌握展开与因式分解。展开是指去掉括号：(x + p)(x + q) = x² + (p + q)x + pq。例如，(x + 3)(x − 5) = x² − 2x − 15。

Factorising is the reverse process: given x² + bx + c, find two numbers whose product is c and whose sum is b. To factorise x² − 2x − 15, we look for two numbers that multiply to −15 and add to −2; these are 3 and −5, giving (x + 3)(x − 5).

因式分解是相反的过程：对于 x² + bx + c，找到两个数，它们的乘积为 c，和为 b。要分解 x² − 2x − 15，我们找两个乘积为 −15、和为 −2 的数；它们是 3 和 −5，得到 (x + 3)(x − 5)。

If the coefficient of x² is not 1, use the “ac method” or trial and error: 2x² + 7x + 3 = (2x + 1)(x + 3).
如果 x² 的系数不是 1，可用”ac 法”或试错法：2x² + 7x + 3 = (2x + 1)(x + 3)。

Special pattern: the difference of two squares, a² − b² = (a − b)(a + b). For example, x² − 25 = (x − 5)(x + 5), and 9x² − 1 = (3x − 1)(3x + 1). Recognising this pattern saves time in exams.

特殊情况：平方差公式 a² − b² = (a − b)(a + b)。例如，x² − 25 = (x − 5)(x + 5)，9x² − 1 = (3x − 1)(3x + 1)。在考试中识别出这一模式可以节省时间。

3. Solving by Factorisation | 因式分解法解方程

If a quadratic can be factorised, this is usually the fastest solution method. The key idea is the zero product property: if A × B = 0, then A = 0 or B = 0. This means each factor can be set to zero independently.

如果一个二次方程可以因式分解，这通常是最快的解法。关键原理是零乘积性质：若 A × B = 0，则 A = 0 或 B = 0。这意味着可以分别令每个因式等于零。

Worked example: Solve x² − 5x + 6 = 0. First factorise: (x − 2)(x − 3) = 0. Then x − 2 = 0 gives x = 2, and x − 3 = 0 gives x = 3. The solution set is x = 2 or x = 3.

示例：解方程 x² − 5x + 6 = 0。先因式分解：(x − 2)(x − 3) = 0。然后令 x − 2 = 0，得 x = 2；令 x − 3 = 0，得 x = 3。解集为 x = 2 或 x = 3。

Always rearrange the equation to the form ax² + bx + c = 0 before factorising. For example, to solve 2x² = 9x − 4, first rewrite as 2x² − 9x + 4 = 0, then factorise: (2x − 1)(x − 4) = 0, giving x = ½ or x = 4.

在因式分解之前，务必先将方程整理为 ax² + bx + c = 0 的形式。例如，解 2x² = 9x − 4 时，先改写为 2x² − 9x + 4 = 0，再因式分解：(2x − 1)(x − 4) = 0，得 x = ½ 或 x = 4。

Do not divide both sides by a common factor involving the unknown, because you may lose a root. For instance, solving x² = 3x by dividing by x gives only x = 3; the correct method is to rearrange to x² − 3x = 0, factorise to x(x − 3) = 0, and obtain x = 0 or x = 3.

不要两边同时除以含有未知数的公因式，否则可能会丢根。例如，解 x² = 3x 时，若两边除以 x 只会得到 x = 3；正确做法是移项得 x² − 3x = 0，分解为 x(x − 3) = 0，得到 x = 0 或 x = 3。

4. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + h)² + k. For the expression x² + bx, we add and subtract (b/2)² to create a perfect square. This method works for any quadratic and is especially useful for finding the vertex of a parabola.

配方法将二次式改写为 (x + h)² + k 的形式。对于 x² + bx，我们加上再减去 (b/2)²，以构造完全平方。这种方法适用于任何二次方程，在求抛物线顶点时尤其有用。

Example: Solve x² + 6x + 2 = 0 by completing the square. Begin with x² + 6x = −2. Add 9 to both sides, because (6/2)² = 9, giving x² + 6x + 9 = 7. The left side is (x + 3)², so (x + 3)² = 7.

示例：用配方法解 x² + 6x + 2 = 0。先将 x² + 6x = −2。两边加上 9，因为 (6/2)² = 9，得到 x² + 6x + 9 = 7。左边为 (x + 3)²，因此 (x + 3)² = 7。

Now take the square root of both sides, remembering the ± symbol: x + 3 = ±√7. Hence x = −3 + √7 or x = −3 − √7. These are exact values, which calculators often cannot display in this simple form.

然后两边开平方，记得加上 ± 符号：x + 3 = ±√7。因此 x = −3 + √7 或 x = −3 − √7。这些是精确值，计算器通常无法以这种简洁形式直接显示。

If the coefficient of x² is not 1, factor it out first. For 2x² + 8x + 3 = 0, write 2(x² + 4x) + 3 = 0, complete the square inside the bracket, and then solve carefully.

若 x² 的系数不是 1，先将其提出。对于 2x² + 8x + 3 = 0，写成 2(x² + 4x) + 3 = 0，在括号内配方，然后仔细求解。

5. The Quadratic Formula | 二次公式

The quadratic formula solves any quadratic equation ax² + bx + c = 0 directly:

二次公式可以直接求解任意二次方程 ax² + bx + c = 0：

x = (−b ± √(b² − 4ac)) / (2a)

This formula is derived by completing the square on the general quadratic. You are expected to memorise it for the IGCSE exam, as it is not provided on the formula sheet in most papers.

该公式通过对一般二次方程配方推导而来。IGCSE 考试要求你记住它，因为在多数试卷中该公式不会附在公式表上。

Worked example: Solve 2x² − 3x − 1 = 0. Here a = 2, b = −3 and c = −1. Substituting gives x = (3 ± √(9 + 8)) / 4 = (3 ± √17) / 4. The two roots are approximately 1.78 and −0.28.

示例：解 2x² − 3x − 1 = 0。这里 a = 2，b = −3，c = −1。代入得 x = (3 ± √(9 + 8)) / 4 = (3 ± √17) / 4。两个根约为 1.78 和 −0.28。

To avoid sign errors, enclose negative values of b in brackets when substituting. Many students lose marks by writing −(−3) incorrectly as −3 rather than +3. Always check your substitution line carefully.

为避免符号错误，代入负数 b 时请加括号。许多学生因把 −(−3) 误写成 −3 而不是 +3 而丢分。务必仔细检查代入步骤。

6. The Discriminant | 判别式

The discriminant is the expression Δ = b² − 4ac, which appears under the square root in the quadratic formula. Its sign determines how many real roots the equation has.

判别式是表达式 Δ = b² − 4ac，它出现在二次公式的根号下。它的符号决定了方程有多少个实数根。

Value of Δ	Number of real roots	Graph interpretation
Δ > 0	Two distinct real roots	Parabola crosses the x-axis twice
Δ = 0	One repeated real root	Parabola touches the x-axis once
Δ < 0	No real roots	Parabola never touches the x-axis

判别式的值 | 实数根个数 | 图像解释

Δ > 0 | 两个不同实数根 | 抛物线与 x 轴相交两次

Δ = 0 | 一个重根 | 抛物线与 x 轴相切一次

Δ < 0 | 无实数根 | 抛物线与 x 轴无交点

Example: For the equation 3x² − 5x + 2 = 0, Δ = 25 − 24 = 1 > 0, so there are two real roots. In contrast, for x² + x + 1 = 0, Δ = 1 − 4 = −3 < 0, so the equation has no real solutions.

示例：对于方程 3x² − 5x + 2 = 0，Δ = 25 − 24 = 1 > 0，所以有两个实数根。相反，对于 x² + x + 1 = 0，Δ = 1 − 4 = −3 < 0，因此该方程没有实数解。

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

The graph of y = ax² + bx + c is a parabola. If a > 0 it opens upward like a smile, with a minimum point; if a < 0 it opens downward like a frown, with a maximum point.

函数 y = ax² + bx + c 的图像是抛物线。若 a > 0，开口向上，有最低点；若 a < 0，开口向下，有最高点。

The roots of the equation ax² + bx + c = 0 are exactly the x-intercepts of the graph. If the equation has no real roots, the graph does not cross the x-axis at all.

方程 ax² + bx + c = 0 的根正是图像与 x 轴的交点横坐标。若方程无实数根，图像就完全不与 x 轴相交。

The vertex (turning point) can be found by completing the square: for y = (x − h)² + k, the vertex is (h, k). The axis of symmetry is the vertical line x = h. For the general form, the axis of symmetry is x = −b/(2a).

顶点（转向点）可通过配方求得：对于 y = (x − h)² + k，顶点为 (h, k)。对称轴是竖直线 x = h。对于一般形式，对称轴为 x = −b/(2a)。

When sketching, always label the roots, the y-intercept, and the vertex. For example, y = x² − 4x + 3 factorises to y = (x − 1)(x − 3), so the x-intercepts are 1 and 3, the y-intercept is 3, and the axis of symmetry is x = 2.

画草图时，务必标出根、y 截距和顶点。例如，y = x² − 4x + 3 分解为 y = (x − 1)(x − 3)，因此 x 截距为 1 和 3，y 截距为 3，对称轴为 x = 2。

8. Word Problems and Applications | 实际应用题

Many IGCSE questions present quadratic equations in context. The first step is to define a variable, translate the information into an equation, solve it, and then check whether each root makes sense in the original context.

许多 IGCSE 题目会以实际情境呈现二次方程。第一步是设未知数，将信息转化为方程，求解，然后检查每个根在原始情境中是否有意义。

Number problem: The product of two consecutive positive integers is 132. Let the integers be n and n + 1. Then n(n + 1) = 132, so n² + n − 132 = 0. Factorising gives (n − 11)(n + 12) = 0, so n = 11 or n = −12. Since the integers are positive, the answer is 11 and 12.

数字问题：两个连续正整数的乘积为 132。设这两个数为 n 和 n + 1。则 n(n + 1) = 132，即 n² + n − 132 = 0。因式分解得 (n − 11)(n + 12) = 0，所以 n = 11 或 n = −12。因为要求正整数，所以答案为 11 和 12。

Area problem: A rectangular garden has length 3 m greater than its width, and its area is 70 m². Let the width be w. Then w(w + 3) = 70, so w² + 3w − 70 = 0. Factorising gives (w + 10)(w − 7) = 0, so w = 7 (rejecting w = −10). The dimensions are 7 m by 10 m.

面积问题：一块矩形花园的长比宽多 3 米，面积为 70 平方米。设宽为 w。则 w(w + 3) = 70，即 w² + 3w − 70 = 0。因式分解得 (w + 10)(w − 7) = 0，所以 w = 7（舍去 w = −10）。尺寸为 7 米 × 10 米。

Motion problem: A ball thrown upward has height h = 20t − 5t². To find when it reaches the ground, set h = 0: 5t(4 − t) = 0, so t = 0 (launch time) or t = 4 seconds. Always interpret your roots in context before writing a final answer.

运动问题：一个向上抛出的球，其高度为 h = 20t − 5t²。要求落地时间，令 h = 0：5t(4 − t) = 0，所以 t = 0（抛出时刻）或 t = 4 秒。在写出最终答案前，务必结合情境解释每一个根。

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

Here are the most frequent errors students make in IGCSE quadratic questions, with advice on how to avoid each one.

以下是学生在 IGCSE 二次方程题目中最常犯的错误，以及如何避免它们的建议。

Forgetting to rearrange the equation to zero before factorising. Always write ax² + bx + c = 0 first.
忘记在因式分解前将方程整理为零。务必先写成 ax² + bx + c = 0。

Losing the ± sign when taking square roots. Remember that x² = 9 gives x = ±3, not x = 3.
开平方时漏掉 ± 符号。记住 x² = 9 给出 x = ±3，而不是 x = 3。

Misapplying the quadratic formula when b is negative. Substitute with brackets: x = (−(−5) ± √(…)) / (2a) = (5 ± √(…)) / (2a).
当 b 为负数时误用二次公式。代入时加括号：x = (−(−5) ± √(…)) / (2a) = (5 ± √(…)) / (2a)。

Dividing both sides by x and losing the root x = 0. Rearrange to zero and factorise instead.
两边除以 x 导致丢失 x = 0 这个根。应移项至零并因式分解。

Exam tips: Show every step of your working. In “solve” questions, method marks are awarded for factorising or substituting into the formula, even if your final answer is wrong. For exact answers, leave surds like √17 rather than giving a rounded decimal unless the question asks for an approximation.

考试技巧：写出每一步过程。在”求解”类题目中，即使最终答案有误，因式分解或代入公式的步骤也能获得步骤分。若要求精确值，请保留 √17 这样的根式，除非题目要求近似值，否则不要给出四舍五入的小数。

10. Practice Questions | 巩固练习

Test your understanding by solving the following questions on paper before checking the answers below.

请先动笔解答以下题目，再核对下方的答案，以检验你的理解。

Solve x² − 7x + 10 = 0.
解方程 x² − 7x + 10 = 0。

Solve 3x² + 2x − 1 = 0 using the quadratic formula.
用二次公式解 3x² + 2x − 1 = 0。

Find the exact solutions of x² + 4x + 1 = 0 by completing the square.
用配方法求 x² + 4x + 1 = 0 的精确解。

Determine the number of real roots of 2x² − 4x + 5 = 0.
判断 2x² − 4x + 5 = 0 的实数根个数。

Answers | 参考答案

1. (x − 2)(x − 5) = 0 → x = 2 or x = 5

1. (x − 2)(x − 5) = 0 → x = 2 或 x = 5

2. x = (−2 ± √(4 + 12)) / 6 = (−2 ± 4) / 6 → x = 1/3 or x = −1

2. x = (−2 ± √(4 + 12)) / 6 = (−2 ± 4) / 6 → x = 1/3 或 x = −1

3. (x + 2)² = 3 → x = −2 ± √3

4. Δ = 16 − 40 = −24 < 0 → no real roots

4. Δ = 16 − 40 = −24 < 0 → 无实数根

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Solving Quadratic Equations | 解二次方程