Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most important topics in the IGCSE Mathematics syllabus. They appear in almost every exam paper and form the foundation for many higher-level topics such as functions, graphs, and calculus. This guide will take you through every method you need to solve them, with clear worked examples and exam-style practice.

二次方程是 IGCSE 数学考纲中最重要的内容之一。几乎每次考试都会出现，并且是函数、图象和微积分等高阶课题的基础。本指南将带你掌握所有解法，包含清晰的例题和考试风格练习。

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

A quadratic equation is an equation where the highest power of the unknown variable is 2. The standard form is written as ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. If a = 0, the equation becomes linear, not quadratic.

二次方程是未知数最高次数为 2 的方程。标准形式为 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。如果 a = 0，方程就变成一次方程，而不是二次方程。

For example, 3x² − 5x + 2 = 0 is quadratic, while 2x + 7 = 0 is linear. The coefficient a controls the shape of the graph, b affects the position of the vertex, and c gives the y-intercept.

例如，3x² − 5x + 2 = 0 是二次方程，而 2x + 7 = 0 是一次方程。系数 a 控制图象的形状，b 影响顶点的位置，c 决定 y 轴截距。

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2. Solving by Factorisation | 因式分解法

Factorisation is the quickest method when the quadratic has simple integer roots. To solve x² + bx + c = 0, find two numbers that multiply to give c and add to give b. Then set each bracket equal to zero.

当二次方程有简单的整数根时，因式分解是最快的方法。解 x² + bx + c = 0，需要找到两个数，相乘等于 c，相加等于 b。然后将每个括号分别置零。

Example: Solve x² − 5x + 6 = 0. We need two numbers that multiply to 6 and add to −5. The numbers are −2 and −3, so we factorise as (x − 2)(x − 3) = 0. Therefore x = 2 or x = 3.

例：解 x² − 5x + 6 = 0。需要两个数相乘等于 6，相加等于 −5。这两个数是 −2 和 −3，因此因式分解为 (x − 2)(x − 3) = 0。所以 x = 2 或 x = 3。

The zero product property states that if the product of two factors is zero, then at least one factor must be zero. This is the key principle behind factorisation.

零积性质指出：如果两个因式的乘积为零，则至少有一个因式为零。这正是因式分解解题的核心原理。

Example with a leading coefficient: Solve 2x² + 7x + 3 = 0. Multiply a and c: 2 × 3 = 6. Find two factors of 6 that add to 7: they are 1 and 6. Split the middle term: 2x² + x + 6x + 3 = 0. Group: x(2x + 1) + 3(2x + 1) = 0. Factorise: (2x + 1)(x + 3) = 0. Hence x = −½ or x = −3.

例（首项系数不为 1）：解 2x² + 7x + 3 = 0。将 a 和 c 相乘：2 × 3 = 6。找到 6 的两个因数，相加等于 7：它们是 1 和 6。拆分中间项：2x² + x + 6x + 3 = 0。分组：x(2x + 1) + 3(2x + 1) = 0。因式分解：(2x + 1)(x + 3) = 0。因此 x = −½ 或 x = −3。

If AB = 0, then A = 0 or B = 0. | 若 AB = 0，则 A = 0 或 B = 0。

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3. The Quadratic Formula | 求根公式法

When factorisation is difficult or impossible, use the quadratic formula. This formula works for every quadratic equation, whether the roots are integers, fractions, or surds. You must memorise it.

当因式分解困难或无法进行时，使用求根公式。该公式适用于所有二次方程，无论根是整数、分数还是根式。你务必牢记此公式。

x = (−b ± √(b² − 4ac)) / (2a)

To use the formula, first rewrite the equation in the standard form ax² + bx + c = 0, then identify a, b and c carefully, paying special attention to negative signs.

使用公式前，先将方程整理为标准形式 ax² + bx + c = 0，然后仔细确定 a、b、c 的值，特别注意负号。

Example: Solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, c = −2. Substitute into the formula:

例：解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。代入公式：

x = (−3 ± √(9 − 4 × 2 × (−2))) / (2 × 2) = (−3 ± √25) / 4 = (−3 ± 5) / 4

This gives x = (−3 + 5)/4 = 0.5 or x = (−3 − 5)/4 = −2. Always simplify your answers fully; check whether the question asks for exact values or decimal approximations.

因此 x = (−3 + 5)/4 = 0.5 或 x = (−3 − 5)/4 = −2。始终将答案化简；注意题目要求精确值还是小数近似值。

On a calculator paper, you can check your answers by substituting them back into the original equation. On a non-calculator paper, the formula still works with surds, as shown by the completing the square method below.

在允许使用计算器的试卷中，可以将答案代回原方程验证。在不允许使用计算器的试卷中，公式法同样适用于根式，如下一节配方法所示。

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4. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + p)² + q. This form is extremely useful for finding the vertex of a parabola and for solving equations without a calculator.

配方法将二次式改写为 a(x + p)² + q 的形式。这种形式非常适用于求抛物线的顶点，以及在没有计算器的情况下解方程。

For the equation x² + bx + c = 0, take half of b, square it, and add and subtract this value to keep the equation balanced.

对于方程 x² + bx + c = 0，取 b 的一半，平方它，然后加减该值以保持方程平衡。

Example: Solve x² + 6x + 2 = 0. Half of 6 is 3, and 3² = 9. Rewrite as x² + 6x + 9 − 9 + 2 = 0, which is (x + 3)² − 7 = 0. Then (x + 3)² = 7, so x + 3 = ±√7, giving x = −3 ± √7.

例：解 x² + 6x + 2 = 0。6 的一半是 3，3² = 9。改写为 x² + 6x + 9 − 9 + 2 = 0，即 (x + 3)² − 7 = 0。于是 (x + 3)² = 7，所以 x + 3 = ±√7，得 x = −3 ± √7。

Notice that the equation of the axis of symmetry is x = −p, and the coordinates of the vertex are (−p, q). For the example above, the vertex is (−3, −7).

注意对称轴方程为 x = −p，顶点坐标为 (−p, q)。以上面的例子为例，顶点为 (−3, −7)。

If the coefficient of x² is not 1, first factor it out. For example, 2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3.

如果 x² 的系数不为 1，先将其提出。例如，2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3。

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5. The Discriminant | 判别式

The discriminant is the expression b² − 4ac inside the square root of the quadratic formula. It tells us how many real roots a quadratic equation has, without solving it fully.

判别式是求根公式中根号内的表达式 b² − 4ac。它可以告诉我们二次方程有多少个实数根，而无需完整求解。

Δ = b² − 4ac

There are three cases to remember:

需要记住以下三种情况：

Discriminant | 判别式	Nature of Roots | 根的性质	Graph Interpretation | 图象含义
Δ > 0	Two distinct real roots | 两个不等实数根	Cuts the x-axis at two points | 与 x 轴有两个交点
Δ = 0	One repeated real root | 两个相等实数根（重根）	Touches the x-axis at one point | 与 x 轴相切于一点
Δ < 0	No real roots | 没有实数根	Does not intersect the x-axis | 与 x 轴不相交

Example: For x² + 2x + 5 = 0, the discriminant is Δ = 4 − 20 = −16 < 0, so the equation has no real roots.

例：对于 x² + 2x + 5 = 0，判别式为 Δ = 4 − 20 = −16 < 0，因此该方程没有实数根。

Examiners often ask you to “find the value of k for which the equation has equal roots.” You set the discriminant equal to zero and solve for k. This is a common IGCSE problem type.

考官经常要求”求 k 的值，使方程有相等的根。”此时令判别式等于零，然后解出 k。这是 IGCSE 中常见的题型。

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6. Word Problems with Quadratics | 二次方程应用题

Many exam questions translate real-world situations into quadratic equations. The most common are area problems, number problems, and projectile motion problems. The key is to define your variable clearly and set up the equation from the given conditions.

许多考题将实际情境转化为二次方程。最常见的是面积问题、数字问题和抛体运动问题。关键是要明确设定未知数，并根据给定条件建立方程。

Example: A rectangle has length 3 cm longer than its width. Its area is 28 cm². Find the dimensions.

例：一个矩形的长比宽长 3 厘米，面积为 28 平方厘米。求其尺寸。

Let the width be w. Then the length is w + 3. The area equation is w(w + 3) = 28, so w² + 3w − 28 = 0. Factorising gives (w + 7)(w − 4) = 0, so w = 4 or w = −7. Since a length cannot be negative, the width is 4 cm and the length is 7 cm.

设宽为 w，则长为 w + 3。面积方程为 w(w + 3) = 28，即 w² + 3w − 28 = 0。因式分解得 (w + 7)(w − 4) = 0，所以 w = 4 或 w = −7。由于长度不能为负数，所以宽为 4 厘米，长为 7 厘米。

Number problem: The product of two consecutive positive integers is 42. Let the integers be n and n + 1. Then n(n + 1) = 42, so n² + n − 42 = 0, which factorises to (n + 7)(n − 6) = 0. Since the numbers are positive, n = 6 and the integers are 6 and 7.

数字问题：两个连续正整数的乘积为 42。设整数为 n 和 n + 1，则 n(n + 1) = 42，即 n² + n − 42 = 0，因式分解为 (n + 7)(n − 6) = 0。由于是正整数，n = 6，这两个整数是 6 和 7。

Projectile motion: A ball is thrown upward with height h metres after t seconds given by h = 20t − 5t². To find when the ball hits the ground, set h = 0 and solve 5t(4 − t) = 0, giving t = 0 (start) or t = 4 seconds.

抛体运动：球被向上抛出，t 秒后高度 h 米由 h = 20t − 5t² 给出。求球何时落地，令 h = 0，解 5t(4 − t) = 0，得 t = 0（起点）或 t = 4 秒。

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7. Sketching Quadratic Graphs | 二次函数图象

A quadratic graph is called a parabola. It has a characteristic U-shape (or ∩-shape when the coefficient of x² is negative). To sketch it, you need four key features: y-intercept, x-intercepts (roots), the vertex, and the axis of symmetry.

二次函数图象称为抛物线。它具有典型的 U 形（当 x² 的系数为负时是 ∩ 形）。画草图需要四个关键特征：y 轴截距、x 轴截距（根）、顶点和对称轴。

• y-intercept | y 轴截距: Set x = 0; this is simply the constant term c.

  y 轴截距：令 x = 0，即为常数项 c。

• x-intercepts | x 轴截距: Solve ax² + bx + c = 0 using any method.

  x 轴截距：用任意方法解 ax² + bx + c = 0。

• Axis of symmetry | 对称轴: The vertical line x = −b/(2a).

  对称轴：竖直直线 x = −b/(2a)。

• Vertex | 顶点: The turning point, found by substituting the axis of symmetry into the equation, or from the completed square form.

  顶点：转折点，将对称轴代入方程求得，或由配方法形式直接读出。

Example: Sketch y = x² − 4x + 3. The y-intercept is (0, 3). Factorising x² − 4x + 3 = (x − 1)(x − 3) gives roots at x = 1 and x = 3. The axis of symmetry is x = 2. Substituting x = 2 gives y = 4 − 8 + 3 = −1, so the vertex is (2, −1). Since a = 1 > 0, the parabola opens upward.

例：画出 y = x² − 4x + 3 的草图。y 轴截距为 (0, 3)。因式分解 x² − 4x + 3 = (x − 1)(x − 3)，根为 x = 1 和 x = 3。对称轴为 x = 2。代入 x = 2 得 y = 4 − 8 + 3 = −1，所以顶点为 (2, −1)。由于 a = 1 > 0，抛物线开口向上。

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8. Common Mistakes to Avoid | 常见错误

Many students lose marks on quadratic questions because of small but repetitive errors. Being aware of these mistakes is the first step to avoiding them in your exam.

许多学生在二次方程题目上丢分，是因为重复犯一些小错误。认识到这些错误是考试中避免它们的第一步。

• Forgetting to rearrange to standard form | 忘记整理为标准形式: You must have all terms on one side with zero on the other before factorising or using the formula.

  忘记整理为标准形式：在因式分解或使用公式前，必须将所有项移到一边，使另一边为零。

• Missing the negative root | 遗漏负根: When you take the square root of both sides, always write ±. The equation x² = 9 has two solutions: x = 3 and x = −3.

  遗漏负根：两边开平方时，务必写上 ±。方程 x² = 9 有两个解：x = 3 和 x = −3。

• Sign errors when b or c is negative | 当 b 或 c 为负数时出现符号错误: In the formula, substitute negative numbers carefully inside brackets, e.g. (−3)² ≠ −3².

  当 b 或 c 为负数时出现符号错误：代入公式时，注意将负数放在括号内，例如 (−3)² ≠ −3²。

• Dividing by a variable | 除以变量: Never divide both sides by x or by any expression containing x, because you may lose the root x = 0.

  除以变量：切勿两边同时除以 x 或任何含有 x 的表达式，因为可能会丢失 x = 0 这个根。

• Forgetting to reject negative answers in word problems | 在应用题中忘记舍弃负数答案: Length, time and count cannot be negative; always interpret your answers in context.

  在应用题中忘记舍弃负数答案：长度、时间和数量不能为负数；务必结合题意解释你的答案。

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9. Practice Questions | 练习

Now it is your turn. Try these five questions without looking at the answers, then check your work carefully step by step.

现在轮到你了。先不看答案尝试以下五道题，然后逐步仔细检查你的解答过程。

1. Solve x² − 7x + 10 = 0 by factorisation. | 用因式分解法解 x² − 7x + 10 = 0。

2. Solve 3x² − 5x − 2 = 0 using the quadratic formula. | 用求根公式解 3x² − 5x − 2 = 0。

3. Solve x² + 8x + 1 = 0 by completing the square, leaving your answer in surd form. | 用配方法解 x² + 8x + 1 = 0，答案保留根式形式。

4. Find the values of k for which the equation kx² + 6x + 3 = 0 has exactly one real root. | 求 k 的值，使方程 kx² + 6x + 3 = 0 恰好有一个实数根。

5. A rectangle has a perimeter of 26 cm and an area of 36 cm². Find its dimensions. | 一个矩形周长为 26 厘米，面积为 36 平方厘米。求其尺寸。

Answers and short solutions | 答案与简要解答:

Question | 题号	Answer | 答案
1	x = 2 or x = 5
2	x = 2 or x = −1/3
3	x = −4 ± √15
4	k = 3
5	4 cm by 9 cm

For question 4, set Δ = 0: 36 − 12k = 0, hence k = 3. For question 5, let length be l and width be w. Then 2l + 2w = 26 and lw = 36. From the first equation, l + w = 13, so w = 13 − l. Substitute: l(13 − l) = 36, giving l² − 13l + 36 = 0, which factorises to (l − 4)(l − 9) = 0. The dimensions are 4 cm and 9 cm.

第 4 题：令 Δ = 0，即 36 − 12k = 0，因此 k = 3。第 5 题：设长为 l，宽为 w，则 2l + 2w = 26 且 lw = 36。由第一个方程得 l + w = 13，所以 w = 13 − l。代入得 l(13 − l) = 36，即 l² − 13l + 36 = 0，因式分解为 (l − 4)(l − 9) = 0。尺寸为 4 厘米和 9 厘米。

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10. Summary | 总结

You now have three main methods for solving quadratic equations. Use factorisation when the roots are simple integers; use the quadratic formula when factorisation is not obvious; use completing the square when you need the vertex form or when working without a calculator.

你现在掌握了三种解二次方程的主要方法。当根是简单整数时使用因式分解法；当因式分解不明显时使用求根公式；当需要顶点形式或在无计算器环境下解题时使用配方法。

Always remember to check the discriminant first to know what type of roots to expect, and always interpret your answers in the context of word problems. With regular practice, solving quadratic equations will become second nature.

始终记得先看判别式，了解根的类型；在应用题中，始终结合题意解释答案。坚持练习，解二次方程将变得得心应手。

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