Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in almost every paper, whether as direct solving questions, graph work, or word problems. This guide covers everything you need: expanding, factorising, solving by three methods, the discriminant, graphs, and exam techniques.

二次方程是 IGCSE 数学中最重要的考点之一。几乎每份试卷都会出现，无论是直接求解、函数图象还是应用题。本指南将涵盖你需要掌握的全部内容：展开、因式分解、三种求解方法、判别式、图象以及考试技巧。

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 form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the variable x is 2, which is why it is called ‘quadratic’ (from the Latin quadratus, meaning square).

一元二次方程是可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 是常数，且 a ≠ 0。变量的最高次数为 2，因此被称为 “quadratic”（源自拉丁语 quadratus，意为 “平方”）。

For example, 3x² − 5x + 2 = 0 is a quadratic equation because a = 3, b = −5 and c = 2. However, equations like 2x + 1 = 0 (linear) or x³ − 1 = 0 (cubic) are not quadratic.

例如，3x² − 5x + 2 = 0 是一个二次方程，其中 a = 3，b = −5，c = 2。但像 2x + 1 = 0（一次方程）或 x³ − 1 = 0（三次方程）这样的方程不是二次方程。

ax² + bx + c = 0 (a ≠ 0)

2. Expanding and Factorising | 展开与因式分解

Before solving quadratics, you must be confident with expanding brackets and factorising. Expanding means removing brackets: (x + 3)(x + 2) = x² + 5x + 6. Each term in the first bracket multiplies each term in the second bracket — use the FOIL method (First, Outer, Inner, Last).

在求解二次方程之前，你必须熟练展开括号和因式分解。展开就是去掉括号：(x + 3)(x + 2) = x² + 5x + 6。第一个括号中的每一项都要与第二个括号中的每一项相乘——使用 FOIL 方法（First 首项，Outer 外项，Inner 内项，Last 末项）。

Factorising is the reverse process. To factorise x² + 5x + 6, find two numbers that multiply to give 6 and add to give 5. The numbers are 2 and 3, so x² + 5x + 6 = (x + 2)(x + 3).

因式分解是相反的过程。要分解 x² + 5x + 6，找到两个数相乘得 6 且相加得 5。这两个数是 2 和 3，所以 x² + 5x + 6 = (x + 2)(x + 3)。

For quadratics with a ≠ 1, such as 2x² + 7x + 3, use the ‘ac method’: multiply a and c (2 × 3 = 6), find two numbers that multiply to 6 and add to 7 (1 and 6), split the middle term, and factorise by grouping:

对于 a ≠ 1 的二次式，如 2x² + 7x + 3，可使用 “ac 方法”：把 a 和 c 相乘（2 × 3 = 6），找到两个数相乘得 6 且相加得 7（即 1 和 6），将中间项拆分，然后分组因式分解：

2x² + 7x + 3 = 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)

Remember these special forms:

记住这些特殊形式：

Difference of two squares: a² − b² = (a + b)(a − b), e.g. x² − 9 = (x + 3)(x − 3).

平方差：a² − b² = (a + b)(a − b)，例如 x² − 9 = (x + 3)(x − 3)。
Perfect square: a² + 2ab + b² = (a + b)², e.g. x² + 6x + 9 = (x + 3)².

完全平方：a² + 2ab + b² = (a + b)²，例如 x² + 6x + 9 = (x + 3)²。

3. Solving by Factorisation | 因式分解法求解

If a quadratic equation can be factorised, solving is quick and elegant. The key principle is the zero product property: if A × B = 0, then either A = 0 or B = 0.

如果一个二次方程可以因式分解，求解就快速而简洁。关键原理是 零乘积性质：如果 A × B = 0，那么 A = 0 或 B = 0。

To solve x² − 5x + 6 = 0, first factorise: (x − 2)(x − 3) = 0. Then set each bracket to zero: x − 2 = 0 gives x = 2, and x − 3 = 0 gives x = 3. The solution set is x = 2 or x = 3.

要求解 x² − 5x + 6 = 0，先因式分解：(x − 2)(x − 3) = 0。然后令每个括号为零：x − 2 = 0 得到 x = 2，x − 3 = 0 得到 x = 3。解集为 x = 2 或 x = 3。

Step-by-step procedure:

分步步骤：

Rearrange the equation so one side is 0, e.g. x² = 3x → x² − 3x = 0.

整理方程使一边为 0，例如 x² = 3x → x² − 3x = 0。
Factorise the other side completely: x(x − 3) = 0.

将另一边彻底因式分解：x(x − 3) = 0。
Set each factor equal to 0 and solve: x = 0 or x = 3.

令每个因式等于 0 并求解：x = 0 或 x = 3。

Note: never divide both sides by x unless you are sure x ≠ 0. Dividing x² = 3x by x gives x = 3, losing the solution x = 0.

注意：除非确定 x ≠ 0，否则切勿两边同除以 x。将 x² = 3x 两边除以 x 会得到 x = 3，从而丢失解 x = 0。

4. Solving by Completing the Square | 配方法求解

Completing the square rewrites a quadratic in the form (x + p)² + q. This method is valuable because it always works, even when factorisation is impossible, and it reveals the turning point of the graph.

配方法将二次式改写为 (x + p)² + q 的形式。这种方法很有价值，因为即使无法因式分解它也始终有效，并且能揭示图象的顶点。

Take x² + 6x + 2 = 0. Halve the coefficient of x (6 ÷ 2 = 3), square it (3² = 9), and write:

以 x² + 6x + 2 = 0 为例。将 x 的系数减半（6 ÷ 2 = 3），取其平方（3² = 9），然后写出：

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

Now solve (x + 3)² − 7 = 0:

现在解 (x + 3)² − 7 = 0：

(x + 3)² = 7 → x + 3 = ±√7 → x = −3 ± √7

When the coefficient of x² is not 1, first factor it out. For 2x² + 8x + 5 = 0, write 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3.

当 x² 的系数不为 1 时，先提取该系数。对于 2x² + 8x + 5 = 0，写为 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3。

The general rule for x² + bx is to add and subtract (b/2)²:

x² + bx 的一般规则是加上并减去 (b/2)²：

x² + bx = (x + b/2)² − (b/2)²

5. The Quadratic Formula | 求根公式

For any quadratic equation ax² + bx + c = 0, the solutions are given by the quadratic formula. This is the most reliable method — it works for every quadratic, provided the equation has real solutions.

对于任意二次方程 ax² + bx + c = 0，其解由求根公式给出。这是最可靠的方法——只要方程有实数解，它对所有二次方程都适用。

x = (−b ± √(b² − 4ac)) / 2a

To use it, identify a, b and c carefully — including their signs — then substitute into the formula. For example, solve 2x² + 3x − 5 = 0. Here a = 2, b = 3, c = −5.

使用时，认真识别 a、b、c——包括它们的符号——然后代入公式。例如，解 2x² + 3x − 5 = 0。这里 a = 2，b = 3，c = −5。

x = (−3 ± √(3² − 4 × 2 × (−5))) / (2 × 2) = (−3 ± √(9 + 40)) / 4 = (−3 ± √49) / 4 = (−3 ± 7) / 4

So x = (−3 + 7)/4 = 1, or x = (−3 − 7)/4 = −2.5. Always check your answer by substituting back into the original equation.

因此 x = (−3 + 7)/4 = 1，或 x = (−3 − 7)/4 = −2.5。务必代回原方程检验答案。

Method	When to use	Advantage
Factorisation	When the quadratic is simple and factorable	Fast; often gives exact integer answers
Completing the square	When you also need the turning point	Reveals vertex form of the parabola
Quadratic formula	For any quadratic; especially with irrational roots	Always works; no factorising needed

6. The Discriminant | 判别式

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

根号内的表达式 b² − 4ac 称为 判别式，记为 Δ。它告诉我们二次方程有多少个实数根，而无需完全求解。

If b² − 4ac > 0, there are two distinct real roots (the graph crosses the x-axis twice).

如果 b² − 4ac > 0，则有两个不同的实数根（图象与 x 轴相交两次）。
If b² − 4ac = 0, there is exactly one real root (a repeated root; the graph touches the x-axis at one point).

如果 b² − 4ac = 0，则恰有一个实数根（重根；图象与 x 轴相切于一点）。
If b² − 4ac < 0, there are no real roots (the graph does not intersect the x-axis).

如果 b² − 4ac < 0，则没有实数根（图象不与 x 轴相交）。

Example: For x² − 4x + 4 = 0, Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0, so there is one repeated root. Indeed, x² − 4x + 4 = (x − 2)² gives x = 2 as the only solution.

例如：对于 x² − 4x + 4 = 0，Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0，因此有一个重根。事实上，x² − 4x + 4 = (x − 2)² 给出唯一解 x = 2。

Exam questions often ask: “Find the range of k such that kx² + 2x + 1 = 0 has two distinct real roots.” Set the discriminant greater than zero and solve the resulting inequality.

考试常问：”求 k 的范围，使得 kx² + 2x + 1 = 0 有两个不同的实数根。” 令判别式大于零并解所得不等式。

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

The graph of y = ax² + bx + c is a parabola. If a > 0, it opens upward (a ‘smiley’ shape) and has a minimum point. If a < 0, it opens downward (a 'frown' shape) and has a maximum point.

y = ax² + bx + c 的图象是抛物线。如果 a > 0，开口向上（”微笑”形），有最小值点。如果 a < 0，开口向下（"皱眉"形），有最大值点。

Key features to identify on a quadratic graph:

二次函数图象上需要识别的关键特征：

x-intercepts: the real roots of ax² + bx + c = 0.

x 截距：ax² + bx + c = 0 的实数根。
y-intercept: the constant term c, since setting x = 0 gives y = c.

y 截距：常数项 c，因为令 x = 0 得到 y = c。
Turning point: the vertex of the parabola, where the function reaches its maximum or minimum.

顶点：抛物线的最高点或最低点，函数在此达到最大值或最小值。
Axis of symmetry: the vertical line x = −b/(2a) that passes through the turning point.

对称轴：穿过顶点的垂直线 x = −b/(2a)。

To sketch a quadratic graph, find the intercepts and the turning point, then draw a smooth curve through these points.

要画出二次函数草图，先求截距和顶点，然后通过这些点画一条平滑曲线。

8. Finding the Turning Point | 求顶点坐标

The turning point can be found by completing the square. If y = (x + p)² + q, then the turning point is at (−p, q). For example, y = (x − 3)² + 4 has its minimum at (3, 4).

可以通过配方法求顶点。如果 y = (x + p)² + q，则顶点在 (−p, q)。例如，y = (x − 3)² + 4 的最小值点为 (3, 4)。

Alternatively, use the formula x = −b/(2a) to find the x-coordinate of the turning point, then substitute back to find y. For y = 2x² − 8x + 1, we have:

另一种方法是使用公式 x = −b/(2a) 求顶点的 x 坐标，然后代回求出 y。对于 y = 2x² − 8x + 1：

x = −(−8)/(2 × 2) = 8/4 = 2 → y = 2(2)² − 8(2) + 1 = 8 − 16 + 1 = −7

So the turning point is (2, −7). Since a = 2 > 0, this is a minimum point.

因此顶点为 (2, −7)。由于 a = 2 > 0，这是一个最小值点。

Remember: the axis of symmetry always passes through the turning point, so the two x-intercepts (if they exist) are equidistant from x = −b/(2a).

记住：对称轴总是穿过顶点，因此两个 x 截距（如果存在）到 x = −b/(2a) 的距离相等。

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

Many exam questions present quadratics in context. Common scenarios include area problems, projectile motion, and number problems. The approach is always the same: define a variable, form a quadratic equation, solve it, and interpret the answers in context.

许多考试题目将二次方程置于实际情境中。常见场景包括面积问题、抛体运动和数字问题。解题方法总是相同的：设变量，建立二次方程，求解，并结合情境解释答案。

Example: A rectangle has length 4 cm more than its width. Its area is 96 cm². Find the rectangle’s dimensions.

例题：一个长方形的长比宽多 4 cm，面积为 96 cm²。求长方形的尺寸。

Let the width be x cm. Then the length is (x + 4) cm. The area equation is x(x + 4) = 96, so x² + 4x − 96 = 0. Factorising gives (x + 12)(x − 8) = 0, so x = −12 or x = 8. Since width cannot be negative, x = 8. Thus the width is 8 cm and the length is 12 cm.

设宽为 x cm，则长为 (x + 4) cm。面积方程为 x(x + 4) = 96，即 x² + 4x − 96 = 0。因式分解得 (x + 12)(x − 8) = 0，所以 x = −12 或 x = 8。由于宽度不能为负，x = 8。因此宽为 8 cm，长为 12 cm。

Always reject negative or impossible solutions in word problems, and remember to write the final answer with the correct units.

在应用题中务必舍弃负数或不合理的解，并记得用正确的单位写出最终答案。

10. Solving Quadratics by Drawing Graphs | 用图象法求解二次方程

Sometimes the exam asks you to solve a quadratic equation by drawing its graph. You plot y = ax² + bx + c for a given range of x, then read off the x-values where the curve crosses the x-axis (where y = 0).

有时考试要求你通过画图来解二次方程。在给定的 x 范围内描绘 y = ax² + bx + c，然后读出曲线与 x 轴交点的 x 值（此时 y = 0）。

To solve a related equation like x² − 2x − 3 = 1, draw the graph of y = x² − 2x − 3 and also the horizontal line y = 1; the x-coordinates of the intersection points are the solutions.

要求解相关方程如 x² − 2x − 3 = 1，画出 y = x² − 2x − 3 的图象以及水平线 y = 1；交点处的 x 坐标就是解。

When plotting a quadratic graph, choose x-values that bracket the turning point, and calculate y for each x. Then join the points with a smooth, symmetrical curve.

画二次函数图象时，选择的 x 值要包含顶点附近的范围，计算每个 x 对应的 y 值，然后用平滑、对称的曲线连接各点。

11. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Students lose marks in quadratic questions for predictable reasons. Here are the most common pitfalls and how to avoid them:

学生在二次方程题目中失分的原因往往可以预测。以下是最常见的陷阱及避免方法：

Forgetting to rearrange to = 0. Always rewrite the equation so the right-hand side is 0 before factorising.

忘记整理为 = 0。因式分解前务必将方程右边整理为 0。
Losing a root. Do not divide by a variable expression unless you have shown it cannot be zero.

丢根。除非已证明变量表达式不可能为零，否则不要除以该表达式。
Sign errors. In the quadratic formula, carefully substitute negative values of b and c. Use brackets: x = (−(−5) ± √((−5)² − 4 × 1 × 6)) / 2.

符号错误。在求根公式中，仔细代入 b 和 c 的负值。使用括号：x = (−(−5) ± √((−5)² − 4 × 1 × 6)) / 2。
Miscalculating the discriminant. Remember that (−5)² = 25, not −25. Take the square before applying the sign.

判别式计算错误。记住 (−5)² = 25，而不是 −25。先算平方再处理符号。
Confusing the turning point with the y-intercept. The turn is at (−b/2a, f(−b/2a)), while the y-intercept is simply c.

混淆顶点与 y 截距。顶点在 (−b/2a, f(−b/2a))，而 y 截距就是 c。

12. Exam Tips and Practice | 考试技巧与练习

To maximise your marks on quadratic questions, follow these strategies:

要在二次方程题目中拿满分，请遵循以下策略：

Read the question carefully — does it ask for exact values or answers correct to 3 significant figures or 2 decimal places?

仔细读题——是要求精确值，还是保留 3 位有效数字或 2 位小数？
Show all working. Even if you use a calculator, the examiner needs to see the method for method marks.

写出全部过程。即使你使用计算器，考官也需要看到方法才能给步骤分。
If a quadratic does not factorise easily, use the quadratic formula immediately — do not waste time trying to factorise.

如果二次式不容易因式分解，立即使用求根公式——不要浪费时间尝试因式分解。
Always check your roots by substituting back into the original equation.

总是代回原方程检验根是否正确。
In graph questions, label the axes, mark the intercepts and turning point clearly, and use a ruler-friendly scale.

在图象题中，标出坐标轴、截距和顶点，并使用便于绘图的刻度。

Practice question: Solve 3x² − 7x + 2 = 0 by the quadratic formula. Answer: a = 3, b = −7, c = 2, so x = (7 ± √(49 − 24)) / 6 = (7 ± 5) / 6, giving x = 2 or x = 1/3.

练习：用求根公式解 3x² − 7x + 2 = 0。答案：a = 3，b = −7，c = 2，所以 x = (7 ± √(49 − 24)) / 6 = (7 ± 5) / 6，得到 x = 2 或 x = 1/3。

Mastering quadratics takes practice, but once you understand the three methods — factorisation, completing the square, and the quadratic formula — you will be able to handle any question the exam throws at you. Pair this with graph-sketching skills and a strong grasp of the discriminant, and this topic becomes a reliable source of marks.

掌握二次方程需要练习，但一旦你理解了三种方法——因式分解法、配方法和求根公式——你就能应对考试中的任何题目。结合图象绘制技巧和对判别式的深刻理解，这个专题将成为你稳定的得分来源。

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Solving Quadratic Equations | 解一元二次方程