Mastering Quadratic Equations | 掌握一元二次方程

📚 Mastering Quadratic Equations | 掌握一元二次方程

Quadratic equations are one of the most heavily tested topics in IGCSE Mathematics. Whether you take the Core or Extended paper, you must be able to solve them accurately, understand their graphs, and apply them to real-world problems. This guide breaks everything down step by step, with the exact methods examiners expect you to use.

一元二次方程是 IGCSE 数学中考查频率最高的知识点之一。无论你参加 Core 还是 Extended 试卷，都必须能够准确求解、理解其图像，并将其应用于实际生活问题。本指南将逐步拆解所有考点，带你掌握阅卷官所期望的标准解法。

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 is 2, which is why it is called “quadratic” — from the Latin word “quadratus”, meaning square.

一元二次方程是能写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数，且 a ≠ 0。变量最高次数为 2，因此称为 “二次”（quadratic），源自拉丁语 “quadratus”，意为”正方形”。

Examples of quadratic equations include:

• x² + 3x − 4 = 0
• 2x² − 5x + 1 = 0
• −x² + 2x = 0 (which is −x² + 2x + 0 = 0)

Examples of non-quadratic equations include:

• x³ + 2x − 1 = 0 (cubic, highest power 3)
• x + 5 = 0 (linear, highest power 1)
• √x + 2 = 9 (involves a square root, not a power of 2)

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2. Standard Form and Key Features | 标准形式与关键特征

The standard form of a quadratic equation is ax² + bx + c = 0. In IGCSE questions, you will frequently need to rearrange a given equation into this form before solving. For example, 2x² + 3 = 7x must first be rewritten as 2x² − 7x + 3 = 0.

一元二次方程的标准形式为 ax² + bx + c = 0。在 IGCSE 考题中，你经常需要先将给定方程整理成这种形式再求解。例如，2x² + 3 = 7x 必须先改写为 2x² − 7x + 3 = 0。

The three coefficients tell us important information:

• a controls the shape of the graph: if a > 0, the parabola opens upwards; if a < 0, it opens downwards.
• b affects the position of the axis of symmetry.
• c is the y-intercept of the graph — where the curve crosses the y-axis.

系数 a 决定图像的开口方向：若 a > 0，抛物线开口向上；若 a < 0，开口向下。

系数 b 影响对称轴的位置；系数 c 是图像与 y 轴的交点（y 截距）。

a ≠ 0 is an essential condition — if a = 0, the equation becomes linear.

a ≠ 0 是必要条件——若 a = 0，方程就退化成为一次方程。

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3. Solving by Factorisation | 因式分解法求解

Factorisation is the fastest method when the quadratic has simple integer factors. The idea is to write the quadratic as a product of two brackets, then set each bracket equal to zero.

因式分解法是当二次方程具有简单整数因子时最快捷的方法。其核心思想是将二次式写成两个括号的乘积，然后令每个括号分别等于零。

Example: Solve x² − 5x + 6 = 0

Step 1: Find two numbers that multiply to give 6 and add to give −5. These are −2 and −3.

Step 2: Write the factorised form: (x − 2)(x − 3) = 0

Step 3: Apply the zero product property: x − 2 = 0 or x − 3 = 0

Step 4: Solve: x = 2 or x = 3

示例：解方程 x² − 5x + 6 = 0

第一步：找到两个数，乘积为 6，且和为 −5。这两个数是 −2 和 −3。

第二步：写成因式形式：(x − 2)(x − 3) = 0

第三步：运用零乘积性质：x − 2 = 0 或 x − 3 = 0

第四步：解得：x = 2 或 x = 3

When the coefficient of x² is not 1, the process requires more care.

当 x² 的系数不为 1 时，过程需要更加仔细。

Example: Solve 2x² − 7x + 3 = 0

Multiply the leading coefficient by c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to −7: these are −1 and −6. Split the middle term: 2x² − x − 6x + 3 = 0. Then factor by grouping: x(2x − 1) − 3(2x − 1) = 0, giving (x − 3)(2x − 1) = 0. Hence x = 3 or x = ½.

示例：解方程 2x² − 7x + 3 = 0

首项系数与 c 相乘：2 × 3 = 6。找到乘积为 6 且和为 −7 的两个数：它们是 −1 和 −6。拆中间项：2x² − x − 6x + 3 = 0。然后分组因式分解：x(2x − 1) − 3(2x − 1) = 0，即 (x − 3)(2x − 1) = 0。因此 x = 3 或 x = ½。

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4. The Quadratic Formula | 求根公式

When factorisation is difficult or impossible, the quadratic formula always works. It is given on the IGCSE formula sheet, but you must know how to substitute into it correctly.

当因式分解难以进行或无法进行时，求根公式永远是可靠的兜底方法。IGCSE 公式表会给出该公式，但你必须学会正确代入。

x = (−b ± √(b² − 4ac)) ⁄ 2a

Example: Solve 2x² − 4x − 3 = 0, giving answers correct to 2 decimal places.

Here a = 2, b = −4, c = −3. Substitute:

x = (4 ± √((−4)² − 4 × 2 × (−3))) ⁄ (2 × 2)

x = (4 ± √(16 + 24)) ⁄ 4 = (4 ± √40) ⁄ 4

√40 ≈ 6.324, so x ≈ (4 + 6.324) ⁄ 4 = 2.581 and x ≈ (4 − 6.324) ⁄ 4 = −0.581.

示例：解方程 2x² − 4x − 3 = 0，答案精确到两位小数。

这里 a = 2，b = −4，c = −3。代入得：

x = (4 ± √((−4)² − 4 × 2 × (−3))) ⁄ (2 × 2)

x = (4 ± √(16 + 24)) ⁄ 4 = (4 ± √40) ⁄ 4

√40 ≈ 6.324，因此 x ≈ (4 + 6.324) ⁄ 4 = 2.581，x ≈ (4 − 6.324) ⁄ 4 = −0.581。

A common error is forgetting the double negative inside the square root: (−4)² − 4(2)(−3) = 16 + 24 = 40, not 16 − 24. Always write down the substitution in full before pressing the calculator.

一个常见错误是忘记根号内的负负得正：(−4)² − 4(2)(−3) = 16 + 24 = 40，而非 16 − 24。务必在按计算器之前完整写出代入过程。

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5. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + p)² + q. This form is especially useful for finding the vertex of a parabola and solving equations when the quadratic formula is not required.

配方法将二次式改写为 a(x + p)² + q 的形式。这种形式特别适用于求抛物线顶点以及在无需使用求根公式时解方程。

Example: Solve x² + 6x + 1 = 0 by completing the square.

Take half of the coefficient of x: 6 ⁄ 2 = 3. Square it: 3² = 9. Rewrite: (x + 3)² − 9 + 1 = 0, which simplifies to (x + 3)² − 8 = 0.

Then (x + 3)² = 8, so x + 3 = ±√8, giving x = −3 ± √8. In decimal form, x ≈ 1.83 or x ≈ −7.83.

示例：用配方法解方程 x² + 6x + 1 = 0

取 x 系数的一半：6 ⁄ 2 = 3。将其平方：3² = 9。改写： (x + 3)² − 9 + 1 = 0，即 (x + 3)² − 8 = 0。

于是 (x + 3)² = 8，所以 x + 3 = ±√8，得 x = −3 ± √8。写成小数约为 x ≈ 1.83 或 x ≈ −7.83。

The vertex form y = a(x − h)² + k tells you that the turning point is at (h, k). For the example above, y = (x + 3)² − 8 has vertex at (−3, −8).

顶点式 y = a(x − h)² + k 告诉我们顶点坐标为 (h, k)。对于上述例子，y = (x + 3)² − 8 的顶点在 (−3, −8)。

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6. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root: Δ = b² − 4ac. 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部分：Δ = b² − 4ac。无需完整解方程，它就能告诉我们该二次方程有多少个实数根。

Value of Δ | Δ 的值	Nature of roots | 根的性质	Graphical meaning | 图像含义
b² − 4ac > 0	Two distinct real roots | 两个不等实数根	Curve crosses x-axis twice | 曲线与 x 轴有两个交点
b² − 4ac = 0	One repeated real root | 一个重根（两相等实数根）	Curve touches x-axis once | 曲线与 x 轴相切
b² − 4ac < 0	No real roots | 无实数根	Curve does not touch x-axis | 曲线与 x 轴无交点

Example: Determine the nature of the roots of x² − 4x + 4 = 0.

Δ = (−4)² − 4(1)(4) = 16 − 16 = 0. Therefore the equation has one repeated root: x = 2.

示例：判断方程 x² − 4x + 4 = 0 的根的性质。

Δ = (−4)² − 4(1)(4) = 16 − 16 = 0。因此方程有一个重根：x = 2。

Exam questions may ask you to find the value of k such that a quadratic has two equal roots. For example, if kx² + 6x + 3 = 0 has one repeated root, set the discriminant equal to zero: 6² − 4(k)(3) = 0, giving 36 − 12k = 0, so k = 3.

考试可能会要求你求 k 的值，使二次方程有两个相等实数根。例如，若 kx² + 6x + 3 = 0 有重根，则令判别式为零：6² − 4(k)(3) = 0，得 36 − 12k = 0，所以 k = 3。

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7. Quadratic Graphs | 二次函数图像

The graph of y = ax² + bx + c is a smooth curve called a parabola. If a > 0, the parabola is U-shaped with a minimum point; if a < 0, it is an upside-down U with a maximum point.

y = ax² + bx + c 的图像是一条平滑曲线，称为抛物线。若 a > 0，抛物线开口朝上，有最小值点；若 a < 0，开口朝下，有最大值点。

To sketch a quadratic graph, you need four key features:

• The y-intercept: substitute x = 0, giving y = c.
• The roots: solve ax² + bx + c = 0.
• The axis of symmetry: x = −b ⁄ (2a).
• The vertex: found using the axis of symmetry, or by completing the square.

绘制二次函数图像需要四个关键特征：

• y 截距：令 x = 0，得 y = c。
• 根：解方程 ax² + bx + c = 0。
• 对称轴：x = −b ⁄ (2a)。
• 顶点：通过对称轴求出，或用配方法求得。

Example: Sketch y = x² − 2x − 3, labelling all intercepts and the vertex.

The y-intercept is (0, −3). The roots solve x² − 2x − 3 = 0, giving (x − 3)(x + 1) = 0, so x = 3 and x = −1. The axis of symmetry is x = 2 ⁄ 2 = 1. Substituting x = 1 gives y = 1 − 2 − 3 = −4, so the vertex is (1, −4). Since a = 1 > 0, the parabola is U-shaped.

示例：画出 y = x² − 2x − 3 的草图，标出所有截距和顶点。

y 截距为 (0, −3)。根：解 x² − 2x − 3 = 0，得 (x − 3)(x + 1) = 0，所以 x = 3 或 x = −1。对称轴为 x = 2 ⁄ 2 = 1。将 x = 1 代入得 y = 1 − 2 − 3 = −4，因此顶点为 (1, −4)。由于 a = 1 > 0，抛物线开口向上。

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8. Roots, Vertex and Axis of Symmetry | 根、顶点与对称轴

The x-coordinate of the vertex is always the midpoint of the two real roots (if they exist). This is a powerful shortcut: for the equation y = x² − 6x + 5, the roots are x = 1 and x = 5, so the vertex lies exactly halfway, at x = 3.

顶点坐标的 x 值始终是两个实数根的中点（如果根存在）。这是一个非常实用的技巧：对于 y = x² − 6x + 5，根为 x = 1 和 x = 5，因此顶点恰好位于中点 x = 3 处。

To find the y-coordinate of the vertex, substitute the x-coordinate back into the original equation: y = 3² − 6(3) + 5 = 9 − 18 + 5 = −4. Thus the vertex is (3, −4).

要求顶点的 y 坐标，将 x 坐标代回原函数：y = 3² − 6(3) + 5 = 9 − 18 + 5 = −4。因此顶点为 (3, −4)。

When the quadratic does not factorise, use the formula x = −b ⁄ (2a) for the axis of symmetry. This formula always works, even for equations with no real roots.

当二次式无法因式分解时，使用公式 x = −b ⁄ (2a) 求对称轴。这个公式总是有效的，即使方程没有实数根也适用。

Vertex form: y = a(x − h)² + k → vertex at (h, k)

顶点式：y = a(x − h)² + k → 顶点为 (h, k)

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9. Word Problems | 应用题

IGCSE quadratic word problems usually involve areas, consecutive numbers, or projectile motion. The key steps are: define the variable, translate the problem into an equation, solve, and then check which solutions are sensible.

IGCSE 二次应用题通常涉及面积、连续整数或抛体运动。关键步骤是：设未知数、将问题转化为方程、求解，然后判断哪些解是合理的。

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

Let the width be x cm. Then the length is (x + 3) cm. The area is x(x + 3) = 28, so x² + 3x − 28 = 0. Factorise: (x + 7)(x − 4) = 0, giving x = −7 or x = 4. Since length cannot be negative, we reject x = −7. Therefore the width is 4 cm and the length is 7 cm.

示例：一个矩形的长比宽多 3 cm，面积为 28 cm²。求该矩形的尺寸。

设宽为 x cm，则长为 (x + 3) cm。面积为 x(x + 3) = 28，即 x² + 3x − 28 = 0。因式分解：(x + 7)(x − 4) = 0，得 x = −7 或 x = 4。由于长度不能为负数，舍去 x = −7。因此宽为 4 cm，长为 7 cm。

Always read the final sentence of the problem. A garden path question might ask for a distance, but a negative solution might correspond to a valid alternative such as distance measured in the opposite direction — however, in most IGCSE contexts, negative lengths and negative times are rejected.

务必仔细阅读题目的最后一句。例如花园小径问题可能要求距离，但负根可能对应相反方向的有效情形——不过在大多数 IGCSE 情境中，负长度和负时间都会被舍去。

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

The examiner reports show that students repeatedly lose marks on quadratics for five main reasons.

考官报告显示，学生在二次方程问题上反复失分主要源于五个原因。

• Not rearranging to standard form: Solving 2x² = 3x + 1 as if it were linear is a fatal error. Always write ax² + bx + c = 0 first.
• Sign errors in the quadratic formula: Write out every substitution. Pay special attention when b or c is negative.
• Incorrect factorisation: Check your factors by expanding them back. (x + 2)(x + 3) = x² + 5x + 6, not x² + 6x + 5.
• Forgetting to solve after factorising: After writing (x − 4)(x + 5) = 0, you must state x = 4 or x = −5. The factorised form alone does not earn full marks.
• Ignoring the context: In word problems, reject solutions that do not make sense, and always include units in your final answer.

• 未化为标准形式：把 2x² = 3x + 1 当作一次方程来解是致命错误。务必先写成 ax² + bx + c = 0。
• 求根公式中的符号错误：完整写出每一次代入。当 b 或 c 为负数时尤其要小心。
• 因式分解错误：通过展开来检查你的因式。(x + 2)(x + 3) = x² + 5x + 6，而不是 x² + 6x + 5。
• 忘记分解后继续求解：写出 (x − 4)(x + 5) = 0 之后，必须说明 x = 4 或 x = −5。仅写出因式形式无法拿到满分。
• 忽略实际情境：在应用题中，要舍去不合理的解，并在最终答案中注明单位。

Additional exam tips: if a question asks for answers “correct to 2 decimal places”, use the quadratic formula or complete the square — factorisation will not produce decimals. If the discriminant is negative, write “no real roots” rather than trying to factorise. Sketch graphs neatly and label the vertex, intercepts, and axis of symmetry; many marks in graph questions are given for correct labelling.

更多应试技巧：如果题目要求答案”精确到小数点后两位”，应使用求根公式或配方法——因式分解不会得到小数结果。若判别式为负，直接写”无实数根”，不要强行分解。画图时要整洁，并标出顶点、截距和对称轴；图像题中很多分数来自正确的标注。

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Quadratic equations appear in every IGCSE paper, sometimes twice. Master the four solving methods — factorisation, the formula, completing the square, and graphical reasoning — and you will handle any question the examiner can ask. Practise with past papers and always check your solutions by substituting them back into the original equation.

一元二次方程在每一份 IGCSE 试卷中都会出现，有时甚至出现两次。熟练掌握四种求解方法——因式分解法、求根公式、配方法和图像分析——你就能轻松应对任何题目。请用历年真题练习，并始终将解代回原方程进行验算。

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