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Quadratic Equations: Solving, Graphing and Applications | 二次方程：求解、图像与应用

2026-08-21 15:51 · 本文部分内容由AI辅助创作

📚 Quadratic Equations: Solving, Graphing and Applications | 二次方程：求解、图像与应用

Quadratic equations appear throughout the IGCSE and A-Level Mathematics syllabus, from factorising simple trinomials to interpreting curved graphs in real-life problems. This revision guide brings together the methods, the graphs and the applications you need in order to answer both core and extended questions confidently.

二次方程贯穿 IGCSE 与 A-Level 数学考纲，从简单的三项式因式分解到结合实际问题的曲线图像都会考查。本复习指南汇总了考试所需的解法、图像与应用，帮助你更自信地应对核心卷和扩展卷题目。

You will meet five main techniques: factorisation, completing the square, the quadratic formula, reading the discriminant, and sketching the parabola. Each technique has a clear purpose, and examiners often test whether you can choose the most efficient one for a given question.

你需要掌握五种主要方法：因式分解、配方法、求根公式、判别式的判断以及抛物线草图。每种方法都有明确用途，考官常常考查你是否能为特定题目选择最高效的方法。

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

lways be written in the form a x² + b x + c = 0, where a, b and c are constants and a ≠ 0.

二次方程是次数为 2 的多项式方程。在一元情形下，它总可以写成 a x² + b x + c = 0 的形式，其中 a、b、c 为常数且 a ≠ 0。

a x² + b x + c = 0, a ≠ 0

r, such as b x + c = 0, which has at most one solution and behaves completely differently.

a ≠ 0 这个条件很关键。如果 a 为零，x² 项就消失了，方程退化为一次方程，例如 b x + c = 0，它最多只有一个解，性质完全不同。

. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.

y = a x² + b x + c 的图像是一条抛物线。若 a > 0，抛物线开口向上并有一个最低点；若 a < 0，抛物线开口向下并有一个最高点。

The magnitude of a controls the width of the curve: a large positive or negative value of |a| one. The constant c always gives the y-intercept, because substituting x = 0 leaves y = c.

a 的大小控制曲线的宽窄：|a| 越大，抛物线越窄；|a| 越小，抛物线越宽而平缓。常数 c 总是给出 y 轴截距，因为代入 x = 0 后得到 y = c。

2. Standard Form and Key Vocabulary | 标准形式与关键术语

− 6 must first be written as x² − 5x + 6 = 0 by subtracting 5x and adding 6 to both sides.

求解前必须先把二次方程整理成标准形式。例如 x² = 5x − 6 要先在两边同时减去 5x、加上 6，写成 x² − 5x + 6 = 0。

The coefficient a is called the leading coefficient, b is the linear coefficient and c is the constant term. Getting these three values correct is the foundation of every method that follows.

系数 a 称为二次项系数（首项系数），b 是一次项系数，c 是常数项。把这三个数值写对，是后面所有方法的基础。

The solutions of a quadratic equation are called its roots. On a graph, the roots are the x-coordinates where the parabola crosses the x-axis. These points are also called x-intercepts or zeros.

二次方程的解称为根。在图像上，根就是抛物线与 x 轴交点的横坐标。这些点也叫 x 轴截距或零点。

Always check that the right-hand side is zero before factorising or using the formula. This is the most common first step in IGCSE quadratic questions and the source of many lost marks.

在因式分解或使用求根公式之前，一定要检查等式右边是否为零。这是 IGCSE 二次方程题中最常见的第一步，也是很多失分点的来源。

3. Solving by Factorisation | 因式分解法

x² − 5x + 6 = 0, you need two numbers that multiply to give c = 6 and add to give b = −5.

当根是整数或简单分数时，因式分解是最快的方法。对于 x² − 5x + 6 = 0，你需要找两个数，使它们的乘积等于 c = 6，和等于 b = −5。

cause (−2) × (−3) = 6 and (−2) + (−3) = −5. Hence the equation becomes (x − 2)(x − 3) = 0.

这两个数是 −2 和 −3，因为 (−2) × (−3) = 6，且 (−2) + (−3) = −5。于是方程化为 (x − 2)(x − 3) = 0。

(x − 2)(x − 3) = 0 → x = 2 or x = 3

t least one factor must be zero. So either x − 2 = 0 or x − 3 = 0, giving x = 2 and x = 3.

零积法则指出：若两个因式的乘积为零，则至少有一个因式为零。所以要么 x − 2 = 0，要么 x − 3 = 0，得到 x = 2 和 x = 3。

ch are 1 and 6), then write 2x² + x + 6x + 3 = 0 and group: x(2x + 1) + 3(2x + 1) = 0, so (2x + 1)(x + 3) = 0 and x = −½ or x = −3.

当 a 不等于 1 时，可拆分中间项再分组分解。对 2x² + 7x + 3 = 0，先算 a 与 c 的乘积得 6，找乘积为 6、和为 7 的两个数（即 1 和 6），再写成 2x² + x + 6x + 3 = 0 并分组：x(2x + 1) + 3(2x + 1) = 0，于是 (2x + 1)(x + 3) = 0，得 x = −½ 或 x = −3。

wo squares such as x² − 25 = (x − 5)(x + 5). Recognising these saves a great deal of time.

还要注意特殊结构：公因式型如 3x² − 12x = 3x(x − 4)；平方差型如 x² − 25 = (x − 5)(x + 5)。识别这些结构可以节省大量时间。

4. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + p)² + q. It is the standard method for finding the vertex exactly, and it is also required when a question asks for a solution in surd form.

配方法把二次式改写为 a(x + p)² + q 的形式。它是精确求顶点的标准方法，当题目要求以根式（无理数）形式给出解时也必须使用。

x² + b x + c, take half of b, square it, and adjust: x² + b x + c = (x + b/2)² + c − b²/4.

对于 x² + b x + c，取 b 的一半、平方后再调整：x² + b x + c = (x + b/2)² + c − b²/4。

x² + 6x = (x + 3)² − 9. The equation becomes (x + 3)² − 9 − 7 = 0, that is (x + 3)² = 16.

例：用配方法解 x² + 6x − 7 = 0。6 的一半是 3，3² = 9，所以 x² + 6x = (x + 3)² − 9。方程变为 (x + 3)² − 9 − 7 = 0，即 (x + 3)² = 16。

x = −7. Notice that you must write the ± symbol; forgetting it loses one of the two roots.

两边开平方得 x + 3 = ±4，所以 x = 1 或 x = −7。注意必须写出 ± 符号，漏掉它就会丢掉两个根中的一个。

If the solutions are not rational, leave them in surd form, for example x = −3 ± √16 becomes x = 1 or x = −7, while x = 2 ± √5 stays as it is. Exact surd answers are normally preferred unless the question specifies a decimal accuracy.

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如果解不是有理数，就用根式保留，例如 x = 2 ± √5 保持原样即可。除非题目规定了小数位数，通常优先给出精确的根式答案。

5. The Quadratic Formula | 求根公式

factorises. It is derived by completing the square on the general form a x² + b x + c = 0.

求根公式可以解任何二次方程，无论它能否因式分解。它正是对一般形式 a x² + b x + c = 0 配方推导出来的。

x = (−b ± √b² − 4ac) / (2a)

For x² − 5x + 6 = 0 we have a = 1, b = −5 and c = 6. Substituting: x = (5 ± √25 − 24) / 2 = (5 ± 1) / 2, so x = 3 or x = 2, matching the factorised answer.

对 x² − 5x + 6 = 0，有 a = 1、b = −5、c = 6。代入得 x = (5 ± √25 − 24) / 2 = (5 ± 1) / 2，所以 x = 3 或 x = 2，与因式分解的结果一致。

Give non-exact answers to the accuracy stated, usually 2 or 3 significant figures, and show the substitution line so that method marks can be awarded even if the arithmetic slips.

非精确答案要按题目要求取有效数字，通常是 2 位或 3 位，并写出代入过程，这样即使算错也能拿到方法分。

6. The Discriminant | 判别式

The expression under the square root, b² − 4ac, is called the discriminant and is often written as Δ. It tells you how many real roots the equation has, without solving it.

根号下的表达式 b² − 4ac 称为判别式，常记作 Δ。它能在不解方程的情况下告诉你方程有几个实根。

Discriminant 判别式Roots 根的情况Graph 图像
Δ > 0Two distinct real roots 两个不同实根Crosses the x-axis twice 与 x 轴交于两点
Δ = 0One repeated root 一个重根Touches the x-axis 与 x 轴相切
Δ < 0No real roots 无实根Never meets the x-axis 与 x 轴不相交

+ kx + 9 = 0 has equal roots. Equal roots mean Δ = 0, so k² − 36 = 0 and k = 6 or k = −6.

典型考题表述：求使 x² + kx + 9 = 0 有相等实根的 k 值。相等实根意味着 Δ = 0，即 k² − 36 = 0，所以 k = 6 或 k = −6。

If a question says "two distinct roots", write Δ > 0; if it says "no real roots", write Δ < 0. Always state the inequality or equation you are solving, not just the final value.

如果题目说 "有两个不同的根"，就写 Δ > 0；如果说 "没有实根"，就写 Δ < 0。务必写出你所解的不等式或方程，而不只是最终数值。

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

ures: the shape, the y-intercept, the x-intercepts and the vertex. A clear sketch with labelled points often carries marks on its own.

画 y = a x² + b x + c 的草图时，要找出四个要素：开口方向、y 轴截距、x 轴截距和顶点。标注清楚的草图本身常常就能得分。

line of symmetry lies halfway between them at x = 2.5, which also equals −b / (2a) = 5/2.

对 y = x² − 5x + 6，y 轴截距是 (0, 6)，x 轴截距是 x = 2 和 x = 3。对称轴位于两者中点 x = 2.5，这也等于 −b / (2a) = 5/2。

it is a minimum because a > 0. The whole parabola is symmetric about the line x = 2.5.

代入 x = 2.5 得 y = 6.25 − 12.5 + 6 = −0.25，所以顶点为 (2.5, −0.25)，因为 a > 0，它是极小值点。整条抛物线关于直线 x = 2.5 对称。

8. Vertex, Axis of Symmetry and Maximum Values | 顶点、对称轴与最值

ts the maximum or minimum value of the function. Its x-coordinate is always x = −b / (2a).

顶点是抛物线最重要的点，因为它代表函数的最大值或最小值。它的横坐标总是 x = −b / (2a)。

Vertex: (−b / (2a), c − b² / (4a))

Alternatively, the completed square form a(x + p)² + q gives the vertex directly as (−p, q). This is usually quicker than substituting into the formula, and it also reveals whether the turning point is a maximum or a minimum.

另一种方法是利用配方形式 a(x + p)² + q，直接读出顶点 (−p, q)。这通常比代入公式更快，而且还能看出转折点是最大值还是最小值。

(x − 2)² − 4] + 5 = 2(x − 2)² − 3. The vertex is (2, −3) and the minimum value of y is −3.

例：y = 2x² − 8x + 5。配方得 y = 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 3。顶点为 (2, −3)，y 的最小值是 −3。

e vertex is at x = −4 / (2 × (−1)) = 2, and y = −4 + 8 + 1 = 5, so the maximum value is 5.

当 a < 0 时，顶点给出的是最大值。对 y = −x² + 4x + 1，顶点横坐标 x = −4 / (2 × (−1)) = 2，纵坐标 y = −4 + 8 + 1 = 5，所以最大值是 5。

9. Applications and Word Problems | 应用题

Most applied quadratic questions follow the same routine: define a variable, form an equation from the given information, rearrange into standard form, solve, then interpret the answer in context and reject any impossible values.

大多数二次方程应用题都遵循同样的流程：设未知数，根据条件列方程，整理成标准形式，求解，再结合情境解释答案并舍去不合理的值。

= 40, so w² + 3w − 40 = 0, which factorises as (w + 8)(w − 5) = 0. Hence w = 5 or w = −8.

例（面积）：一个矩形的长比宽多 3 cm，面积是 40 cm²。设宽为 w，则 w(w + 3) = 40，即 w² + 3w − 40 = 0，因式分解得 (w + 8)(w − 5) = 0，所以 w = 5 或 w = −8。

th is 8 cm. Always write the rejection explicitly, because it earns a mark in most mark schemes.

长度不可能为负，因此舍去 w = −8。宽为 5 cm，长为 8 cm。一定要明确写出舍去过程，因为多数评分标准会为此给分。

ground when h = 0: 5t(4 − t) = 0, so t = 0 or t = 4. The ball is in the air for 4 seconds.

例（抛体运动）：某球被抛出后，t 秒时的高度为 h = 20t − 5t² 米。落地时 h = 0：5t(4 − t) = 0，得 t = 0 或 t = 4，即球在空中飞行 4 秒。

he greatest height is 20 metres. Profit and revenue problems work in exactly the same way.

最大高度出现在顶点处，位于两根的中点 t = 2。此时 h = 20(2) − 5(4) = 40 − 20 = 20，所以最大高度为 20 米。利润与收入类问题的处理方式完全相同。

10. Choosing a Method and Exam Technique | 方法选择与考试技巧

Choose factorisation first when the coefficients are small integers; use the formula when factorisation fails or when the question asks for significant figures; use completing the square when the vertex, the maximum value or an exact surd answer is required.

系数是较小整数时优先尝试因式分解；分解不出来或题目要求有效数字时用求根公式；需要顶点、最值或精确根式答案时用配方法。