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Quadratic Equations for IGCSE Mathematics | IGCSE 数学二次方程精讲

2026-08-21 15:46 · 本文部分内容由AI辅助创作

📚 Quadratic Equations for IGCSE Mathematics | IGCSE 数学二次方程精讲

Quadratic equations appear in nearly every IGCSE Mathematics paper, from solving simple factorised forms to sketching graphs and applying quadratics in real-world problems. Whether you are sitting the Cambridge 0580, the Edexcel 4MA1 or the Oxford AQA 9260 specification, the same core skills are tested again and again: rearranging into standard form, choosing an efficient method, interpreting the discriminant, reading the graph, and communicating your answer with the precision the mark scheme demands. This revision guide brings together the essential methods, key formulas, graph interpretation, exam technique and the most common errors in one place, so that you can revise the whole topic in a single sitting and then return to individual sections before a mock.

二次方程几乎出现在每一套 IGCSE 数学试卷中，从简单的因式分解求解，到图像草图，再到二次函数在实际问题中的应用，都属于高频考点。无论你参加的是剑桥 0580、爱德思 4MA1 还是牛津 AQA 9260，考查的核心能力始终相同：整理成标准形式、选择高效解法、解读判别式、读取图像信息，并按照评分标准的要求规范书写答案。本复习指南把核心方法、关键公式、图像解读、应试技巧和常见错误集中讲解，帮助考生用一次通读建立完整框架，之后再做专项查漏补缺。

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

d a ≠ 0. The highest power of the unknown x is 2, which gives the equation its name; the word itself comes from the Latin quadratus, meaning 'made square' the axis of symmetry, and c gives the y-intercept of the graph. At IGCSE level you will meet quadratics written in many disguises: expanded, factorised, completed-square, or embedded in a word problem about areas and speeds.

一元二次方程是指可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 是常数且 a ≠ 0。未知数 x 的最高次数为 2，因此称为二次方程。系数 a 决定抛物线开口方向和形状，b 影响对称轴的位置，c 则给出图像与 y 轴的交点。在 IGCSE 阶段，二次方程会以多种形式出现：展开式、因式分解式、配方式，或隐藏在面积、速度等文字题中，所以识别"它是不是二次方程"是解题的第一步。

becomes linear, so the condition a ≠ 0 is essential; a question may deliberately test whether you notice that a term has vanished and the equation is no longer quadratic.

例如 x² + 5x + 6 = 0、2x² − 3x + 1 = 0 和 x² − 9 = 0 都是二次方程。如果 a = 0，方程就退化为一次方程，因此 a ≠ 0 这个条件非常重要；有些题目会故意让二次项系数消失，考查你能不能意识到此时已经不能用二次方程的结论。

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

erms. For example, 3x² = 5x − 2 becomes 3x² − 5x + 2 = 0, and x(x + 4) = 12 becomes x² + 4x − 12 = 0. Working in standard form makes it far easier to read off a, b and c, which you need for the formula and the discriminant. A sign slip at this stage is one of the most expensive mistakes in the whole paper, because every later line inherits the error.

在求解之前，必须先把方程整理成标准形式 ax² + bx + c = 0。把所有项移到同一侧，通常移到左边，并合并同类项。例如 3x² = 5x − 2 要写成 3x² − 5x + 2 = 0，x(x + 4) = 12 要写成 x² + 4x − 12 = 0。标准形式便于直接读出 a、b、c，而这两个（三个）数值是使用求根公式和判别式的前提。这一步的符号错误代价极高，因为后面每一步都会继承这个错误。

Key vocabulary you should be able to define precisely:

以下术语必须能够准确表述：

3. Solving by Factorisation | 因式分解法求解

ply to give c and add to give b. For x² + 5x + 6 = 0, the pair 2 and 3 works, so the equation becomes (x + 2)(x + 3) = 0. Since a product is zero only when one factor is zero, x + 2 = 0 or x + 3 = 0, giving x = −2 or x = −3. Always set each bracket equal to zero explicitly if the question is worth more than one mark.

因式分解通常是最快的方法，当数字比较"友好"时，也是考官最希望看到的方法。思路是找到两个数，它们的乘积等于 c、和等于 b。对于 x² + 5x + 6 = 0，2 和 3 满足条件，于是方程变为 (x + 2)(x + 3) = 0。因为乘积为零当且仅当某个因式为零，所以 x + 2 = 0 或 x + 3 = 0，得到 x = −2 或 x = −3。如果题目分值超过 1 分，务必显式写出"令每个括号等于零"这一步。

2 and −1, then split the middle term: 2x² − 2x − x + 1 = 0, giving 2x(x − 1) − 1(x − 1) = 0, so (2x − 1)(x − 1) = 0 and x = ½ or x = 1. The difference of two squares is a special case worth memorising: x² − 9 = (x − 3)(x + 3), so x² − 9 = 0 has roots x = 3 and x = −3.

当 a ≠ 1 时，观察法需要调整。对于 2x² − 3x + 1 = 0，先算 a 与 c 的乘积得 2，再找两个数乘积为 2、和为 −3，即 −2 和 −1，然后拆分中间项：2x² − 2x − x + 1 = 0，得到 2x(x − 1) − 1(x − 1) = 0，即 (2x − 1)(x − 1) = 0，所以 x = ½ 或 x = 1。平方差是必须牢记的特殊情形：x² − 9 = (x − 3)(x + 3)，因此 x² − 9 = 0 的根是 x = 3 和 x = −3。

4. The Quadratic Formula | 求根公式

When factorisation fails, or when the coefficients are awkward decimals or surds, use the quadratic formula. It solves every quadratic equation that has real roots and is printed on most IGCSE formula sheets, but you must still know how to substitute correctly. Write down a, b and c first, then substitute in one clean line, then evaluate the discriminant, then take the square root.

当因式分解行不通，或者系数是讨厌的小数、根式时，就要使用求根公式。它能解出所有有实数根的二次方程，多数 IGCSE 试卷的公式表上也会给出，但正确代入仍是你自己的责任。建议先写出 a、b、c 的值，再一次性代入，然后计算判别式，最后开平方。

x = ( −b ± √b² − 4ac ) ÷ 2a

For 2x² + 7x − 4 = 0 we have a = 2, b = 7 and c = −4, so b² − 4ac = 49 + 32 = 81 and √81 answers, and how the negative constant term turns the subtraction into an addition. Round only at the very end: if the question says 'give your answers correct to 3 significant figures' carry full accuracy through the working and round once, or you may lose the accuracy mark.

例如 2x² + 7x − 4 = 0 中，a = 2、b = 7、c = −4，所以 b² − 4ac = 49 + 32 = 81，√81 = 9。于是 x = (−7 ± 9) ÷ 4，得到 x = ½ 或 x = −4。注意 ± 会给出两个答案，而负的常数项会让减法变成加法。只在最后一步取近似：如果题目要求"答案保留三位有效数字"，就要在过程中保持完整精度，最后只舍入一次，否则可能丢掉精度分。

5. Completing the Square | 配方法

Completing the square rewrites a quadratic as a perfect square plus a constant, which is indispensable for finding turning points and for proving that a quadratic is always positive. For an expression x² + bx + c, take half of b, square it, add and subtract it, then factorise the first three terms.

配方法把二次式改写成一个完全平方加上一个常数，这在求顶点和证明二次式恒为正时不可或缺。对于 x² + bx + c，取 b 的一半、平方，加上再减去它，然后把前三项写成完全平方。

x² + bx + c = (x + b/2)² + c − (b/2)²

eached when x = −3, so the turning point is (−3, −8). If the coefficient of x² is not 1, factor it out of the first two terms before completing the square, as in 2x² − 8x + 3 = 2(x² − 4x) + 3 = 2[(x − 2)² − 4] + 3 = 2(x − 2)² − 5.

以 x² + 6x + 1 为例，6 的一半是 3，3² = 9，所以 x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8。由于 (x + 3)² 永远非负，该式的最小值为 −8，在 x = −3 时取到，因此顶点是 (−3, −8)。如果 x² 的系数不是 1，要先把系数从前两项中提出来，例如 2x² − 8x + 3 = 2(x² − 4x) + 3 = 2[(x − 2)² − 4] + 3 = 2(x − 2)² − 5。

6. The Discriminant and the Number of Roots | 判别式与根的个数

The discriminant is the expression under the square root in the quadratic formula. Its sign alone tells you how many real solutions the equation has, which is exactly what many exam questions ask for, often without requiring you to solve anything.

判别式就是求根公式中根号下的表达式。仅凭它的符号就能判断方程有多少个实数解，很多考题正是这样提问的，往往并不需要你真正把方程解出来。

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Δ = b² − 4ac

Value of ΔNumber of real rootsGraph behaviour
Δ > 0Two distinct real rootsCurve crosses the x-axis twice
Δ = 0One repeated rootCurve just touches the x-axis
Δ < 0No real rootsCurve never meets the x-axis

o real solutions and the graph lies entirely above the x-axis. A very common exam style is 'find the values of k for which x² + kx + 9 = 0 has equal roots' state the condition you are using before solving, because that is often a mark by itself.

例如 x² − 4x + 5 = 0 的判别式 Δ = 16 − 20 = −4 < 0，所以没有实数解，图像完全位于 x 轴上方。常见考法是"求 k 的值，使 x² + kx + 9 = 0 有两个相等的根"。令 Δ = 0 得 k² − 36 = 0，所以 k = 6 或 k = −6。解之前先写出所依据的条件，因为这一句本身往往就是一分。

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

The larger the magnitude of a, the narrower the curve. The y-intercept is always c, found by putting x = 0, and the x-intercepts are the roots of the equation ax² + bx + c = 0. A sketch needs only the shape, the intercepts and the turning point clearly labelled.

y = ax² + bx + c 的图像是一条抛物线。若 a > 0，开口向上，有最低点；若 a < 0，开口向下，有最高点。a 的绝对值越大，图像越"瘦"。y 轴截距恒为 c，只需令 x = 0 即可得到；x 轴截距则是方程 ax² + bx + c = 0 的根。草图只需画对开口方向，并清楚标注两个截距和顶点。

ck into the equation to find the y-coordinate of the vertex. For y = x² − 6x + 5 the axis is x = 3, the vertex is (3, −4), and the roots are x = 1 and x = 5, so the sketch is fully determined by three numbers.

对称轴是位于两个根正中间的竖直线，总可以写成 x = −b ÷ 2a。求出这个 x 值后，代回原式即可得到顶点的 y 坐标。比如 y = x² − 6x + 5，对称轴是 x = 3，顶点是 (3, −4)，根是 x = 1 和 x = 5，一张完整的图像就由这三个数字确定了。

8. Roots, Intercepts and the Turning Point | 根、截距与顶点

e y-intercept is the single point where x = 0, so it is simply the value of c. The turning point is the vertex, an ordered pair (x, y), and it is where the curve changes direction.

考生常把这三个概念混在一起，要在脑中分清：根是使 y = 0 的 x 值，通过解方程得到；y 轴截距是 x = 0 时的那一个点，直接就是 c；顶点是一个有序数对 (x, y)，是曲线改变方向的位置。

If a question asks you to 'use your graph to solve x² − 6x + 5 = 2' s the curve, and give those readings to the accuracy requested, usually one decimal place.

如果题目要求"利用图像解 x² − 6x + 5 = 2"，就不要用代数方法求解。画出（或使用）直线 y = 2，读出它与抛物线交点的 x 值，并按题目要求给出精度，通常是一位小数。

9. Quadratic Inequalities | 二次不等式

ola, then read off the region that satisfies the inequality. Solving x² − x − 6 = 0 gives x = 3 and x = −2, and since the parabola opens upwards, the curve dips below the x-axis between the roots. The solution is therefore −2 < x < 3.

求解形如 x² − x − 6 < 0 的二次不等式分三步：先解对应方程得到临界值，再画出抛物线草图，最后读出满足不等式的区间。解 x² − x − 6 = 0 得 x = 3 和 x = −2；因为抛物线开口向上，在两个根之间曲线位于 x 轴下方，所以解集是 −2 < x < 3。

Take care with the direction of the sign. If the inequality had been x² − x − 6 > 0, the answer would be x < −2 or x > 3, written as two separate ranges. A useful memory aid is 'outside for greater, inside for less' when a > 0. If the coefficient a is negative, multiply through by −1 and reverse the inequality sign before you begin, which avoids almost every sign error.

一定要注意不等号方向。若题目是 x² − x − 6 > 0，答案就是 x < −2 或 x > 3，必须写成两个区间。当 a > 0 时，可以记成"大于取两边，小于取中间"。如果 a 是负数，先在两边同乘 −1 并改变不等号方向再开始，这样可以避免几乎所有符号错误。

10. Simultaneous Equations: One Linear, One Quadratic | 联立方程：一次与二次

When a straight line meets a parabola, the intersection points are found by solving the linear equation and the quadratic equation simultaneously. Rearrange the linear equation to make x or y the subject, substitute it into the quadratic, and solve the resulting quadratic equation. The number of intersection points equals the number of real roots of that quadratic, which links neatly back to the discriminant.

当直线与抛物线相交时，交点就是这两个方程联立的解。先把一次方程整理成 x 或 y 的表达式，代入二次方程，再解得到的一元二次方程。交点的个数等于这个二次方程实数根的个数，这正好与判别式联系起来。

−1. Substituting back gives the points (2, 3) and (−1, 0). Always give your final answer as coordinate pairs, and always substitute into the simpler, linear equation to save time and reduce error.

例如 y = x + 1 与 y = x² − 1，代入得 x + 1 = x² − 1，即 x² − x − 2 = 0，所以 x = 2 或 x = −1，回代得到交点 (2, 3) 和 (−1, 0)。最终答案一定要写成坐标对，并且优先代回较简单的一次方程，既省时间又能减少错误。

11. Real-World Applications and Word Problems | 实际应用与文字题

IGCSE papers frequently wrap quadratics inside a context: the area of a rectangle, the height of a projectile, profit maximisation, or a number problem. The mathematics never changes, but the setting adds two extra demands. First, you must translate the words into an equation. Second, you must decide whether every algebraic solution makes physical sense.

IGCSE 试题常把二次方程放在实际背景中：矩形面积、抛体高度、利润最大化或数字问题。数学本身没有变化，但情境带来两个额外要求：一是把文字翻译成方程，二是判断每个代数解在现实中是否合理。

A length cannot be negative, so the width is 5 cm and the length 8 cm. This rejection step is a marked part of the answer; state the reason explicitly rather than simply deleting the negative root.

例如：一个矩形的长比宽多 3 cm，面积是 40 cm²。设宽为 x，则 x(x + 3) = 40，即 x² + 3x − 40 = 0，(x + 8)(x − 5) = 0，所以 x = 5 或 x = −8。长度不能为负，因此宽是 5 cm，长是 8 cm。这个"舍去"步骤是有分的，要明确写出理由，而不是直接把负根划掉。

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

Most lost marks in this topic come from a small number of repeated slips rather than from genuine misunderstanding. Reading through the list below before a mock is one of the highest-value uses of ten minutes you can make.

这个主题上丢分，多数并非源于不理解，而是反复踩进同样的陷阱。考前花十分钟过一遍下面的清单，是性价比极高的复习方式。