Mastering Quadratic Equations | 二次方程完全攻略

📚 Mastering Quadratic Equations | 二次方程完全攻略

Quadratic equations appear in almost every IGCSE Mathematics paper. Whether you are aiming for a grade 4 or a grade 9, understanding how to solve and interpret equations of the form ax² + bx + c = 0 is essential. This guide covers every method you need, explains when to use each one, and highlights the common traps that cost students marks.

二次方程几乎出现在每一份 IGCSE 数学试卷中。无论你的目标是4分还是9分，理解并求解形如 ax² + bx + c = 0 的方程都是至关重要的。本指南涵盖你所需的每一种解法，说明每种方法的适用时机，并指出学生最常失分的陷阱。

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

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

二次方程是未知变量最高次数为2的多项式方程。其一般形式为 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。若 a = 0，方程就变成一次方程，而不是二次方程。

Examples of quadratic equations include x² – 5x + 6 = 0, 2x² + 3x – 1 = 0, and x² = 9. The first two are already in standard form, while the third can be rearranged to x² – 9 = 0.

二次方程的例子包括 x² – 5x + 6 = 0、2x² + 3x – 1 = 0 和 x² = 9。前两个已是标准形式，第三个可以改写为 x² – 9 = 0。

Notice that a quadratic equation always contains an x² term. It may or may not contain an x term or a constant term. For example, x² – 4 = 0 has no x term, and x² – 3x = 0 has no constant term, but both are perfectly valid quadratics.

注意，二次方程必定含有 x² 项，可能含有也可能不含有 x 项或常数项。例如，x² – 4 = 0 没有 x 项，x² – 3x = 0 没有常数项，但它们都是完全合法的二次方程。

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

Before solving any quadratic equation, you should rearrange it into the standard form ax² + bx + c = 0. This means all terms are on one side, and the other side is zero. Expanding brackets and collecting like terms are often necessary first steps.

在求解任何二次方程之前，你应先将其整理成标准形式 ax² + bx + c = 0。这意味着所有项都在等号一侧，另一侧为零。展开括号并合并同类项通常是必要的第一步。

For the equation 2x² – 4x = 6, subtract 6 from both sides to obtain 2x² – 4x – 6 = 0. Here a = 2, b = -4 and c = -6. Remember that the signs of b and c are part of their values.

对于方程 2x² – 4x = 6，两边同时减去6，得到 2x² – 4x – 6 = 0。此时 a = 2，b = -4，c = -6。切记 b 和 c 的正负号是其数值的一部分。

If an equation contains brackets, such as x(x + 3) = 10, expand first to get x² + 3x = 10, then rearrange to x² + 3x – 10 = 0. If it contains fractions, multiply through by the common denominator to clear them before solving.

如果方程含有括号，如 x(x + 3) = 10，先展开得到 x² + 3x = 10，再整理为 x² + 3x – 10 = 0。如果方程含有分数，先乘以公分母将其化为整数，再进行求解。

3. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when it works. You need to find two numbers whose product equals the constant term c and whose sum equals the coefficient b. For x² + 7x + 12 = 0, look for two numbers multiplying to 12 and adding to 7. The numbers 3 and 4 work, so the equation becomes (x + 3)(x + 4) = 0.

因式分解法是可行时最快捷的方法。你需要找到两个数，使它们的乘积等于常数项 c，和等于一次项系数 b。对于 x² + 7x + 12 = 0，寻找两个乘积为12、和为7的数。3 和 4 满足条件，因此方程化为 (x + 3)(x + 4) = 0。

Since the product is zero, at least one factor must be zero. This gives x + 3 = 0 or x + 4 = 0, so x = -3 or x = -4. Always write both solutions clearly, and then substitute them back to check.

由于乘积为零，至少有一个因式为零。由此得到 x + 3 = 0 或 x + 4 = 0，因此 x = -3 或 x = -4。务必清楚写出两个解，并代回原方程进行检验。

When the coefficient of x² is not 1, such as 2x² + 5x – 3 = 0, the process is slightly harder. One way is to find two numbers whose product is a × c = 2 × (-3) = -6 and whose sum is b = 5. The numbers 6 and -1 work. Rewrite the middle term: 2x² + 6x – x – 3 = 0, then factor by grouping.

当 x² 的系数不为1时，例如 2x² + 5x – 3 = 0，过程稍微复杂。一种方法是寻找两个数，使它们的乘积等于 a × c = 2 × (-3) = -6，和为 b = 5。6 和 -1 满足条件。改写中间项：2x² + 6x – x – 3 = 0，然后分组因式分解。

Group the first two terms and the last two terms: 2x(x + 3) – 1(x + 3) = 0, which gives (2x – 1)(x + 3) = 0. The solutions are x = 1/2 or x = -3. With practice, you can write this directly without rewriting the middle term.

将前两项和后两项分组：2x(x + 3) – 1(x + 3) = 0，得到 (2x – 1)(x + 3) = 0。解为 x = 1/2 或 x = -3。多加练习后，你可以直接写出结果，而不必重新改写中间项。

4. Solving by Completing the Square | 配方法

Completing the square is a powerful technique that also helps you find the turning point of a quadratic graph. The idea is to rewrite x² + bx as a perfect square plus or minus a constant. For x² + 6x + 2 = 0, halve 6 to get 3, then write (x + 3)² – 9 + 2 = 0.

配方法是一种强大的技巧，它还能帮助你找到二次函数图像的顶点。其核心思想是把 x² + bx 改写成一个完全平方加上或减去某个常数。对于 x² + 6x + 2 = 0，将6取半得到3，然后写成 (x + 3)² – 9 + 2 = 0。

Simplify to (x + 3)² – 7 = 0. Then add 7 to both sides: (x + 3)² = 7. Take the square root of both sides, remembering the ± sign: x + 3 = ±√7. Finally, subtract 3 to get x = -3 ± √7.

化简得 (x + 3)² – 7 = 0。两边同时加7： (x + 3)² = 7。两边开平方，切记 ± 符号：x + 3 = ±√7。最后两边减去3，得到 x = -3 ± √7。

The general rule is that x² + bx = (x + b/2)² – (b/2)². For 2x² – 8x + 1 = 0, first divide by 2 to make the coefficient of x² equal to 1: x² – 4x + 1/2 = 0. Then complete the square: (x – 2)² – 4 + 1/2 = 0.

一般规则是 x² + bx = (x + b/2)² – (b/2)²。对于 2x² – 8x + 1 = 0，先除以2使 x² 的系数变为1：x² – 4x + 1/2 = 0。然后配方：(x – 2)² – 4 + 1/2 = 0。

This simplifies to (x – 2)² = 7/2, so x – 2 = ±√(7/2), and therefore x = 2 ± √(7/2). Leave your answer in surd form unless the question asks for a decimal approximation.

这化简为 (x – 2)² = 7/2，因此 x – 2 = ±√(7/2)，所以 x = 2 ± √(7/2)。除非题目要求小数近似值，否则保留根式形式即可。

5. The Quadratic Formula | 求根公式

The quadratic formula works for every quadratic equation, including those that cannot be factorised. It is derived from completing the square and is given as follows:

求根公式适用于所有二次方程，包括那些无法因式分解的方程。它由配方法推导而来，公式如下：

x = (-b ± √(b² – 4ac)) / 2a

To use the formula, write down the values of a, b and c from the standard form, then substitute them carefully. For 2x² + 3x – 5 = 0, we have a = 2, b = 3 and c = -5. Substituting gives x = (-3 ± √(3² – 4 × 2 × (-5))) / (2 × 2).

使用公式时，先从标准形式中写出 a、b、c 的值，然后仔细代入。对于 2x² + 3x – 5 = 0，a = 2，b = 3，c = -5。代入得 x = (-3 ± √(3² – 4 × 2 × (-5))) / (2 × 2)。

Calculate the discriminant first: b² – 4ac = 9 – (-40) = 49. Then x = (-3 ± 7) / 4, which gives x = 1 or x = -5/2. The formula is reliable, but a single sign error can ruin the whole answer, so check your substitution twice.

先计算判别式：b² – 4ac = 9 – (-40) = 49。然后 x = (-3 ± 7) / 4，得到 x = 1 或 x = -5/2。公式本身可靠，但一个符号错误就可能毁掉整个答案，所以请复查两次代入过程。

On the IGCSE exam, you may be asked to give answers correct to 2 or 3 decimal places. Use your calculator carefully and write down more digits than needed before rounding, so that your final answer is accurate.

在 IGCSE 考试中，题目可能要求答案精确到2位或3位小数。请仔细使用计算器，在四舍五入前多写下几位数字，以确保最终答案准确。

6. The Discriminant: How Many Roots? | 判别式：方程有几个根

The expression b² – 4ac is called the discriminant, usually denoted by the letter Δ. It tells you how many real roots the equation has without solving it. If Δ > 0, there are two distinct real roots. If Δ = 0, there is exactly one repeated root. If Δ < 0, there are no real roots.

表达式 b² – 4ac 称为判别式，通常用希腊字母 Δ 表示。它能在不求解方程的情况下告诉你方程有几个实数根。若 Δ > 0，方程有两个不相等的实数根；若 Δ = 0，方程有一个重根；若 Δ < 0，方程没有实数根。

For the equation x² – 6x + 9 = 0, the discriminant is (-6)² – 4 × 1 × 9 = 36 – 36 = 0, so the equation has one repeated root. Indeed, x² – 6x + 9 = (x – 3)², giving x = 3 twice.

对于方程 x² – 6x + 9 = 0，判别式为 (-6)² – 4 × 1 × 9 = 36 – 36 = 0，所以方程有一个重根。事实上，x² – 6x + 9 = (x – 3)²，得到 x = 3 这个二重根。

Examiners often ask questions such as “Find the values of k for which the equation kx² + 4x + k = 0 has two distinct roots.” Here you use the condition b² – 4ac > 0 and solve the resulting inequality. With a = k, b = 4 and c = k, we get 16 – 4k² > 0, so -2 < k < 2.

考官经常出这样的题目：”求 k 的值，使方程 kx² + 4x + k = 0 有两个不相等的根。” 此时你使用条件 b² – 4ac > 0 并解所得不等式。代入 a = k，b = 4，c = k，得到 16 – 4k² > 0，因此 -2 < k < 2。

7. Solving Word Problems | 解应用题

Quadratic equations often arise from geometry, area, and motion problems. The key is to translate the words into an equation carefully. Always define the unknown variable clearly at the start, and check whether both solutions make sense in the context.

二次方程常出现在几何、面积和运动问题中。关键是把文字仔细地翻译成方程。开始时应清楚定义未知变量，并检查两个解在题目语境中是否都合理。

A rectangle has length 3 cm longer than its width, and its area is 28 cm². Let the width be w cm. Then the length is w + 3 cm, so w(w + 3) = 28. Expanding gives w² + 3w – 28 = 0, which factorises to (w + 7)(w – 4) = 0, so w = 4 or w = -7.

一个矩形的长比宽长3厘米，面积为28平方厘米。设宽为 w 厘米，则长为 w + 3 厘米，所以 w(w + 3) = 28。展开得 w² + 3w – 28 = 0，因式分解为 (w + 7)(w – 4) = 0，因此 w = 4 或 w = -7。

Since a width cannot be negative, reject w = -7. The width is 4 cm and the length is 7 cm. Always conclude with a sentence that answers the original question, and mention any rejected solution with a reason.

由于宽度不能为负数，舍去 w = -7。宽为4厘米，长为7厘米。最后务必用一句话回答原题的问题，并说明舍弃某个解的理由。

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

A quadratic function y = ax² + bx + c produces a curve called a parabola. If a > 0, the parabola opens upwards and has a minimum turning point. If a < 0, it opens downwards and has a maximum turning point. The x-intercepts of the graph are exactly the roots of the equation ax² + bx + c = 0.

二次函数 y = ax² + bx + c 的图像是一条称为抛物线的曲线。若 a > 0，抛物线开口向上，有最小值点；若 a < 0，抛物线开口向下，有最大值点。图像与 x 轴的交点正是方程 ax² + bx + c = 0 的根。

The y-intercept is found by setting x = 0, giving y = c. The axis of symmetry is the vertical line x = -b/(2a), and the turning point lies on this axis. If you complete the square to write y = a(x – h)² + k, then the turning point is at (h, k).

令 x = 0，可得 y 轴截距为 c。对称轴是竖直线 x = -b/(2a)，顶点位于这条对称轴上。如果你用配方法把函数写成 y = a(x – h)² + k，那么顶点坐标为 (h, k)。

For y = x² – 4x + 3, completing the square gives y = (x – 2)² – 1. The turning point is (2, -1), the y-intercept is 3, and the roots are x = 1 and x = 3. Sketch the graph by plotting these key points and drawing a smooth symmetric curve through them.

对于 y = x² – 4x + 3，配方得到 y = (x – 2)² – 1。顶点为 (2, -1)，y 轴截距为3，根为 x = 1 和 x = 3。画出这些关键点，并通过它们画一条平滑对称的曲线，即可完成草图。

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

The most common mistake is forgetting the ± sign when taking square roots. From (x – 3)² = 16, you must write x – 3 = ±4, giving x = 7 or x = -1. Writing only x = 7 loses half of the solutions and typically loses half of the marks.

最常见的错误是开平方时忘记 ± 符号。从 (x – 3)² = 16 出发，你必须写出 x – 3 = ±4，得到 x = 7 或 x = -1。只写 x = 7 会丢失一半的解，通常也会丢失一半的分数。

A second common error is dividing both sides by a variable. If you divide x² = 3x by x, you get x = 3 and miss the solution x = 0. Instead, rearrange to x² – 3x = 0, then factorise as x(x – 3) = 0. Never divide by a variable that could be zero.

第二个常见错误是两边同时除以变量。如果你把 x² = 3x 两边除以 x，会得到 x = 3，从而漏掉解 x = 0。正确做法是移项得 x² – 3x = 0，再因式分解为 x(x – 3) = 0。永远不要除以可能为零的变量。

Another issue is sign errors when substituting negative values into the quadratic formula. For 3x² – 2x – 1 = 0, b = -2, so -b = 2, not -2. Also, the term -4ac becomes -4 × 3 × (-1) = 12. Carefully write each substitution on a separate line to reduce errors.

另一个问题是代入负数到求根公式时出现符号错误。对于 3x² – 2x – 1 = 0，b = -2，所以 -b = 2，而不是 -2。同时，-4ac 应计算为 -4 × 3 × (-1) = 12。请把每一步代入单独写一行，以减少错误。

Finally, do not forget to rearrange the equation into standard form before applying the formula. If you use the formula directly on 2x² + 5x = 3 with c = 3, you will get wrong answers because the correct value is c = -3 after rearranging.

最后，在使用公式前不要忘记把方程整理为标准形式。如果你在 2x² + 5x = 3 上直接套用公式并取 c = 3，你会得到错误答案，因为移项后正确的 c = -3。

10. Exam Tips and Revision Strategy | 考试技巧与复习策略

In the IGCSE exam, read the question carefully to see which method is expected. If the question says “solve by factorisation”, you will not get full marks for using the quadratic formula alone. If it says “give your answers correct to 2 decimal places”, use the formula or completing the square and round at the end.

在 IGCSE 考试中，请仔细阅读题目，判断期望使用哪种方法。如果题目要求”用因式分解法求解”，仅使用求根公式将无法获得满分。如果题目要求”答案精确到2位小数”，请使用公式法或配方法，并在最后进行四舍五入。

Always show your working neatly. Even if your final answer is wrong due to a small slip, you can still earn method marks. For example, writing the correct standard form ax² + bx + c = 0 and substituting correctly into the formula will gain credit even if the final simplification is incorrect.

务必整洁地展示你的解题过程。即使最终答案因小失误而错误，你仍能获得方法分。例如，写出正确的标准形式 ax² + bx + c = 0 并正确代入公式，即使最后化简出错，也能获得相应步骤分。

When revising, practise each method separately before mixing them. Create a checklist: (1) can it be factorised? (2) is b even enough for completing the square? (3) otherwise use the formula. Check your answers by substituting them back into the original equation. Over time, this process becomes automatic.

复习时，先分别练习每种方法，再进行混合练习。制作一个检查清单：(1) 能否因式分解？(2) b 是否为偶数，适合配方法？(3) 否则使用求根公式。把解代回原方程进行检验。久而久之，这个过程就变得自然而然。

Finally, practise with past papers under timed conditions. Quadratic equations are guaranteed to appear, so mastering them is one of the most efficient ways to raise your grade. Aim to solve each problem in under two minutes, and always double-check your discriminant before finalising your answer.

最后，在限时条件下练习历年真题。二次方程一定会出现，因此掌握它是提高成绩最有效的方法之一。力求在两分钟内解出每道题，并在确定最终答案前反复检查你的判别式。

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Mastering Quadratic Equations | 二次方程完全攻略