📚 Solving Quadratic Equations | 解一元二次方程
Quadratic equations are one of the most important topics in IGCSE Mathematics, appearing in almost every paper across both Foundation and Higher tiers. Mastering this topic will help you solve problems in algebra, graphs, coordinate geometry, and even calculus later on.
二次方程是 IGCSE 数学中最重要的话题之一，在基础级和高级试卷中几乎都会出现。掌握这一专题将帮助你在代数、函数图像、坐标几何甚至后续微积分中解决各类问题。
1. What Is a Quadratic Equation? | 什么是二次方程
A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the unknown variable is 2. The most general form is written as ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.
二次方程是最高次数为 2 的多项式方程，即未知数的最高次幂为 2。其最一般的形式写作 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。
3. The condition a ≠ 0 is essential: if a = 0, the equation becomes linear, not quadratic.
例如，2x² − 5x + 3 = 0 是一个二次方程，其中 a = 2，b = −5，c = 3。条件 a ≠ 0 至关重要：若 a = 0，方程退化为一次方程，就不再是二次方程了。
2. Solving by Factorisation | 因式分解法
h bracket equal to zero. For example, x² − 5x + 6 = (x − 2)(x − 3) = 0, so x = 2 or x = 3.
当二次方程具有简单的整数根时，因式分解通常是最快的方法。其思路是将 ax² + bx + c 改写为两个括号的乘积，然后令每个括号为零。例如，x² − 5x + 6 = (x − 2)(x − 3) = 0，因此 x = 2 或 x = 3。
inal expression. Here, 2x² + 7x + 3 = (2x + 1)(x + 3), giving solutions x = −½ and x = −3.
对于 a ≠ 1 的二次方程，例如 2x² + 7x + 3，需要寻找两个二项式使其乘积等于原式。这里，2x² + 7x + 3 = (2x + 1)(x + 3)，得到解 x = −½ 和 x = −3。
(2x + 1)(x + 3) = 0 → x = −½ or x = −3
Always expand your brackets to check your factorisation before writing the final answer. A common trap is missing the negative signs when one bracket contains a negative term.
在写出最终答案前，务必展开括号检验因式分解是否正确。常见的陷阱是当括号内含负项时漏掉负号。
3. Solving by Completing the Square | 配方法
a parabola. For x² + 6x + 2, the process is x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7.
配方法将二次式改写为 a(x + p)² + q 的形式。当因式分解不明显时，这种方法尤其有用，并且能清楚地联系到抛物线的顶点。对于 x² + 6x + 2，过程为 x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7。
3)² − 7 = 0, rearrange to (x + 3)² = 7, then take the square root of both sides: x + 3 = ±√7, so x = −3 ± √7. Remember the ± sign—taking only the positive root is a classic error.
要解 (x + 3)² − 7 = 0，将其变形为 (x + 3)² = 7，然后两边开平方：x + 3 = ±√7，因此 x = −3 ± √7。切记 ± 号——只取正根是常见错误。
(x + 3)² = 7 → x = −3 + √7 or x = −3 − √7
square. For example, 2x² − 4x + 1 = 2(x² − 2x) + 1 = 2[(x − 1)² − 1] + 1 = 2(x − 1)² − 1.
当 a ≠ 1 时，先将 a 从 x² 项和 x 项中提出，再进行配方。例如，2x² − 4x + 1 = 2(x² − 2x) + 1 = 2[(x − 1)² − 1] + 1 = 2(x − 1)² − 1。
4. Solving by the Quadratic Formula | 求根公式法
The quadratic formula works for every quadratic equation, whether the roots are rational, irrational, or even complex. It is derived directly from completing the square and is stated as follows:
求根公式适用于所有二次方程，无论根是有理数、无理数还是复数。该公式由配方法直接推导得出，表述如下：
x = (−b ± √b² − 4ac) / 2a
b² − 4ac = 25 − 4 × 3 × (−2) = 25 + 24 = 49, so x = (−5 ± 7) / 6, giving x = ⅓ or x = −2.
对方程 3x² + 5x − 2 = 0，将 a = 3，b = 5，c = −2 代入公式。判别式 b² − 4ac = 25 − 4 × 3 × (−2) = 25 + 24 = 49，故 x = (−5 ± 7) / 6，得到 x = ⅓ 或 x = −2。
Always calculate the discriminant first as a separate step. This keeps your working organised and helps you check whether the roots are real before computing them.
务必先将判别式作为单独步骤计算出。这样能使解题过程更有条理，并帮助你在求根前判断根是否为实数。
5. The Discriminant: b² − 4ac | 判别式 b² − 4ac
ac, tells us the nature of the roots without actually solving the equation. This is a frequently tested concept in IGCSE papers, often as a short-answer question.
判别式写作 Δ = b² − 4ac，无需真正解方程即可判断根的性质。这是 IGCSE 考试中的高频考点，常以简答题形式出现。
If Δ > 0, the equation has two distinct real roots.
若 Δ > 0，方程有两个不相等的实数根。
If Δ = 0, the equation has exactly one repeated real root.
若 Δ = 0，方程有两个相等的实数根，即一个重根。
If Δ < 0, the equation has no real roots (two complex roots).
若 Δ < 0，方程无实数根（有两个复数根）。
Δ = b² − 4ac → positive: two roots; zero: one root; negative: no real roots
oots. In contrast, x² − 6x + 9 = 0 has Δ = 36 − 36 = 0, so it has exactly one root, x = 3.
例如，x² + 2x + 5 = 0 的判别式 Δ = 4 − 20 = −16 < 0，因此无实数根。相比之下，x² − 6x + 9 = 0 的判别式 Δ = 36 − 36 = 0，因此只有一个根 x = 3。
6. Graphs of Quadratic Functions | 二次函数图像
int. The roots of the equation ax² + bx + c = 0 are exactly the x-intercepts of the graph.
二次函数 y = ax² + bx + c 的图像是一条抛物线。当 a > 0 时，抛物线开口向上，有最小值点；当 a < 0 时，开口向下，有最大值点。方程 ax² + bx + c = 0 的根正是抛物线与 x 轴交点的横坐标。
Δ > 0: the graph crosses the x-axis at two points.
Δ > 0：图像与 x 轴有两个交点。
Δ = 0: the graph touches the x-axis at one point (the vertex lies on the axis).
Δ = 0：图像与 x 轴相切于一点（顶点在 x 轴上）。
Δ < 0: the graph does not intersect the x-axis at all.
Δ < 0：图像与 x 轴没有交点。
ertex coordinates can be found by substituting this x-value into the original equation. This is a direct connection between algebra and graph sketching.
抛物线的对称轴是竖直线 x = −b / 2a，将这一 x 值代入原方程即可求得顶点坐标。这正是代数与图像绘制之间的直接联系。
7. Sum and Product of Roots | 根与系数的关系
quently tested in examination questions and are derived from expanding (x − α)(x − β) = 0.
对于根为 α 和 β 的二次方程 ax² + bx + c = 0，无需解方程即可得到两个重要关系。它们由展开 (x − α)(x − β) = 0 推导而来，在考试中经常出现。
Sum of roots: α + β = −b / a
Product of roots: α × β = c / a
You can verify this by solving: the roots are 1 and 3, and indeed 1 + 3 = 4 and 1 × 3 = 3.
例如，方程 2x² − 8x + 6 = 0 的两根之和为 8 / 2 = 4，两根之积为 6 / 2 = 3。你可以通过求解验证：两根为 1 和 3，确实 1 + 3 = 4，1 × 3 = 3。
These formulas are especially useful when you are asked to find the new equation whose roots are related to α and β, such as α² and β², without computing the roots themselves.
当题目要求构造一个以 α 和 β 的衍生形式（如 α² 和 β²）为根的新方程时，这些公式尤为有用，无需直接求出原根。
8. Solving Quadratic Inequalities | 二次不等式
− 6 = 0 are x = 3 and x = −2, so the critical points divide the line into three intervals.
解二次不等式（如 x² − x − 6 > 0）时，先求根，再在数轴上测试各区间。方程 x² − x − 6 = 0 的根为 x = 3 和 x = −2，因此临界点将数轴分为三个区间。
ward, the expression is positive outside the interval between the roots and negative between them. Hence x² − x − 6 > 0 gives x < −2 or x > 3, and x² − x − 6 < 0 gives −2 < x < 3.
由于抛物线 y = x² − x − 6 开口向上，表达式在两根之外为正，两根之间为负。因此 x² − x − 6 > 0 的解为 x < −2 或 x > 3，而 x² − x − 6 < 0 的解为 −2 < x < 3。
x² − x − 6 > 0 → x < −2 or x > 3
equalities (≥ or ≤). Writing the final answer in set notation is also acceptable in IGCSE.
注意：在数轴上，严格不等式（> 或 <）用空心圆圈，含等号的不等式（≥ 或 ≤）用实心圆圈。在 IGCSE 中，也可以用集合符号表示最终答案。
9. Word Problems Involving Quadratics | 二次方程应用题
a is 28 cm². If the width is x cm, the equation is x(x + 3) = 28, giving x² + 3x − 28 = 0.
许多 IGCSE 应用题将实际情境转化为二次方程。典型题型是矩形问题：一个矩形的长比宽长 3 cm，面积为 28 cm²。若宽为 x cm，则方程为 x(x + 3) = 28，即 x² + 3x − 28 = 0。
egative, the width is 4 cm and the length is 7 cm. Always reject solutions that do not make sense in the context of the problem.
因式分解得 (x + 7)(x − 4) = 0，因此 x = −7 或 x = 4。由于长度不能为负，所以宽为 4 cm，长为 7 cm。务必舍去在问题情境中无意义的解。
oduct of consecutive integers, and the Pythagorean theorem with algebraic side lengths. In every case, define your variable clearly at the start.
常见应用场景包括抛体运动（h = ut − ½gt²）、连续整数乘积以及含代数边长的勾股定理问题。在每种情况下，都要在开始时明确设出未知数。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
on is 2x² = 3x + 5, you must first bring all terms to one side to obtain 2x² − 3x − 5 = 0.
考试中最常见的错误之一是忘记先将方程整理成 ax² + bx + c = 0 的形式再进行因式分解。若方程为 2x² = 3x + 5，必须先将所有项移到一边，得到 2x² − 3x − 5 = 0。
Always write down the values of a, b and c before using the formula.
使用求根公式前，务必先写出 a、b、c 的值。
Do not cancel a variable factor such as x from both sides—you may lose the root x = 0.
不要从两边约去含未知数的因子（如 x），否则可能丢失根 x = 0。
Check your answers by substituting them back into the original equation.
将答案代回原方程检验。
Give answers to 3 significant figures unless the question states otherwise, and leave surds in exact form where required.
若无特别说明，答案保留 3 位有效数字；题目要求时，根式保留精确形式。
Time management matters: factorise first, then use the formula only when necessary. Showing clear, structured working earns you method marks even if your final answer has a minor arithmetic slip.
时间管理也很重要：优先尝试因式分解，仅在必要时使用求根公式。清晰、规范的解题过程可以帮你获得方法分，即使最终答案有细微计算错误。
11. Quick Reference Summary | 快速要点总结
The table below summarises the key facts you should memorise before entering the examination hall. Revise these points thoroughly and practise applying them to past-paper questions.
下表总结了进入考场前应熟记的关键要点。请认真复习这些内容，并通过真题加以练习。
| Method | 方法 | When to Use | 适用情境 | Key Formula | 关键公式 |
|---|---|---|
| Factorisation 因式分解 | Simple integer roots 简单整数根 | (px + q)(rx + s) = 0 |
| Completing the Square 配方法 | Finding the vertex 求顶点 | a(x + p)² + q |
| Quadratic Formula 求根公式 | Always works 万能方法 | x = (−b ± √b² − 4ac) / 2a |
| Discriminant 判别式 | Nature of roots 判断根的性质 | Δ = b² − 4ac |
Consistent practice is the key to success. Identify which method you find most challenging and target your revision accordingly.
持续练习是取得成功的关键。找出你最薄弱的解法并针对性地复习。
12. Practice Questions | 巩固练习
Try these questions without looking at the solutions, then check your answers carefully. They cover all the methods discussed in this article.
请先独立完成以下练习，再仔细核对答案。这些题目覆盖了本文讨论的所有方法。
Solve 2x² + 5x − 3 = 0 by factorisation.
用因式分解法解方程 2x² + 5x − 3 = 0。
Solve x² − 4x − 1 = 0 by completing the square. Leave your answer in surd form.
用配方法解方程 x² − 4x − 1 = 0，答案保留根式形式。
Find the value of k such that x² + kx + 9 = 0 has exactly one real root.
求 k 的值，使方程 x² + kx + 9 = 0 有且仅有一个实数根。
The product of two consecutive positive odd integers is 143. Find the two integers.
两个连续正奇数的乘积为 143，求这两个整数。
Answers: ① x = ½ or x = −3; ② x = 2 ± √5; ③ k = ±6; ④ 11 and 13
If you obtained all four answers correctly, you are well prepared for quadratic equation questions. If not, review the relevant section above and try again.
如果你四题全部正确，说明你对二次方程这部分已准备充分。若有错误，请回顾对应章节并重新尝试。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图