📚 Solving Quadratic Equations | 解二次方程
Quadratic equations appear in almost every IGCSE Mathematics paper, whether in the non-calculator paper or the calculator paper. They are also the essential bridge between linear algebra and the more advanced topics you will meet at A Level. This revision guide covers the definition, the three main solving methods, the discriminant, graphs, word problems and common pitfalls, with worked examples throughout.
二次方程几乎是每份 IGCSE 数学试卷都会出现的考点，无论是否允许使用计算器，你都可能遇到它。它也是从一次方程过渡到 A Level 高阶内容的重要桥梁。本复习指南将系统讲解二次方程的定义、三种主要解法、判别式、图象、应用题以及常见的失分陷阱，并配有完整的例题示范。
1. What Is a Quadratic Equation | 什么是二次方程
A quadratic equation is an equation in which the highest power of the unknown variable is 2. It is written in a standard form, and identifying this form is always the first step in solving any quadratic problem.
二次方程是指未知数最高次数为 2 的方程。它通常写成标准形式，而识别标准形式是解一切二次方程问题的第一步。
ax² + bx + c = 0, a ≠ 0
abola opens. Examples of quadratic equations include x² − 4x + 3 = 0 and 2x² + 5x − 3 = 0.
其中 a、b、c 都是常数，且 a 不能为 0。如果 a = 0，方程就会退化为一次方程 bx + c = 0。a 被称为最高次项系数，它决定抛物线的开口方向。例如 x² − 4x + 3 = 0 和 2x² + 5x − 3 = 0 都是二次方程。
2. Rearranging into Standard Form | 化简为标准形式
Many questions give a quadratic in a messy form, for example with terms on both sides of the equals sign. The safest approach is to bring every term to the left-hand side so that the right-hand side becomes 0, with the x² term first, then the x term, then the constant term.
很多题目给出的二次方程形式较乱，例如等号两边都有项。最稳妥的方法是把所有项移到等号左边，使右边变为 0，并按照 x² 项、x 项、常数项的顺序排列。
−2 and c = −5. Always check the signs carefully when moving terms across the equals sign.
例如：解 3x² = 2x + 5。两边同时减去 2x + 5，得到 3x² − 2x − 5 = 0，因此 a = 3，b = −2，c = −5。移项时一定要仔细检查正负号。
3. Solving by Factorisation | 因式分解法
Factorisation is usually the fastest method when the quadratic has simple integer factors. It relies on the zero product property: if two expressions multiply to give 0, then at least one of them must equal 0.
当二次方程含有简单的整数因子时，因式分解通常是最快的方法。它的依据是零积性质：如果两个表达式的乘积为 0，那么其中至少有一个必须等于 0。
If (x − p)(x − q) = 0, then x = p or x = q.
r x = 3. Check by substituting: 4 − 10 + 6 = 0 and 9 − 15 + 6 = 0. Both roots are correct.
例题：解 x² − 5x + 6 = 0。寻找两个数，它们相乘为 6、相加为 −5，这两个数是 −2 和 −3。于是 (x − 2)(x − 3) = 0，所以 x = 2 或 x = 3。代入检验：4 − 10 + 6 = 0，9 − 15 + 6 = 0，均成立。
ving x = 1/2 or x = −3. Always expand your brackets to check the factorisation is correct.
当 x² 的系数不是 1 时，因式分解需要更细心。例如 2x² + 5x − 3 = 0 可分解为 (2x − 1)(x + 3) = 0，因此 x = 1/2 或 x = −3。分解后一定要展开括号检验是否正确。
4. Special Cases: Difference of Two Squares and Perfect Squares | 特殊形式：平方差与完全平方
Two special patterns appear frequently in exams. The first is the difference of two squares:
考试中有两种常见特殊形式。第一种是平方差公式：
a² − b² = (a − b)(a + b)
, giving x = 3/2 or x = −3/2. Notice that the quadratic formula is not needed here at all.
例如 4x² − 9 = 0 可写成 (2x)² − 3² = 0，即 (2x − 3)(2x + 3) = 0，因此 x = 3/2 或 x = −3/2。注意这种情况下根本不需要使用求根公式。
. It occurs when the quadratic can be written as (x + p)² or (x − p)². Recognising these patterns saves time and reduces sign errors.
第二种是完全平方三项式，例如 x² − 6x + 9 = (x − 3)²。它要求二次式能写成 (x + p)² 或 (x − p)² 的形式。识别这些模式能节省时间并减少符号错误。
5. Solving by the Quadratic Formula | 求根公式法
When a quadratic cannot be factorised easily, use the quadratic formula. You will be given this formula in the formula sheet, but you must know exactly how to substitute and simplify.
当二次式不容易因式分解时，应使用求根公式。公式会在公式表中给出，但你必须清楚如何代入和化简。
x = ( −b ± √b² − 4ac ) / 2a
d example: solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, c = −2. Substituting gives x = (−3 ± √9 + 16) / 4 = (−3 ± √25) / 4 = (−3 ± 5) / 4. Therefore x = 2/4 = 1/2 or x = −8/4 = −2.
例题：解 2x² + 3x − 2 = 0。此时 a = 2，b = 3，c = −2。代入得 x = (−3 ± √9 + 16) / 4 = (−3 ± √25) / 4 = (−3 ± 5) / 4。于是 x = 2/4 = 1/2 或 x = −8/4 = −2。
teger, the formula produces surds or decimals. For example, x² + 2x − 4 = 0 gives x = −1 ± √5, which you may leave in surd form unless the question asks for a decimal.
当答案不是整数时，公式会给出根式或小数。例如 x² + 2x − 4 = 0 得到 x = −1 ± √5，除非题目要求保留小数，否则可以保留根号形式。
6. Solving by Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q. This method is particularly useful because it also gives the vertex of a parabola directly.
配方法将二次式改写为 (x + p)² + q 的形式。这种方法特别有用，因为它还能直接给出抛物线的顶点坐标。
x + 3)² − 9 + 2 = 0, which simplifies to (x + 3)² − 7 = 0. Then (x + 3)² = 7, so x + 3 = ±√7 and x = −3 ± √7.
例题：解 x² + 6x + 2 = 0。取 6 的一半得 3，于是写成 (x + 3)² − 9 + 2 = 0，化简为 (x + 3)² − 7 = 0。因此 (x + 3)² = 7，x + 3 = ±√7，解得 x = −3 ± √7。
If the coefficient of x² is not 1, divide both sides of the equation by that coefficient before completing the square. Completing the square also proves why the quadratic formula works, which is a common exam question for higher tier students.
如果 x² 的系数不是 1，应先将方程两边除以该系数，再进行配方。配方法同样能推导出求根公式，这也是高阶拓展考试中常见的证明题。
7. The Discriminant | 判别式
The expression b² − 4ac inside the quadratic formula is called the discriminant, often written as Δ. It tells us how many real roots a quadratic has without solving it fully.
求根公式中的 b² − 4ac 被称为判别式，常用 Δ 表示。它能在不完全求解的情况下告诉我们二次方程有多少个实数根。
Δ = b² − 4ac
| Value of Δ | Number of Real Roots | Graphical Meaning |
| Δ > 0 | Two distinct real roots | Curve crosses the x-axis twice |
| Δ = 0 | One repeated real root | Curve touches the x-axis once |
| Δ < 0 | No real roots | Curve never touches the x-axis |
one repeated root. For x² − 4x + 5 = 0, Δ = 16 − 20 = −4 < 0, so there are no real roots.
对于 x² − 4x + 3 = 0，Δ = 16 − 12 = 4 > 0，所以有两个根；对于 x² − 4x + 4 = 0，Δ = 16 − 16 = 0，所以有一个重根；对于 x² − 4x + 5 = 0，Δ = 16 − 20 = −4 < 0，所以没有实数根。
8. Quadratic Graphs and the Roots | 二次函数图象与根
ph of y = ax² + bx + c is a parabola, opening upwards when a > 0 and downwards when a < 0.
从图象上看，二次方程的实数根就是抛物线与 x 轴交点的横坐标。y = ax² + bx + c 的图象是一条抛物线，当 a > 0 时开口向上，当 a < 0 时开口向下。
e vertex lies on this line. This fact is useful in many coordinate geometry problems and links directly to completing the square.
抛物线的对称轴是竖直线 x = −b/(2a)，顶点也位于这条直线上。这个结论在平面直角坐标几何问题中非常有用，并与配方法紧密相关。
If the discriminant is positive, the parabola crosses the x-axis twice; if it is zero, the vertex is exactly on the x-axis; if it is negative, the parabola lies entirely above or below the x-axis.
若判别式为正，抛物线与 x 轴相交两次；若判别式为零，顶点恰好落在 x 轴上；若判别式为负，抛物线完全位于 x 轴上方或下方。
9. Word Problems | 实际应用题
Exam questions often disguise quadratics inside geometry or motion contexts. The key is to translate the wording into an equation, solve it, and then check whether both answers make sense in the original situation.
考试题常把二次方程隐藏在几何或运动情境中。关键在于把文字翻译成方程，解出答案后再检查两个解在原情境中是否都合理。
−9 or x = 5. Since a length cannot be negative, the width is 5 cm and the length is 9 cm.
例题：一个长方形的长为 (x + 4) cm，宽为 x cm，面积为 45 cm²。根据面积可得 x(x + 4) = 45，即 x² + 4x − 45 = 0。因式分解得 (x + 9)(x − 5) = 0，因此 x = −9 或 x = 5。由于长度不能为负数，所以宽为 5 cm，长为 9 cm。
o t = 1 or t = 3. The ball passes the 15 m height on the way up and again on the way down.
抛体问题也可能产生两个有效答案。若球的高度满足 h = 20t − 5t²，令 20t − 5t² = 15，得到 t² − 4t + 3 = 0，所以 t = 1 或 t = 3。球在上升和下落过程中各经过 15 m 高度一次。
10. Common Mistakes and How to Avoid Them | 常见错误与应对方法
o the formula, write it inside brackets. For example, if b = −5, then −b means +5, not −5.
正负号错误
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