📚 Quadratic Equations: Factorisation, Formula and Graphs | 二次方程：因式分解、公式与图像
Quadratic equations sit at the centre of IGCSE and A-Level algebra. They connect pure algebra, coordinate geometry and real-world modelling, and they appear in almost every exam paper. In this revision guide you will learn how to write a quadratic in standard form, solve it by factorisation, by completing the square and by the quadratic formula, use the discriminant to describe its roots, and read all of this information from its graph.
二次方程是 IGCSE 与 A-Level 代数的核心内容。它把纯代数、坐标几何与现实建模连接起来，几乎出现在每一份试卷中。在本篇复习指南中，你将学会如何把二次方程写成标准形式，如何用因式分解法、配方法和二次公式求解，如何用判别式判断根的情况，以及如何从图像中读出所有这些信息。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is an equation in one variable in which the highest power of the variable is 2. Its general form is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The condition a ≠ 0 is essential: if a were 0 the x² term would vanish and the equation would become linear, not quadratic.
二次方程是只含一个未知数、且未知数最高次数为 2 的方程。它的一般形式是 ax² + bx + c = 0，其中 a、b、c 为常数且 a ≠ 0。a ≠ 0 这一条件至关重要：如果 a 等于 0，x² 项就会消失，方程退化为一次方程，而不再是一次二次方程。
0, but they are still quadratic because the x² term is present. A quadratic equation always has exactly two roots if you count repeated roots, although those roots may be equal or may not be real numbers.
常见的例子包括 x² − 5x + 6 = 0、3x² + 2x − 1 = 0、4x² − 9 = 0 和 x² = 16。注意后两个例子中 b = 0，但由于 x² 项存在，它们仍然是二次方程。如果把重根也计算在内，二次方程总是恰有两个根，只不过这两个根可能相等，也可能不是实数。
A useful way to picture a quadratic is to think of a hidden linear factor. Every quadratic expression that factorises neatly was built by multiplying two brackets together, for example (x − 2)(x − 3). Recognising this structure is what makes factorisation feel natural rather than like guessing.
理解二次式的一个好方法，是把它看作"隐藏的一次因式"。每一个能整齐分解的二次式，都是两个括号相乘的结果，例如 (x − 2)(x − 3)。看穿这一结构，因式分解就不再像靠猜，而是顺理成章的操作。
2. Standard Form and Key Terms | 标准形式与关键术语
called standard form. In this form a is the coefficient of x², b is the coefficient of x, and c is the constant term. Before you use any method, always rearrange the equation so that one side is exactly zero, because every standard technique depends on that.
把二次方程写成 ax² + bx + c = 0 的形式称为标准形式。在这种形式中，a 是 x² 的系数，b 是 x 的系数，c 是常数项。在使用任何方法之前，一定要先把方程整理成一边恰好为 0，因为所有标准方法都以此为前提。
right at this stage prevents almost all later errors, especially in the quadratic formula.
例如，要解 5x + 2 = 3x²，应首先改写为 3x² − 5x − 2 = 0。此时各系数一目了然：a = 3，b = −5，c = −2。在这一步把符号写对，几乎可以避免后面所有的错误，尤其是在代入二次公式时。
l equation and the rearranged equation. If the two sides agree, your rearrangement is very likely correct.
检查整理是否正确的一个快捷方法，是把一个简单的值（例如 x = 0 或 x = 1）分别代入原方程和整理后的方程。如果两边结果一致，那么整理基本就是正确的。
| Term | 项 | Meaning | 含义 | Example in 3x² − 5x − 2 = 0 |
|---|---|---|
| ax² | 二次项 | The quadratic term, a ≠ 0 | 二次项，a ≠ 0 | 3x², so a = 3 |
| bx | 一次项 | The linear term, may be 0 | 一次项，可以等于 0 | −5x, so b = −5 |
| c | 常数项 | The constant term, may be 0 | 常数项，可以等于 0 | −2, so c = −2 |
3. Solving by Factorisation | 因式分解法求解
Factorisation is often the fastest method when the quadratic has simple integer or fractional roots. The logic rests on the zero product property: if the product of two factors is zero, then at least one of the factors must be zero.
当二次方程具有简单的整数根或分数根时，因式分解通常是最快的方法。其依据是零积性质：如果两个因式的乘积为 0，那么其中至少有一个因式等于 0。
The standard procedure is:
标准步骤如下：
- Move all terms to one side so the equation equals 0.
把所有项移到同一边，使方程等于 0。 - Factorise the quadratic expression into two brackets.
把二次式分解成两个括号的乘积。 - Set each bracket equal to 0.
令每个括号分别等于 0。 - Solve the two resulting linear equations.
解得到的两个一次方程。 - Check each answer in the original equation.
把每个答案代回原方程检验。
− 3)(x − 4) = 0, so x = 3 or x = 4. Both answers check correctly in the original equation.
例题：解 x² − 7x + 12 = 0。需要找两个数，它们的乘积为 +12，和为 −7，即 −3 和 −4。因此 x² − 7x + 12 = (x − 3)(x − 4) = 0，所以 x = 3 或 x = 4。两个答案代回原方程都成立。
to +5, which are +6 and −1. Split the middle term, then factor by grouping to obtain (2x − 1)(x + 3) = 0, giving x = ½ or x = −3.
当 x² 的系数不为 1 时，可用 ac 法。对 2x² + 5x − 3 = 0，乘积 ac = 2 × (−3) = −6，需要两个数乘积为 −6、和为 +5，即 +6 和 −1。拆开中间项后分组分解，得到 (2x − 1)(x + 3) = 0，所以 x = ½ 或 x = −3。
. A perfect square trinomial such as x² − 10x + 25 = (x − 5)² gives a repeated root x = 5.
平方差公式值得牢记：a² − b² = (a + b)(a − b)。例如 4x² − 9 = 0 可写成 (2x + 3)(2x − 3) = 0，所以 x = ±3/2。完全平方式如 x² − 10x + 25 = (x − 5)² 则给出重根 x = 5。
4. Completing the Square | 配方法
ays works, it never depends on the roots being neat, and it is the gateway to the vertex form used in graph work. It is also the fastest route to the exact answer when roots are surds.
配方法把二次方程改写成 (x + p)² + q = 0 的形式。它总是可行，不依赖根是否整齐，而且是通往图像题中顶点形式的必经之路。当根是无理数时，它也是求精确答案最快的途径。
The key identity is:
核心恒等式是：
x² + bx + c = (x + b/2)² − (b/2)² + c
on becomes (x + 3)² − 9 + 1 = 0, that is (x + 3)² = 8. Taking square roots gives x + 3 = ±√8 = ±2√2, so x = −3 ± 2√2, approximately x = −0.17 or x = −5.83.
例题：解 x² + 6x + 1 = 0。6 的一半是 3，所以 x² + 6x = (x + 3)² − 9。方程变为 (x + 3)² − 9 + 1 = 0，即 (x + 3)² = 8。开平方得 x + 3 = ±√8 = ±2√2，所以 x = −3 ± 2√2，约为 x = −0.17 或 x = −5.83。
lete the square on x² − 4x to obtain (x − 2)² − 4 + 7/3 = 0, so (x − 2)² = 5/3 and x = 2 ± √5/3.
当 a 不为 1 时，先把整个方程除以 a，或把 a 从二次项和一次项中提出。对 3x² − 12x + 7 = 0，除以 3 得 x² − 4x + 7/3 = 0，再对 x² − 4x 配方，得到 (x − 2)² − 4 + 7/3 = 0，即 (x − 2)² = 5/3，所以 x = 2 ± √5/3。
Remember that you can always verify a completed-square result by expanding the brackets back out. If the expansion returns the original expression, the form is correct.
记住：把配方的结果重新展开，就可以随时检验。如果展开后得到原式，说明配方正确。
5. The Quadratic Formula | 二次公式
neral form ax² + bx + c = 0 and should be memorised exactly, including the ± sign and the denominator 2a.
二次公式可以解任何二次方程，无论系数多么难处理。它由一般形式 ax² + bx + c = 0 配方推导而来，必须一字不差地背下来，包括 ± 号和分母 2a。
x = (−b ± √b² − 4ac) / 2a
2x² − 4x − 3 = 0. Here a = 2, b = −4 and c = −3. Then b² − 4ac = 16 + 24 = 40, so x = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10 x ≈ 2.58 or x ≈ −0.58. Always simplify the surd and, if asked, round only at the very end.
例题：解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。于是 b² − 4ac = 16 + 24 = 40，所以 x = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2。数值上 x ≈ 2.58 或 x ≈ −0.58。要尽量把根式化简，并且只有在最后一步才按要求取近似值。
bstituting the wrong signs is the single most frequent source of lost marks in this topic.
一个常见陷阱是忘记先整理。如果题目给出 x² = 5x + 3，必须先改写为 x² − 5x − 3 = 0，再读出 a = 1，b = −5，c = −3。代入时符号写错，是这一主题中最常见的失分原因。
If the question says "give your answers to 3 significant figures", use the formula and round at the end. If it says "leave your answers in surd form", simplify the square root and cancel common factors instead of going decimal.
如果题目要求"答案保留 3 位有效数字"，就用二次公式并在最后取近似值。如果题目要求"保留根式形式"，则应化简根式并约分，而不要化成小数。
6. The Discriminant | 判别式
t tells you how many real roots the equation has before you do any further work, which is why examiners love asking about it.
二次公式中根号下的表达式称为判别式，记作 Δ = b² − 4ac。它能在进一步计算之前就告诉你方程有多少个实根，因此考试特别爱考。
| Value of Δ | Δ 的取值 | Nature of roots | 根的性质 | Graph behaviour | 图像表现 |
|---|---|---|
| Δ > 0 | Two distinct real roots | 两个不同实根 | Curve cuts the x-axis twice | 曲线与 x 轴交于两点 |
| Δ = 0 | One repeated real root | 一个重实根 | Curve touches the x-axis | 曲线与 x 轴相切 |
| Δ < 0 | No real roots | 没有实根 | Curve does not meet x-axis | 曲线与 x 轴不相交 |
− 7x + 12 has Δ = 49 − 48 = 1, a perfect square, confirming the roots 3 and 4 are rational.
此外，如果 a、b、c 都是整数且 Δ 是完全平方数，那么根是有理数，该二次式可以在整数范围内分解。例如 x² − 7x + 12 的 Δ = 49 − 48 = 1，是完全平方数，说明根 3 和 4 是有理数。
t x² + kx + 9 = 0 has a repeated root. Set Δ = k² − 36 = 0, so k = 6 or k = −6. The phrase "repeated root", "tangent" or "equal roots" is your signal to set Δ = 0.
判别式题常常含未知常数。例如，求使 x² + kx + 9 = 0 有重根的 k。令 Δ = k² − 36 = 0，得 k = 6 或 k = −6。题目中出现"重根""相切""两根相等"等字眼，就是令 Δ = 0 的信号。
7. Graphs of Quadratic Functions | 二次函数图像
. If a > 0 the parabola opens upwards and has a minimum point; if a < 0 it opens downwards and has a maximum point. The larger the value of |a|, the narrower the curve.
y = ax² + bx + c 的图像是一条抛物线。若 a > 0，抛物线开口向上，有最小值点；若 a < 0，开口向下，有最大值点。|a| 越大，曲线越窄。
Key features of a parabola:
抛物线的关键特征：
- The y-intercept is at (0, c).
与 y 轴交点坐标为 (0, c)。 - The roots of ax² + bx + c = 0 are the x-intercepts of the curve.
方程 ax² + bx + c = 0 的根就是曲线与 x 轴交点的横坐标。 - The axis of symmetry is the vertical line x = −b / 2a.
对称轴是竖直直线 x = −b / 2a。 - The vertex lies on the axis of symmetry, at x = −b / 2a.
顶点位于对称轴上，横坐标为 x = −b / 2a。 - The curve is symmetric, so points equidistant from the axis have equal y-values.
曲线关于对称轴对称，因此到对称轴距离相等的点，函数值相等。
re (5, 0) and (−1, 0). The axis of symmetry is x = 2, midway between the roots, and substituting gives the minimum point (2, −9).
要画出 y = x² − 4x − 5 的草图：a = 1 > 0，开口向上；c = −5，与 y 轴交于 (0, −5)；分解得 (x − 5)(x + 1) = 0，与 x 轴交于 (5, 0) 和 (−1, 0)。对称轴为 x = 2，正好在两根中点，代入得最小值点 (2, −9)。
Exam questions frequently ask you to use a graph to solve an inequality, such as x² − 4x − 5 < 0. Reading the curve, y is negative between the roots, so the solution is −1 < x < 5. Sketching first makes the inequality almost automatic.
考试常要求利用图像解不等式，例如 x² − 4x − 5 < 0。观察曲线，y 在两根之间为负，所以解为 −1 < x < 5。先画草图，解不等式几乎就是顺水推舟。
8. Vertex Form and Maximum or Minimum Values | 顶点式与最大最小值
k, where (h, k) is the vertex of the parabola. When a > 0, k is the minimum value of y; when a < 0, k is the maximum value. This is the standard method for optimisation questions that ask for the greatest area, the least cost or the highest point.
配方把二次函数化为顶点式 y = a(x − h)² + k，其中 (h, k) 是抛物线的顶点。当 a > 0 时，k 是 y 的最小值；当 a < 0 时，k 是 y 的最大值。对于求最大面积、最低成本或最高点之类的优化问题，这是标准方法。
tex is (2, −3), and since a = 1 > 0 the minimum value of y is −3, occurring when x = 2.
例题：把 y = x² − 4x + 1 化为顶点式。−4 的一半是 −2，所以 y = (x − 2)² − 4 + 1 = (x − 2)² − 3。顶点为 (2, −3)。因为 a = 1 > 0，所以 y 的最小值为 −3，在 x = 2 处取得。
− 5 = −2(x − 2)² + 3. The vertex is (2, 3) and, because a < 0, the maximum value is 3.
若要讨论最大值，以 y = −2x² + 8x − 5 为例。从前两项提出 −2：y = −2(x² − 4x) − 5 = −2[(x − 2)² − 4] − 5 = −2(x − 2)² +
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