📚 Mastering Quadratic Equations for IGCSE Mathematics | IGCSE 数学二次方程完全攻略
A quadratic equation is one of the most important topics in IGCSE Mathematics. It appears in algebra, graphs, and real-life modelling, and it is often tested in both Paper 1 and Paper 2. Understanding how to solve quadratic equations by factorising, completing the square, and using the quadratic formula gives you a reliable toolkit for many exam questions.
二次方程是 IGCSE 数学中最重要的内容之一。它出现在代数、图像和实际生活建模中，并经常在 Paper 1 和 Paper 2 中考查。掌握因式分解法、配方法和二次公式法来解二次方程，能为你应对许多考试题目提供一套可靠的工具。
This guide follows the standard IGCSE syllabus sequence: definition, standard form, four solving methods, the discriminant, graphs, root relationships, modelling questions, and finally the mistakes that cost candidates the most marks. Work through it section by section and use the summary table at the end as a last-minute revision sheet.
本攻略按照 IGCSE 考纲的常规顺序展开：定义、标准式、四种解法、判别式、图像、根的关系、建模题，最后是考生最容易丢分的错误。请逐节学习，并把文末的总结表当作考前最后的复习清单。
1. What Is a Quadratic Equation? | 什么是二次方程？
d function graphs as a parabola, a smooth U-shaped curve that either opens upwards or downwards. Typical examples include x² + 5x + 6 = 0 and 2x² − 3x + 1 = 0.
二次方程是含有一个未知数且最高次数为 2 的多项式方程。由于未知数的最高次幂是 2，因此它的相关函数图像是一条抛物线，即一条平滑的 U 形曲线，开口向上或向下。典型例子包括 x² + 5x + 6 = 0 和 2x² − 3x + 1 = 0。
is cubic, not quadratic. In IGCSE questions, you must usually write the equation in standard form before solving it, because the solving methods all assume that shape.
像 3x + 1 = 0 这样的方程就不是二次方程，因为 x 的最高次数是 1，它是线性方程。同理，x³ − 2x = 0 是三次方程，也不是二次方程。在 IGCSE 题目中，你通常需要先把方程写成标准形式再求解，因为所有解法都默认方程处于这种形式。
A quadratic equation can have two distinct real roots, one repeated real root, or no real roots at all. Deciding which case applies is part of the skill being tested, and the discriminant in Section 5 handles that decision quickly.
二次方程可能有两个不同的实根、一个重根，或者没有实根。判断属于哪种情况本身就是考点之一，而第 5 节的判别式可以让你快速做出判断。
2. Standard Form and Coefficients | 标准式与系数
nstants and a ≠ 0. This is called the standard form. If a were 0, the x² term would disappear and the equation would no longer be quadratic.
任何二次方程都可以写成 ax² + bx + c = 0，其中 a、b 和 c 是常数，且 a ≠ 0。这称为标准式。如果 a = 0，x² 项就会消失，方程就不再是二次方程。
rectly is essential for using the quadratic formula and the discriminant, because a single sign error changes the answer completely.
例如，在 3x² − 5x + 2 = 0 中，a = 3，b = −5，c = 2。正确读取符号对于使用二次公式和判别式至关重要，因为一个符号错误就会彻底改变答案。
Many exam questions give the equation in a disguised form. You must rearrange it yourself before identifying the coefficients.
很多考题给出的方程是"伪装"过的形式。你必须先自己整理，才能识别出各个系数。
- 5x² = 4x + 1 becomes 5x² − 4x − 1 = 0, so a = 5, b = −4, c = −1. | 5x² = 4x + 1 整理为 5x² − 4x − 1 = 0，所以 a = 5，b = −4，c = −1。
- x(x − 7) = 2 becomes x² − 7x − 2 = 0, so a = 1, b = −7, c = −2. | x(x − 7) = 2 展开整理为 x² − 7x − 2 = 0，所以 a = 1，b = −7，c = −2。
- (x + 3)(x − 5) = 0 is already factorised but expands to x² − 2x − 15 = 0. | (x + 3)(x − 5) = 0 已经是因式形式，展开后为 x² − 2x − 15 = 0。
- 3x² = 12 becomes 3x² − 12 = 0, so b = 0; the equation has no x term. | 3x² = 12 整理为 3x² − 12 = 0，此时 b = 0，方程没有 x 项。
| Equation | a | b | c |
| 3x² − 5x + 2 = 0 | 3 | −5 | 2 |
| x² + 5x + 6 = 0 | 1 | 5 | 6 |
| 2x² − 3x + 1 = 0 | 2 | −3 | 1 |
| x² − 9 = 0 | 1 | 0 | −9 |
3. Solving by Factorising | 因式分解法求解
Factorising is usually the fastest method, and it is the one IGCSE examiners expect you to try first when the numbers are friendly. The key idea is the zero product rule: if the product of two factors is zero, then at least one of the factors must be zero.
因式分解通常是最快的方法，当数字比较"友好"时，这也是 IGCSE 考官希望你首先尝试的方法。核心思想是零积法则：如果两个因式的乘积为零，那么至少有一个因式必须为零。
If (x − p)(x − q) = 0, then x = p or x = q.
若 (x − p)(x − q) = 0，则 x = p 或 x = q。
add to 5, so the equation becomes (x + 2)(x + 3) = 0 and the roots are x = −2 and x = −3.
当 a = 1 时，寻找两个数，它们相乘得 c、相加得 b。在 x² + 5x + 6 = 0 中，2 和 3 相乘得 6、相加得 5，因此方程变为 (x + 2)(x + 3) = 0，根为 x = −2 和 x = −3。
: (2x − 1)(x − 1) = 2x² − 2x − x + 1 = 2x² − 3x + 1. Always expand mentally to confirm your factors.
当 a ≠ 1 时，必须考虑二次项系数。在 2x² − 3x + 1 = 0 中，因式分解为 (2x − 1)(x − 1) = 0，得到 x = ½ 和 x = 1。展开检验：(2x − 1)(x − 1) = 2x² − 2x − x + 1 = 2x² − 3x + 1。一定要在脑中展开一次来确认因式正确。
Two special patterns appear constantly in papers:
试卷中反复出现两种特殊模式：
- Difference of two squares: x² − 9 = 0 gives (x − 3)(x + 3) = 0, so x = 3 or x = −3. | 平方差：x² − 9 = 0 给出 (x − 3)(x + 3) = 0，所以 x = 3 或 x = −3。
- Common factor: 3x² − 12x = 0 gives 3x(x − 4) = 0, so x = 0 or x = 4. | 公因式：3x² − 12x = 0 给出 3x(x − 4) = 0，所以 x = 0 或 x = 4。
the root x = 0 and is one of the most common lost-mark errors in IGCSE. Factorise instead.
千万不要为了解 3x² − 12x = 0 而在两边同时除以 x。这会丢掉根 x = 0，是 IGCSE 中最常见的丢分错误之一。应改用因式分解。
4. The Quadratic Formula | 二次公式
When factorisation is not obvious, the quadratic formula always works. It is given on the IGCSE formula sheet in most specifications, but you should memorise it anyway to save time and to avoid looking it up under pressure.
当因式分解不明显时，二次公式永远有效。大多数 IGCSE 考纲都会在公式表上给出它，但你仍应背下来，以节省时间并避免在紧张时翻查公式表。
x = (−b ± √b² − 4ac) ÷ (2a)
4 × 3 × 2 = 25 − 24 = 1. Then x = (5 ± 1) ÷ 6, which gives x = 6 ÷ 6 = 1 or x = 4 ÷ 6 = ⅔.
例题：解 3x² − 5x + 2 = 0。此处 a = 3，b = −5，c = 2，所以 b² − 4ac = (−5)² − 4 × 3 × 2 = 25 − 24 = 1。于是 x = (5 ± 1) ÷ 6，得到 x = 6 ÷ 6 = 1 或 x = 4 ÷ 6 = ⅔。
ed example with rounding: solve x² + 4x − 1 = 0. Here b² − 4ac = 16 + 4 = 20, so x = (−4 ± √20) ÷ 2 = −2 ± √5, giving x = 0.236 or x = −4.236 to three significant figures.
含取近似值的例题：解 x² + 4x − 1 = 0。此处 b² − 4ac = 16 + 4 = 20，所以 x = (−4 ± √20) ÷ 2 = −2 ± √5，得到 x = 0.236 或 x = −4.236（保留三位有效数字）。
- Substitute a, b and c inside brackets, especially when b or c is negative: (−5)² is 25, but −5² is −25. | 把 a、b、c 代入时务必加括号，尤其当 b 或 c 为负数时：(−5)² 等于 25，而 −5² 等于 −25。
- Calculate the discriminant first, then take the square root; this reduces calculator slips. | 先算判别式，再开平方，可减少计算器操作失误。
- Unless the question says otherwise, give answers to 3 significant figures, or leave them as exact surds. | 除非题目另有说明，答案保留三位有效数字，或写成精确的根式形式。
- Write down the formula, the substitution, and then the answer; method marks are awarded for the substitution line. | 先写公式、再写代入过程、最后写答案；代入这一步是有方法分的。
5. The Discriminant and the Number of Roots | 判别式与根的个数
The expression b² − 4ac is called the discriminant, often written as Δ. It tells you how many real roots a quadratic equation has without solving it, which is exactly what many exam questions ask.
表达式 b² − 4ac 称为判别式，通常记作 Δ。它能在不解方程的情况下告诉你方程有多少个实根，而这正是许多考题所问的内容。
Δ = b² − 4ac
- If Δ > 0, there are two distinct real roots, and the parabola crosses the x-axis twice. | 若 Δ > 0，有两个不同的实根，抛物线与 x 轴相交两次。
- Δ = 0, there is one repeated real root, and the parabola touches the x-axis at its vertex. | 若 Δ = 0，有一个重根，抛物线在顶点处与 x 轴相切。
- If Δ < 0, there are no real roots, and the parabola never meets the x-axis. | 若 Δ < 0，没有实根，抛物线与 x 轴永不相交。
= 4 − 4 = 0, so there is a repeated root at x = −1, since the equation is (x + 1)² = 0. For x² + x + 1 = 0, Δ = 1 − 4 = −3 < 0, so there are no real roots.
例如：对于 x² + 5x + 6 = 0，Δ = 25 − 24 = 1 > 0，故有两个不同实根，与因式分解 (x + 2)(x + 3) = 0 一致。对于 x² + 2x + 1 = 0，Δ = 4 − 4 = 0，故有重根 x = −1，因为方程为 (x + 1)² = 0。对于 x² + x + 1 = 0，Δ = 1 − 4 = −3 < 0，故没有实根。
= −6. Always give both values; many candidates give only the positive one and lose a mark.
非常常见的考题形式是：求 k 的值，使 x² + kx + 9 = 0 有相等的根。令 Δ = 0 得 k² − 36 = 0，所以 k = 6 或 k = −6。一定要写出两个值；很多考生只写正数，从而丢掉一分。
6. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q. It is the method behind the quadratic formula, and it is the only reliable way to find the vertex of a parabola or to prove that a quadratic expression is always positive.
配方法把二次式改写为 (x + p)² + q 的形式。它是二次公式的来源，也是求抛物线顶点、或证明某个二次式恒为正的唯一可靠方法。
x² + bx + c = (x + b ÷ 2)² − (b ÷ 2)² + c
(x + 3)² − 9 + 1 = (x + 3)² − 8. Solving (x + 3)² − 8 = 0 gives (x + 3)² = 8, so x + 3 = ±√8 = ±2√2 and x = −3 ± 2√2.
例题：把 x² + 6x + 1 化为配方式。6 的一半是 3，3² = 9，所以 x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8。解 (x + 3)² − 8 = 0 得 (x + 3)² = 8，于是 x + 3 = ±√8 = ±2√2，x = −3 ± 2√2。
When a ≠ 1, first take the coefficient a out of the first two terms. For 2x² − 3x + 1, write 2(x² − 1.5x) + 1, then complete the square inside the bracket to get 2(x − 0.75)² − 0.125. Working with fractions avoids rounding errors.
当 a ≠ 1 时，先把系数 a 从前两项中提出来。对于 2x² − 3x + 1，写成 2(x² − 1.5x) + 1，然后在括号内配方，得到 2(x − 0.75)² − 0.125。使用分数可以避免取整误差。
- The minimum (or maximum) value of the expression is q, occurring at x = −p. | 表达式的最小值（或最大值）是 q，在 x = −p 处取得。
- The vertex of the curve y = (x + p)² + q is the point (−p, q). | 曲线 y = (x + p)² + q 的顶点是点 (−p, q)。
- The line of symmetry is the vertical line x = −p. | 对称轴是竖直线 x = −p。
- If q > 0 and the coefficient of x² is positive, the expression is always positive for all real x. | 若 q > 0 且 x² 的系数为正，则该表达式对所有实数 x 恒为正。
7. Graphs, Roots and the Parabola | 图像、根与抛物线
. 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，抛物线开口向上并有最小值点；若 a < 0，则开口向下并有最大值点。|a| 越大，曲线越窄。
Three features carry most of the marks in graph questions:
图像题中主要得分点集中在三个特征上：
- Roots: the x-intercepts of the curve are exactly the solutions of ax² + bx + c = 0. | 根：曲线与 x 轴交点的横坐标，正好是方程 ax² + bx + c = 0 的解。
- y-intercept: substituting x = 0 gives y = c, so the curve crosses the y-axis at (0, c). | y 轴截距：代入 x = 0 得 y = c，因此曲线在 (0, c) 处与 y 轴相交。
- Line of symmetry: the axis of symmetry is x = −b ÷ (2a), and the vertex lies on it. | 对称轴：对称轴为 x = −b ÷ (2a)，顶点就在这条直线上。
+ 6 = k, draw or imagine the horizontal line y = k and count intersections: two, one, or none depending on how high the line sits relative to the vertex.
你也可以通过读图来解二次方程，即从给定的草图上读出与 x 轴的交点。如果题目问 x² + 5x + 6 = k 有多少个解，就画（或想象）水平线 y = k 并数交点个数：两个、一个或零个，取决于这条线相对于顶点的高度。
8. Sum and Product of Roots | 两根之和与两根之积
t are given by two short formulas that follow directly from expanding a(x − α)(x − β) = 0.
若二次方程 ax² + bx + c = 0 的根为 α 和 β，那么它们的和与积由两个简短公式给出，这两个公式可直接由 a(x − α)(x − β) = 0 展开得到。
α + β = −b ÷ a and αβ = c ÷ a
al roots are 1 and ½, and indeed 1 + ½ = 1.5 and 1 × ½ = 0.5, which confirms the formulas.
例题：对于 2x² − 3x + 1 = 0，α + β = 3 ÷ 2 = 1.5，αβ = 1 ÷ 2 = 0.5。实际根为 1 和 ½，确实 1 + ½ = 1.5、1 × ½ = 0.5，验证了公式。
and the product is −10, so the equation is x² − (−3)x + (−10) = 0, that is x² + 3x − 10 =
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