📚 Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与求根公式
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, coordinate geometry, area problems and many real-life modelling questions. This revision guide covers factorising, completing the square and the quadratic formula, with worked examples, the discriminant, common mistakes and exam tips.
二次方程是 IGCSE 数学中最重要的主题之一。它们出现在代数、坐标几何、面积问题以及许多现实建模题中。本复习指南涵盖因式分解、配方法和求根公式，并提供示例、判别式、常见错误与考试技巧。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants (ordinary numbers) and a ≠ 0. The condition a ≠ 0 is essential because if a were zero, the equation would become linear, not quadratic, and it would have at most one solution.
二次方程是任何可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数且 a ≠ 0。条件 a ≠ 0 至关重要，因为如果 a 为零，方程就变成一次方程，而不是二次方程，最多只有一个解。
The highest power of x in a quadratic is 2, which is why it is called "quadratic" (from the Latin word for square). The values of x that make the equation true are called the roots, or solutions, of the equation.
二次方程中 x 的最高次数是 2，这就是它被称为"二次"的原因。使方程成立的 x 值称为方程的根，也叫解。
Before using any method, always rearrange the equation so that one side is exactly zero. For example, 3x² = 5x + 2 must first be written as 3x² − 5x − 2 = 0.
在使用任何方法之前，务必先把方程整理成一边恰好为零。例如 3x² = 5x + 2 必须先写成 3x² − 5x − 2 = 0。
2. The Graph of a Quadratic: Parabolas and Roots | 二次函数的图像：抛物线与根
The graph of y = ax² + bx + c is a parabola (a smooth U-shaped curve). If a > 0 the parabola opens upwards, and if a < 0 it opens downwards. The roots of the equation ax² + bx + c = 0 are the x-intercepts of this graph, that is, the points where the curve crosses the x-axis.
y = ax² + bx + c 的图像是一条抛物线（光滑的 U 形曲线）。如果 a > 0，抛物线开口向上；如果 a < 0，抛物线开口向下。方程 ax² + bx + c = 0 的根就是该图像的 x 轴截距，也就是曲线与 x 轴相交的点。
Every parabola has a vertical line of symmetry. The axis of symmetry passes through the turning point (the vertex) and has the equation x = −b ÷ 2a. The vertex is the minimum point when a > 0 and the maximum point when a < 0.
每条抛物线都有一条竖直对称轴。对称轴经过顶点，其方程为 x = −b ÷ 2a。当 a > 0 时，顶点是最小值点；当 a < 0 时，顶点是最大值点。
This graph picture tells you how many real roots to expect: two intercepts mean two roots, one touching point means one repeated root, and no intercept means no real roots.
这幅图像可以告诉你应有几个实根：两个交点表示两个根，只接触一点表示一个重根，没有交点表示没有实根。
3. Three Main Methods at a Glance | 三种主要方法一览
IGCSE students are expected to solve quadratic equations by factorising, by completing the square and by using the quadratic formula. The syllabus also asks you to recognise when each method is the quickest and most reliable.
IGCSE 学生需要掌握三种解二次方程的方法：因式分解、配方法和求根公式。考纲还要求你判断哪种方法最快、最可靠。
| Method | Best used when | Answer type |
| Factorising | Coefficients are small integers and factorise neatly | Exact (often integers or fractions) |
| Completing the square | You need the vertex, or b is even and a = 1 | Exact, with surds |
| Quadratic formula | Factorising fails, or the numbers are messy | Exact surds or decimals |
In an exam you normally choose the method that the question hints at. If it says "solve by factorising", do not use the formula, because method marks may be lost.
考试中通常要选择题目暗示的方法。如果题目写着"用因式分解求解"，就不要用求根公式，否则可能丢分。
4. Method 1: Solving by Factorising | 方法一：用因式分解求解
Factorising relies on a simple but powerful idea: if two numbers multiply to give zero, then at least one of them must be zero. In symbols, if PQ = 0 then P = 0 or Q = 0. This is called the zero product property.
因式分解依赖于一个简单而有力的思想：如果两个数相乘等于零，那么其中至少有一个数为零。用符号表示：若 PQ = 0，则 P = 0 或 Q = 0。这称为零积性质。
For a monic quadratic x² + bx + c = 0, look for two numbers that multiply to give c and add to give b. Then write (x + p)(x + q) = 0 and set each bracket equal to zero.
对于首项系数为 1 的二次式 x² + bx + c = 0，找出两个数，使它们的乘积为 c、和为 b。然后写成 (x + p)(x + q) = 0，再令每个括号等于零。
Worked example 1: Solve x² − 5x + 6 = 0. We need two numbers multiplying to +6 and adding to −5, namely −2 and −3. So (x − 2)(x − 3) = 0, giving x = 2 or x = 3.
例 1：解 x² − 5x + 6 = 0。需要两个数，乘积为 +6、和为 −5，即 −2 和 −3。因此 (x − 2)(x − 3) = 0，得到 x = 2 或 x = 3。
Worked example 2: Solve 2x² + 7x − 4 = 0 bers that multiply to −8 and add to +7 are +8 and −1. Split the middle term and factorise in pairs: 2x² + 8x − x − 4 = 2x(x + 4) − 1(x + 4) = (2x − 1)(x + 4). So x = 1/2 or x = −4.
例 2：解 2x² + 7x − 4 = 0。先算 a 与 c 的乘积：2 × (−4) = −8。乘积为 −8、和为 +7 的两个数是 +8 和 −1。拆开中间项并分组分解：2x² + 8x − x − 4 = 2x(x + 4) − 1(x + 4) = (2x − 1)(x + 4)。所以 x = 1/2 或 x = −4。
5. Special Cases: Common Factors and the Difference of Two Squares | 特殊情形：公因式与平方差
Always look for a common factor first. The equation 3x² − 12x = 0 factorises as 3x(x − 4) = 0, so x = 0 or x = 4. Never divide both sides by x, because that destroys the root x = 0.
务必先找公因式。方程 3x² − 12x = 0 可分解为 3x(x − 4) = 0，所以 x = 0 或 x = 4。千万不要两边同除以 x，因为那会丢掉根 x = 0。
The difference of two squares is another pattern to memorise: a² − b² = (a + b)(a − b). For instance, x² − 16 = 0 gives (x + 4)(x − 4) = 0, so x = ±4, and 4x² − 25 = 0 gives (2x + 5)(2x − 5) = 0, so x = ±5/2.
平方差是另一个必须记住的模式：a² − b² = (a + b)(a − b)。例如 x² − 16 = 0 给出 (x + 4)(x − 4) = 0，所以 x = ±4；而 4x² − 25 = 0 给出 (2x + 5)(2x − 5) = 0，所以 x = ±5/2。
Not every quadratic factorises with integers. If you cannot find a pair of numbers after a reasonable effort, switch to the quadratic formula instead of guessing.
并非所有二次式都能用整数分解。如果合理尝试后仍找不到那对数，就改用求根公式，不要盲目猜测。
6. Method 2: Completing the Square | 方法二：配方法
Completing the square rewrites a quadratic so that x appears only once, inside a squared bracket. For a monic expression, the key identity is x² + bx = (x + b/2)² − (b/2)². Adding the square of half the coefficient of x, then subtracting it, keeps the expression unchanged.
配方法把二次式改写成 x 只出现一次、且在一个平方括号内的形式。对于首项系数为 1 的式子，关键恒等式是 x² + bx = (x + b/2)² − (b/2)²。加上一次项系数一半的平方，再减去它，式子保持不变。
x² + bx + c = (x + b/2)² − (b/2)² + c
Worked example 3: Solve x² − 6x + 1 = 0 by completing the square. Half of −6 is −3, so write (x − 3)² − 9 + 1 = 0, that is (x − 3)² − 8 = 0. Hence (x − 3)² = 8, so x − 3 = ±√8 and x = 3 ± 2√2.
例 3：用配方法解 x² − 6x + 1 = 0。−6 的一半是 −3，所以写成 (x − 3)² − 9 + 1 = 0，即 (x − 3)² − 8 = 0。因此 (x − 3)² = 8，x − 3 = ±√8，得 x = 3 ± 2√2。
Completing the square always works, even when the equation cannot be factorised, and it gives exact answers with surds when the question asks for them.
配方法总是可行，即使方程无法因式分解；当题目要求精确答案时，它能给出带根号的精确值。
7. Completing the Square for the Vertex | 用配方法求顶点
When a quadratic is written in the completing-the-square form y = (x + p)² + q, the vertex is at (−p, q). If q > 0 the vertex is a minimum point when a > 0, and it gives the minimum value of the function.
当二次式写为配方形式 y = (x + p)² + q 时，顶点在 (−p, q)。当 a > 0 时，顶点是最小值点，同时也是函数的最小值。
Worked example 4: Write y = x² − 8x + 5 in completed-square form and state the minimum point. Half of −8 is −4, so y = (x − 4)² − 16 + 5 = (x − 4)² − 11. The minimum point is (4, −11) and the minimum value is −11.
例 4：把 y = x² − 8x + 5 写成配方形式，并写出最小值点。−8 的一半是 −4，故 y = (x − 4)² − 16 + 5 = (x − 4)² − 11。最小值点为 (4, −11)，最小值为 −11。
If a is not 1, take the coefficient a out of the first two terms first. For y = 2x² + 12x + 7: y = 2(x² + 6x) + 7 = 2[(x + 3)² − 9] + 7 = 2(x + 3)² − 11, so the vertex is (−3, −11).
若 a 不为 1，先把前两项中的 a 提出来。对 y = 2x² + 12x + 7：y = 2(x² + 6x) + 7 = 2[(x + 3)² − 9] + 7 = 2(x + 3)² − 11，所以顶点为 (−3, −11)。
8. Method 3: The Quadratic Formula | 方法三：求根公式
The quadratic formula is obtained by completing the square on the general equation ax² + bx + c = 0. It solves every quadratic equation with real roots and is the safest method when factorising is difficult.
求根公式是通过对一般方程 ax² + bx + c = 0 配方得到的。它能解出所有有实根的二次方程，在因式分解困难时是最稳妥的方法。
x = (−b ± √b² − 4ac) / 2a
the whole expression b² − 4ac. Do not drop the ± sign, because it produces the two roots.
先写出 a、b、c，再仔细代入，注意整个分子都要除以 2a，并记住根号覆盖整个 b² − 4ac。不要漏掉 ± 号，因为它给出了两个根。
Worked example 5: Solve 3x² − 4x − 2 = 0. Here a = 3, b = −4, c = −2, so b² − 4ac = 16 + 24 = 40. Then x = (4 ± √40) / 6 = (4 ± 2√10) / 6 = (2 ± √10) / 3, giving two exact roots.
例 5：解 3x² − 4x − 2 = 0。这里 a = 3，b = −4，c = −2，故 b² − 4ac = 16 + 24 = 40。于是 x = (4 ± √40) / 6 = (4 ± 2√10) / 6 = (2 ± √10) / 3，得到两个精确根。
9. The Discriminant | 判别式
The expression under the square root, Δ = b² − 4ac, is called the discriminant. Its sign tells you how many real roots the equation has before you solve it.
根号下的表达式 Δ = b² − 4ac 称为判别式。它的符号可以在求解之前告诉你方程有多少个实根。
- If Δ > 0, there are two distinct real roots and the parabola crosses the x-axis twice.
- there is exactly one repeated real root and the parabola touches the x-axis at the vertex.
- If Δ < 0, there are no real roots and the parabola never meets the x-axis.
- If Δ is a perfect square, the roots are rational, so the equation factorises with integers.
- 若 Δ > 0，有两个不同的实根，抛物线与 x 轴交于两点。
- 若 Δ = 0，恰有一个重实根，抛物线在顶点处与 x 轴相切。
- 若 Δ < 0，没有实根，抛物线与 x 轴永不相交。
- 若 Δ 是完全平方数，则根为有理数，方程可以用整数因式分解。
10. Worked Exam Examples | 考试例题精讲
Example A: Solve 2x² + 3x − 5 = 0 sing the formula, b² − 4ac = 9 + 40 = 49, which is a perfect square, so exact roots exist: x = (−3 ± 7) / 4, giving x = 1 or x = −5/2. As a check, factorising gives (2x + 5)(x − 1) = 0, which agrees.
例 A：解 2x² + 3x − 5 = 0。用公式，b² − 4ac = 9 + 40 = 49，是完全平方数，故有精确根：x = (−3 ± 7) / 4，得 x = 1 或 x = −5/2。检验：因式分解得 (2x + 5)(x − 1) = 0，结果一致。
Example B: The length of a rectangle is 3 cm more than its width and the area is 40 cm². Find the width. Let the width be w cm, so w(w + 3) = 40, hence w² + 3w − 40 = 0. Factorising: (w + 8)(w − 5) = 0, so w = 5 or w = −8. Since a length cannot be negative, the width is 5 cm.
例 B：一个矩形的长比宽多 3 cm，面积为 40 cm²。求宽。设宽为 w cm，则 w(w + 3) = 40，即 w² + 3w − 40 = 0。因式分解：(w + 8)(w − 5) = 0，所以 w = 5 或 w = −8。由于长度不能为负，宽为 5 cm。
Example C: A projectile height is modelled by h = −5t² + 20t + 1. To find when h = 16, solve −5t² + 20t − 15 = 0, that is t² − 4t + 3 = 0, so (t − 1)(t − 3) = 0 and t =
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