📚 Solving Quadratic Equations | 解一元二次方程
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, coordinate geometry, and word problems across both Paper 1 and Paper 2. Mastering the different methods of solving quadratics will give you a strong foundation for Additional Mathematics and A Level study.
一元二次方程是 IGCSE 数学中最核心的考点之一，在代数、坐标几何以及应用题中频繁出现，覆盖 Paper 1 和 Paper 2。熟练掌握解二次方程的各种方法，将为你学习附加数学（Additional Maths）和 A Level 打下坚实基础。
1. What Is a Quadratic Equation? | 什么是一元二次方程
c are real numbers and a ≠ 0. If a = 0, the equation becomes linear rather than quadratic.
一元二次方程是指未知数最高次数为 2 的方程。其一般形式为 ax² + bx + c = 0，其中 a、b、c 为实数，且 a ≠ 0。若 a = 0，方程就会退化为一次方程，而不属于二次方程。
ax² + bx + c = 0, where a ≠ 0
In IGCSE exams, quadratics may appear in three equivalent forms. You should be able to recognise and convert between them.
在 IGCSE 考试中，二次方程可以以三种等价形式出现，你需要能够识别并在它们之间相互转换。
| Form 形式 | Expression 表达式 | Use 用途 |
|---|---|---|
| General 一般式 | ax² + bx + c = 0 | Using the formula and discriminant 使用公式和判别式 |
| Factorised 因式分解式 | a(x − p)(x − q) = 0 | Reading roots directly 直接读出根 |
| Completed square 配方式 | a(x − h)² + k = 0 | Finding the turning point 求顶点 |
2. Solving by Factorisation | 因式分解法
Factorisation is often the quickest method when a quadratic has simple integer roots. The key idea is 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时，因式分解通常是最快捷的方法。其核心是零积性质：若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
Worked example: Solve x² − 7x + 12 = 0.
例题：解方程 x² − 7x + 12 = 0。
) + (−4) = −7. Hence the equation factorises as (x − 3)(x − 4) = 0, giving x = 3 or x = 4.
我们需要找到两个数，它们的乘积为 12，和为 −7。这两个数就是 −3 和 −4，因为 (−3) × (−4) = 12，且 (−3) + (−4) = −7。因此原方程可分解为 (x − 3)(x − 4) = 0，解得 x = 3 或 x = 4。
Follow these steps when factorising:
因式分解时请按以下步骤操作：
Rearrange the equation so that all terms are on one side and the other side is 0. | 将方程整理为一边为 0 的标准形式。
Factorise the quadratic expression fully. | 将二次表达式完全因式分解。
Set each factor equal to 0 and solve the resulting linear equations. | 令每个因式等于 0，并解出相应的一次方程。
Write down both solutions clearly. | 清楚写出两个解。
3. Special Cases: Common Factor and Difference of Two Squares | 特殊情况：公因式与平方差
Two special patterns appear frequently in IGCSE papers and allow quick factorisation.
有两种特殊模式在 IGCSE 考卷中经常出现，可以快速完成因式分解。
Pattern 1 — Common factor: Look for a factor shared by every term before trying any other method.
模式一——提公因式：在尝试其他方法之前，先检查每一项是否有公共因子。
3x² − 12x = 0 → 3x(x − 4) = 0, so x = 0 or x = 4
Pattern 2 — Difference of two squares: Any expression of the form a² − b² factorises as (a − b)(a + b).
模式二——平方差公式：任何形如 a² − b² 的表达式都可分解为 (a − b)(a + b)。
a² − b² = (a − b)(a + b)
a quadratic with no constant term, such as 2x² − 8x = 0, always has x = 0 as one solution.
例如，x² − 25 = 0 可化为 (x − 5)(x + 5) = 0，因此 x = 5 或 x = −5。注意，没有常数项的二次方程，如 2x² − 8x = 0，一定有一个解是 x = 0。
4. The Quadratic Formula | 求根公式
solutions are given by the following formula. You must memorise this formula for the exam.
当因式分解困难甚至无法进行时，应使用求根公式。对于任意形如 ax² + bx + c = 0 的方程，解由下面的公式给出。你必须牢记这个公式以应对考试。
x = (−b ± √b² − 4ac) / 2a
Worked example: Solve 2x² − 3x − 5 = 0 using the quadratic formula.
例题：用求根公式解方程 2x² − 3x − 5 = 0。
Here a = 2, b = −3 and c = −5. Substitute these values carefully into the formula:
这里 a = 2，b = −3，c = −5。将这些值仔细代入公式：
x = (3 ± √((−3)² − 4 × 2 × (−5))) / (2 × 2) = (3 ± √9 + 40) / 4 = (3 ± 7) / 4
ound negative values of b and c when substituting; this avoids the most common sign error.
因此 x = (3 + 7) / 4 = 2.5，x = (3 − 7) / 4 = −1。代入 b 和 c 为负值时务必加上括号，这是避免最常见符号错误的有效方法。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x − h)² + k. This method is especially useful when the roots are irrational and when you are asked to find the minimum or maximum point of a curve.
配方法将二次式改写为 a(x − h)² + k 的形式。当根为无理数时，或题目要求求曲线的最低点或最高点时，这种方法尤为有用。
x² + bx = (x + b/2)² − (b/2)²
Worked example: Solve x² + 6x − 4 = 0 by completing the square.
例题：用配方法解方程 x² + 6x − 4 = 0。
= x² + 6x + 9, we must subtract 9 to keep the expression unchanged. The equation becomes:
取 6 的一半，即 3，写成 (x + 3)²。因为 (x + 3)² = x² + 6x + 9，所以必须减去 9 才能保持式子不变。方程变为：
(x + 3)² − 9 − 4 = 0 → (x + 3)² = 13 → x + 3 = ±√13 → x = −3 ± √13
You may leave the answer in surd form unless the question asks for a decimal. If a ≠ 1, first factor out a from the x² and x terms before completing the square.
除非题目要求小数答案，否则你可以保留根式形式。如果 a ≠ 1，先提取 x² 项和 x 项的公共系数 a，再进行配方。
6. The Discriminant | 判别式
ells you how many real roots a quadratic equation has, without solving the equation fully.
判别式是求根公式中根号内的表达式，记作 Δ = b² − 4ac。它可以在不完全解方程的情况下判断一个二次方程有多少个实数根。
| Discriminant 判别式 | Number of Real Roots 实数根个数 | Graph Interpretation 图象含义 |
|---|---|---|
| Δ > 0 | Two distinct real roots 两个不同的实数根 | Curve crosses the x-axis twice 曲线与 x 轴有两个交点 |
| Δ = 0 | One repeated root 两个相等的实数根（重根） | Curve touches the x-axis once 曲线与 x 轴相切 |
| Δ < 0 | No real roots 无实数根 | Curve does not meet the x-axis 曲线与 x 轴无交点 |
Additionally, if a, b and c are integers and Δ is a perfect square, the quadratic factorises with integer factors. This is a useful check before spending time on the formula.
此外，如果 a、b、c 为整数且 Δ 是完全平方数，则该二次式可以用整数因子分解。这是一个有用的预判方法，可以避免盲目套用公式。
7. Graphs of Quadratic Functions | 二次函数的图象
ymmetric curve called a parabola. If a > 0, the parabola opens upwards and the curve is U-shaped; if a < 0, it opens downwards and is ∩-shaped. The larger the value of |a|, the steeper and narrower the parabola.
函数 y = ax² + bx + c 的图象是一条平滑对称的曲线，称为抛物线。当 a > 0 时，抛物线开口向上，呈 U 形；当 a < 0 时，开口向下，呈 ∩ 形。|a| 的值越大，抛物线越窄越陡。
When sketching a quadratic graph, you should identify three sets of key features:
画二次函数图象草图时，你需要确定三类关键特征：
Roots: the x-values where the curve crosses the x-axis, found by solving y = 0. | 根：曲线与 x 轴交点的 x 坐标，通过解 y = 0 求得。
y-intercept: the point (0, c), where the curve meets the y-axis. | y 轴截距：点 (0, c)，即曲线与 y 轴的交点。
Turning point: the minimum point when a > 0, or the maximum point when a < 0. | 顶点：当 a > 0 时为最低点，当 a < 0 时为最高点。
Remember that if the equation has no real roots, the parabola never touches the x-axis and you must rely on the y-intercept and turning point to sketch it.
注意，若方程无实数根，抛物线不会接触 x 轴，此时你需要依靠 y 轴截距和顶点来绘制草图。
8. Axis of Symmetry and Turning Point | 对称轴与顶点
of symmetry. For y = ax² + bx + c, the axis of symmetry is given by the following formula.
每条抛物线都关于一条竖直直线对称，这条直线称为对称轴。对于 y = ax² + bx + c，对称轴由下面的公式给出。
x = −b / (2a)
a(x − h)² + k, then the turning point is simply (h, k) and the axis of symmetry is x = h.
顶点始终位于对称轴上。若二次函数写成配方式 y = a(x − h)² + k，则顶点直接为 (h, k)，对称轴为 x = h。
Example: s a minimum point at (3, 5) because a = 2 > 0, and its axis of symmetry is the line x = 3.
例题：曲线 y = 2(x − 3)² + 5 的最低点为 (3, 5)，因为 a = 2 > 0，其对称轴为直线 x = 3。
(2a) back into the original equation. This is a common exam question worth several marks.
若要从一般式求顶点纵坐标，将 x = −b / (2a) 代回原方程即可。这是常见的考试题型，分值通常不小。
9. Solving from Graphs and Real-Life Applications | 图象法与实际应用
4x + 3. To solve x² − 4x + 3 = 1, draw the horizontal line y = 1 on the same axes and read
Published by TutorHao | IGCSE 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