📚 Solving Quadratic Equations: Factorisation, Formula & Graphs | 二次方程求解：因式分解、求根公式与图象
Quadratic equations form one of the most important strands of the Edexcel IGCSE Mathematics syllabus. Questions on expanding, factorising, solving and interpreting quadratic expressions appear in both Paper 1 and Paper 2, and a secure grasp of this topic supports later work on sequences, functions, calculus and coordinate geometry.
二次方程是 Edexcel IGCSE 数学考纲中最重要的主线之一。有关展开、因式分解、求解和解释二次表达式的题目在试卷 1 和试卷 2 中都会出现，扎实掌握这一主题还能为后续学习数列、函数、微积分和坐标几何打下坚实基础。
This guide takes you through every essential skill step by step: from the standard form, through factorisation and the quadratic formula, to the discriminant, graphs, inequalities and real-life applications. Work through each example yourself, and you will be well prepared for any quadratic question the exam can offer.
本指南将带你循序渐进地掌握每一个必备技能：从标准形式、因式分解和求根公式，到判别式、图象、不等式与实际应用。请自己动手把每个例子演算一遍，这样无论考试中出现怎样的二次方程题目，你都能从容应对。
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 standard form below, where a, b and c are real constants and a is not zero.
二次方程是指可以写成如下标准形式的方程，其中 a、b、c 是实数常数，且 a 不等于零。
ax² + bx + c = 0, a ≠ 0
The reason a cannot be zero is simple: if a = 0, the term ax² disappears and the equation becomes linear rather than quadratic.
a 不能为零的原因很简单：如果 a = 0，ax² 这一项就消失了，方程会从二次降为一次（线性）方程。
Here are three examples. x² − 7x + 10 = 0 is quadratic because it contains x². The equation 2x² + 5 = 0 is also quadratic, even though it has no x term; here b = 0. However, x³ + 2x = 0 is a cubic equation, not quadratic, so the methods in this article do not apply to it.
请看三个例子。x² − 7x + 10 = 0 含有 x² 项，所以是二次方程；2x² + 5 = 0 也是二次方程，尽管它没有 x 项，此时 b = 0。但 x³ + 2x = 0 是三次方程而不是二次方程，因此本文的方法对它并不适用。
Identifying the coefficients a, b and c correctly is the very first step in every solution. For example, in 3x² − 4x + 1 = 0 we have a = 3, b = −4 and c = 1. A common mistake is to forget the negative sign on b, so always write the equation down with terms lined up in the order of descending powers of x.
正确识别系数 a、b、c 是求解的第一步。例如，在 3x² − 4x + 1 = 0 中，a = 3，b = −4，c = 1。常见的错误是漏掉 b 的负号，所以务必先把方程按 x 的降幂顺序写好。
2. Expanding and Factorising Quadratics | 展开与因式分解二次式
Solving and graphing quadratics both depend on fluency with expanding and factorising. To expand (x + 3)(x + 4), multiply every term in the first bracket by every term in the second.
求解和作图都依赖于熟练的展开与因式分解。要展开 (x + 3)(x + 4)，只需把第一个括号中的每一项乘以第二个括号中的每一项。
(x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12
Factorising is the reverse process. For x² + bx + c, look for two numbers that multiply to give c and add to give b. The table below shows four common patterns you should recognise instantly.
因式分解是展开的逆过程。对于 x² + bx + c，我们需要找到两个数，使它们的乘积等于 c，和等于 b。下表展示了四种你必须一眼识别的常见模式。
| Expression 表达式 | Factorised form 因式分解形式 |
| x² + 5x + 6 | (x + 2)(x + 3) |
| x² − 5x + 6 | (x − 2)(x − 3) |
| x² − 9 | (x − 3)(x + 3) |
| x² + 6x + 9 | (x + 3)² |
For x² − 9, the two numbers would be 3 and −3: they multiply to −9 and add to 0. This special case, a difference of two squares, appears frequently, so memorise the pattern x² − p² = (x − p)(x + p).
对于 x² − 9，两个数是 3 和 −3：它们相乘得 −9，相加得 0。这种特殊情况叫做平方差，考试中经常出现，请记住公式 x² − p² = (x − p)(x + p)。
When the coefficient of x² is not 1, you may need to use the ‘two brackets’ method with trial and error. For example, 2x² + 5x + 2 factorises as (2x + 1)(x + 2), because the x terms give 4x + x = 5x. Write out the possible bracket pairs and check the middle term.
当 x² 的系数不是 1 时，你可能需要用试凑的“双括号”法。例如，2x² + 5x + 2 = (2x + 1)(x + 2)，因为 x 项为 4x + x = 5x。把可能的括号组合写出来并检查中间项即可。
3. Solving by Factorisation | 用因式分解法求解
Factorisation is the fastest method of solving a quadratic, provided the expression factorises neatly with integers. The key idea is the zero product property: if the product of two numbers is zero, then at least one of them must be zero.
因式分解法是求解二次方程最快的方法，前提是表达式能用整数漂亮地分解。其核心思想是零积性质：如果两个数的乘积为零，那么其中至少有一个数为零。
If AB = 0, then A = 0 or B = 0.
若 AB = 0，则 A = 0 或 B = 0。
Consider x² − 5x + 6 = 0. We need two numbers that multiply to 6 and add to −5, namely −2 and −3. Therefore the equation becomes (x − 2)(x − 3) = 0.
以 x² − 5x + 6 = 0 为例。我们需要找到两个数相乘得 6、相加得 −5，即 −2 和 −3。于是原方程变为 (x − 2)(x − 3) = 0。
Applying the zero product property gives x − 2 = 0 or x − 3 = 0, so x = 2 or x = 3. You should always check your answers by substituting them back into the original equation: 2² − 10 + 6 = 0 and 3² − 15 + 6 = 0.
应用零积性质可得 x − 2 = 0 或 x − 3 = 0，因此 x = 2 或 x = 3。你应当随时把答案代回原方程验算：2² − 10 + 6 = 0，3² − 15 + 6 = 0，均成立。
Write your final answer using ‘or’, not ‘and’, because x cannot equal both values at the same time. In an exam, always show the factorised line before giving the solutions, because the method mark depends on it.
最终答案要用“或 (or)”而不是“且 (and)”来连接，因为 x 不可能同时等于这两个值。考试中务必写出因式分解后的那一步，因为它决定你是否能拿到方法分。
4. Solving by Completing the Square | 用配方法求解
Not every quadratic factorises with integers. Completing the square is a powerful algebraic technique that rewrites x² + px as a perfect square plus a constant.
并非每个二次式都能用整数因式分解。配方法是一种强大的代数技巧，它把 x² + px 改写成完全平方加上一个常数的形式。
x² + px = (x + p/2)² − (p/2)²
For instance, x² + 6x = (x + 3)² − 9, because expanding (x + 3)² gives x² + 6x + 9, so we subtract 9 to keep the expression unchanged.
例如，x² + 6x = (x + 3)² − 9，因为展开 (x + 3)² 得到 x² + 6x + 9，所以要减去 9 才能保持原式不变。
Let us solve x² + 6x − 7 = 0 by completing the square. First we deal with x² + 6x: this becomes (x + 3)² − 9. The full equation is therefore:
下面我们用配方法求解 x² + 6x − 7 = 0。先把 x² + 6x 处理成 (x + 3)² − 9，于是原方程变为：
(x + 3)² − 9 − 7 = 0, so (x + 3)² = 16
Taking square roots of both sides gives x + 3 = ±4, so x = −3 + 4 = 1 or x = −3 − 4 = −7. Remember that √16 has two values, +4 and −4; forgetting the negative root is a very common lost mark.
两边同时开平方得 x + 3 = ±4，于是 x = −3 + 4 = 1 或 x = −3 − 4 = −7。切记 √16 有两个值：+4 和 −4；漏掉负根是非常常见的失分点。
Completing the square is especially useful later when you need to find the turning point of a quadratic graph, because it gives the vertex directly. It also produces the quadratic formula, which we meet next.
配方法在后文求二次函数图象的顶点时尤为有用，因为它能直接给出顶点坐标。它还能推导出接下来要讲的求根公式。
5. The Quadratic Formula | 求根公式
When a quadratic does not factorise neatly, the quadratic formula solves any equation of the form ax² + bx + c = 0 in one step. You are given this formula in the Edexcel IGCSE formula booklet, but you must know how to substitute into it correctly.
当二次式无法整齐分解时，求根公式可以一步解出任何 ax² + bx + c = 0 形式的方程。Edexcel IGCSE 公式簿中会提供该公式，但你必须清楚如何正确代入。
x = (−b ± √(b² − 4ac)) / 2a
Worked example: solve 2x² + 3x − 2 = 0. Here a = 2, b = 3 and c = −2. First calculate the discriminant part inside the square root:
示例：解方程 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。先计算根号内的判别式部分：
b² − 4ac = 3² − 4 × 2 × (−2) = 9 + 16 = 25
Substituting into the formula gives:
代入公式得到：
x = (−3 ± √25) / 4 = (−3 ± 5) / 4
Hence x = (−3 + 5)/4 = ½ or x = (−3 − 5)/4 = −2. Notice that a negative value for c makes −4ac positive; handling the signs carefully is essential.
因此 x = (−3 + 5)/4 = ½ 或 x = (−3 − 5)/4 = −2。注意当 c 为负时，−4ac 为正；细心处理符号至关重要。
When using the formula, always write down the substitution line in full before simplifying. This will earn you method marks even if you make a small arithmetic slip later.
使用公式时，务必先完整写出代入的那一步再化简。这样即使后面出现小的计算错误，你仍能获得方法分。
6. The Discriminant | 判别式
The expression b² − 4ac inside the quadratic formula is called the discriminant, usually denoted Δ. It tells you how many real roots a quadratic equation has, without solving it fully.
求根公式根号内的表达式 b² − 4ac 称为判别式，通常记作 Δ。它能在不求全解的情况下告诉我们二次方程有几个实数根。
The table below summarises the three possible cases. Remember that the graph of y = ax² + bx + c crosses the x-axis at the roots of ax² + bx + c = 0.
下表总结了三种可能的情形。请记住，y = ax² + bx + c 的图象与 x 轴的交点正是方程 ax² + bx + c = 0 的根。
| Discriminant 判别式 | Real roots 实数根 | Graph behaviour 图象特征 |
| Δ > 0 | Two distinct roots 两个不同实数根 | Crosses the x-axis twice 与 x 轴相交于两点 |
| Δ = 0 | One repeated root 一个重根 | Touches the x-axis once 与 x 轴相切于一点 |
| Δ < 0 | No real roots 无实数根 | Never meets the x-axis 与 x 轴无交点 |
For x² − 4x + 3 = 0, we have Δ = (−4)² − 4 × 1 × 3 = 16 − 12 = 4, which is positive, so the equation has two distinct roots. Indeed (x − 1)(x − 3) = 0 gives x = 1 and x = 3.
对于 x² − 4x + 3 = 0，Δ = (−4)² − 4 × 1 × 3 = 16 − 12 = 4，大于零，因此方程有两个不同的实数根。事实上 (x − 1)(x − 3) = 0 给出 x = 1 和 x = 3。
For x² + 2x + 5 = 0, we get Δ = 4 − 20 = −16 < 0, so there are no real roots. The graph never touches the x-axis. In questions that say 'show that the equation has no real roots', calculating Δ is the expected method. 对于 x² + 2x + 5 = 0，Δ = 4 − 20 = −16 < 0，因此没有实数根，图象与 x 轴永不相交。遇到“证明方程无实数根”的题目时，计算 Δ 正是所期望的方法。
7. Quadratic Graphs and Intersections | 二次函数图象与交点
A quadratic function y = ax² + bx + c produces a parabola. If a > 0, the parabola opens upwards like a smile; if a < 0, it opens downwards like a frown. 二次函数 y = ax² + bx + c 的图象是一条抛物线。当 a > 0 时，抛物线开口向上，形如微笑；当 a < 0 时，开口向下，形如皱眉。
The key points of a quadratic graph are the roots (where y = 0), the y-intercept (where x = 0) and the turning point. Consider y = x² − 4x + 3.
二次函数图象的关键点是根（y = 0 处）、y 轴截距（x = 0 处）和顶点。请看 y = x² − 4x + 3。
The y-intercept is found by setting x = 0, which gives y = 3. So the graph passes through (0, 3).
y 轴截距通过令 x = 0 求得，即 y = 3，所以图象经过点 (0, 3)。
The roots are found by solving x² − 4x + 3 = 0. Since this factorises as (x − 1)(x − 3) = 0, the graph crosses the x-axis at x = 1 and x = 3.
根通过解 x² − 4x + 3 = 0 求得。由于它可以分解为 (x − 1)(x − 3) = 0，因此图象在 x = 1 和 x = 3 处穿过 x 轴。
Because the coefficient of x² is positive, the graph opens upwards, so the curve dips below the x-axis between x = 1 and x = 3 and rises above it elsewhere. This ‘sign behaviour’ is crucial for solving quadratic
Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply