📚 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 run through the whole of the IGCSE and A-Level mathematics syllabus, from pure algebraic manipulation to graph sketching and real-world modelling. This revision guide explains what a quadratic equation is, how to solve one by factorising, by completing the square and by the quadratic formula, what the discriminant tells you, how the graph relates to the algebra, and which mistakes cost marks most often in exams.
二次方程贯穿 IGCSE 与 A-Level 数学考纲的始终，从纯代数运算到图像绘制、再到现实建模都有涉及。本复习指南将解释什么是二次方程，如何用因式分解法、配方法和求根公式求解，判别式能告诉我们什么，图像与代数之间如何对应，以及考场上最常丢分的错误。
1. What Is a Quadratic Equation? | 什么是二次方程？
pear in mechanics, sequences and modelling questions. Once all terms are collected on one side, every quadratic equation can be written in the pattern a x² + b x + c = 0 with a not equal to zero.
二次方程是指未知数的最高次数为 2 的方程。未知数通常写作 x，但在力学、数列和建模题中也会出现 t、y、n 或 h 等字母。把所有项移到同一边之后，任何二次方程都可以写成 a x² + b x + c = 0 的形式，其中 a 不等于零。
A quadratic equation is fundamentally different from a linear equation because it can have up to two solutions. Depending on the coefficients, there may be two distinct real solutions, one repeated solution, or no real solutions at all. Each solution is also called a root, and the roots are exactly the values of x that make the expression equal to zero.
二次方程与一次方程有本质区别，因为它最多可以有两个解。根据系数的不同，它可能有两个不同的实数解、一个重根解，或者完全没有实数解。每个解也称为一个根，而根正是使该表达式等于零的 x 的取值。
of the quadratic expression. This single idea is the bridge between factorising, graph sketching and the remainder theorem you meet later in A-Level.
根与因式密切相关。如果 x = p 是一个根，那么 (x - p) 就是该二次式的一个因式。这一个想法把因式分解、图像绘制和你在 A-Level 后续学到的余数定理连接起来。
2. Standard Form and Coefficients | 标准形式与系数
is the coefficient of x², b is the coefficient of x, and c is the constant term. The condition a is not equal to 0 is essential; if a were zero the equation would become linear. Before solving anything, always rearrange the equation so that one side is zero and the terms are written in descending powers of x.
标准形式为 a x² + b x + c = 0，其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。条件 a 不等于 0 至关重要；如果 a 为零，方程就退化为一次方程。在求解之前，务必把方程整理为一边为零、各项按 x 的降幂排列。
kipping the rearrangement will spread into every later step. For example, 3x² = 5 - 2x must first become 3x² + 2x - 5 = 0 before a = 3, b = 2 and c = -5 can be used.
整理不是可做可不做的步骤。求根公式和判别式都直接从标准形式中读取 a、b、c，因此跳过整理而产生的符号错误会一路传导到之后的每一步。例如 3x² = 5 - 2x 必须先整理为 3x² + 2x - 5 = 0，然后才能取 a = 3、b = 2、c = -5。
| Equation | 方程 | a | b | c |
|---|---|---|---|
| 3x² + 5x - 2 = 0 | 3 | 5 | -2 |
| -x² + 4x + 1 = 0 | -1 | 4 | 1 |
| x² - 7x = 0 | 1 | -7 | 0 |
| 2x² = 8 | 2 | 0 | -8 |
Notice that b or c can be zero, and that a can be negative. Only a is forbidden from being zero. A missing term simply means its coefficient is zero, and it must still be included when you substitute into the formula.
注意 b 或 c 都可以为零，a 也可以是负数，只有 a 不允许为零。缺少某一项只意味着它的系数为零，代入公式时仍然要把它算进去。
3. Solving by Factorising | 因式分解法
ers is zero, at least one of them must be zero. So if the expression becomes (p x + q)(r x + s) = 0, then p x + q = 0 or r x + s = 0, giving two simple linear equations to solve.
当二次式能分解为整数因式时，因式分解法是最快的方法。它依赖零乘积法则：如果两个数的乘积为零，那么其中至少有一个为零。因此若表达式化为 (p x + q)(r x + s) = 0，则 p x + q = 0 或 r x + s = 0，从而得到两个简单的一次方程来求解。
For the simple case x² + b x + c, look for two integers p and q whose product is c and whose sum is b. The factorisation is then immediate.
对于 x² + b x + c 这种简单情形，寻找两个整数 p 和 q，使它们的乘积为 c、和为 b，因式分解便立刻得到。
x² + (p + q) x + p q = (x + p)(x + q)
n factorise in pairs. For example, 2x² + 7x + 3 has a c = 6, and 6 + 1 = 7, so split 7x into 6x + x to get 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3), which gives (2x + 1)(x + 3).
对于 a 不等于 1 的一般情形 a x² + b x + c，使用 a 乘 c 法。找到两个数，使它们的乘积为 a c、和为 b，把中间项拆开，再分组提取公因式。例如 2x² + 7x + 3 中 a c = 6，而 6 + 1 = 7，于是把 7x 拆成 6x + x，得 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3)，从而得到 (2x + 1)(x + 3)。
such as 9x² - 16 = (3x - 4)(3x + 4). Also remember that not every quadratic factorises over the integers; when no integer pair works, switch to the formula rather than losing time.
一定要先检查有没有公因式，并留意平方差公式 a² - b² = (a - b)(a + b)，它经常出现在诸如 9x² - 16 = (3x - 4)(3x + 4) 这样的考题中。还要记住，并非每个二次式都能在整数范围内分解；当找不到整数对时，应改用求根公式，而不是浪费时间。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x - h)² + k. It is the method examiners expect when a question asks for the turning point, the maximum or minimum value, or a proof that an expression is always positive. It is also the tool used to derive the quadratic formula itself.
配方法把二次式改写为 a(x - h)² + k 的形式。当题目要求求顶点、最大值或最小值，或者证明某表达式恒为正时，考官期待的就是这种方法。它也是推导求根公式本身所用的工具。
For a monic expression x² + b x + c, half the coefficient of x, square it, and subtract it again so the value is unchanged. The rule is:
对于首项系数为 1 的表达式 x² + b x + c，取 x 的系数的一半、平方、再减去它，使式子的值保持不变。规则是：
x² + b x + c = (x + b ÷ 2)² + c - b² ÷ 4
When a is not 1, first take a out as a factor from the x terms only, complete the square inside the bracket, then multiply the constant back out and add it to the outer constant. Forgetting to multiply that constant back is one of the most common errors in this topic.
当 a 不等于 1 时，先把 a 从含 x 的项中提出来（只提含 x 的项），在括号内配方，再把常数乘回 a 并加到括号外的常数上。忘记把常数乘回去是本主题最常见的错误之一。
the completed square equal to zero also solves the equation: (x + 3)² = 8 gives x = -3 ± 2√2.
在顶点式 a(x - h)² + k 中，顶点坐标为 (h, k)。因此 x² + 6x + 1 化为 (x + 3)² - 8，最小值点位于 (-3, -8)。令配方后的式子等于零也能解方程：(x + 3)² = 8，得 x = -3 ± 2√2。
5. Deriving the Quadratic Formula | 推导求根公式
de every term by a to make the leading coefficient 1, then move the constant to the right-hand side.
求根公式并非凭空而来，它是对一般方程配方得到的。从 a x² + b x + c = 0（a 不等于 0）出发，每一项除以 a 使首项系数为 1，再把常数移到右边。
b ÷ a, which is b² ÷ (4a²), to both sides; the left side becomes a perfect square (x + b ÷ 2a)²; combine the fractions on the right to get (b² - 4a c) ÷ (4a²); take the square root of both sides; and finally subtract b ÷ 2a.
推导过程如下：除以 a 得 x² + (b ÷ a)x + c ÷ a = 0；移项得 x² + (b ÷ a)x = -c ÷ a；两边同时加上 (b ÷ a) 一半的平方，即 b² ÷ (4a²)；左边成为完全平方 (x + b ÷ 2a)²；右边通分得 (b² - 4a c) ÷ (4a²)；两边开平方；最后减去 b ÷ 2a。
The result is the quadratic formula that solves every quadratic equation with real coefficients.
所得结果就是能够求解一切实系数二次方程的求根公式。
x = (-b ± √b² - 4a c) ÷ (2a)
Because the derivation is a standard proof question, learn the sequence of steps rather than just the final line. Examiners award marks for the intermediate lines, especially for adding b² ÷ (4a²) to both sides and for combining the fractions correctly.
由于这段推导本身是标准的证明题，要记住步骤顺序而不只是最后一行。考官会给出中间步骤的分，尤其是两边加上 b² ÷ (4a²) 以及正确通分这两步。
6. Using the Quadratic Formula | 使用求根公式
Step one is to rearrange the equation into standard form and check that a is not zero. Skipping this step is the single biggest source of wrong signs in exam scripts.
第一步是把方程整理为标准形式，并确认 a 不为零。跳过这一步是考卷上出现符号错误的最大原因。
ng. Writing a = 3, b = -7 and c = 2 on the page makes arithmetic slips far easier to spot.
第二步是先明确写出 a、b、c 及其符号，然后再代入。把 a = 3、b = -7、c = 2 写在卷面上，会让计算失误更容易被发现。
be written as -(-7), which equals 7. The denominator is 2a, not a, so 2 times 3 equals 6.
第三步是小心代入，负数要加括号。若 b = -7，则 -b 要写成 -(-7)，结果为 7。分母是 2a 而不是 a，所以 2 乘 3 等于 6。
Step four is to simplify the square root and then the whole fraction. Simplify √40 as 2√10, then divide numerator and denominator by 2 where possible. Leave the answer in surd form unless the question asks for a decimal.
第四步是化简根号，再化简整个分数。把 √40 化为 2√10，然后尽可能把分子分母同时除以 2。除非题目要求小数，否则答案应保留根式形式。
Step five is to round correctly when decimals are required. Three significant figures is the usual IGCSE standard, and if the question says two decimal places, give exactly two. Never round in the middle of the calculation; keep full accuracy until the final line.
第五步是在需要小数时正确取近似值。IGCSE 通常要求三位有效数字；若题目要求两位小数，就精确给出两位。绝不要在计算中途取近似，要保留完整精度直到最后一行。
7. The Discriminant | 判别式
The expression under the square root, b² - 4a c, is called the discriminant and is often written as Δ. It decides how many real roots the equation has, without the need to solve it. This makes it the key tool for questions about the number of intersections between a curve and a line, or about finding an unknown constant.
根号下的表达式 b² - 4a c 叫做判别式，通常写作 Δ。它无需解方程就能决定方程有多少个实根。因此，它是处理曲线与直线交点个数、或求未知常数这类问题的关键工具。
| Discriminant | 判别式 | Nature of roots | 根的性质 | Graph | 图像 |
|---|---|---|
| Δ > 0 | Two distinct real roots | 两个不同实根 | Two x-axis crossings | 与 x 轴有两个交点 |
| Δ = 0 | One repeated root | 一个重根 | Touches the x-axis | 与 x 轴相切 |
| Δ < 0 | No real roots | 无实根 | No x-axis crossing | 与 x 轴无交点 |
e but not a perfect square, the roots are irrational surds that come in conjugate pairs of the form m ± √n.
当 Δ 是正完全平方数且 a、b、c 为整数时，根是有理数，表达式可在整数范围内因式分解。当 Δ 为正但不是完全平方数时，根是无理根式，并以 m ± √n 的共轭形式成对出现。
tangency, such as a line meeting a curve exactly once, are solved in exactly the same way.
典型考题措辞例如：求使 x² + k x + 9 = 0 有相等实根的 k 值。令 Δ = k² - 36 = 0，得 k = 6 或 k = -6。切线类问题，例如直线与曲线恰好只有一个交点，也用完全相同的方法处理。
8. Graphs of Quadratic Functions | 二次函数图像
The graph of y = a x² + b x +
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