📚 Solving Quadratic Equations by Factorising and the Quadratic Formula | 二次方程的因式分解与求根公式求解
rt of the algebra section of IGCSE and A Level Mathematics. They appear in straightforward "solve for x" be rearranged into the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. Mastering the two main methods - factorising and the quadratic formula - gives you a reliable toolkit for the exam, because between them they solve every quadratic you will ever meet.
二次方程是 IGCSE 与 A Level 数学代数部分的核心内容。它们既出现在直接的"求解 x"题目中，也出现在涉及面积、速度与利润的应用题里，还出现在关于抛物线的图像题中。二次方程是任何可以整理成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数且 a ≠ 0。掌握因式分解与求根公式这两种主要方法，就能为你提供一套可靠的考试工具，因为这两者合起来足以解决你遇到的任何二次方程。
1. Recognising Quadratic Equations | 识别二次方程
A quadratic equation has a highest power of x equal to 2. The standard form is written as shown below. If the coefficient of x² is zero, the equation becomes linear, not quadratic, which is why the condition a ≠ 0 matters. The letters a, b and c stand for numbers: a is the coefficient of x², b is the coefficient of x, and c is the constant term.
二次方程中 x 的最高次数为 2。其标准形式如下所示。如果 x² 的系数为零，方程就退化为一次方程而非二次方程，这正是 a ≠ 0 这个条件重要的原因。字母 a、b、c 都代表数字：a 是 x² 的系数，b 是 x 的系数，c 是常数项。
a x² + b x + c = 0, where a ≠ 0
rearranged to 2x² − 4x − 6 = 0, and dividing every term by 2 gives the simpler equivalent x² − 2x − 3 = 0. Look out for disguised quadratics too: x(x + 5) = 24, (x + 1)² = 9 and 3/x = x + 2 can all be turned into standard form with a little algebra.
在 IGCSE 题目中，你可能需要先展开括号或合并同类项，才能把方程写成标准形式。例如，2(x² − 3) = 4x 可以整理为 2x² − 4x − 6 = 0，再把每一项都除以 2，就得到更简洁的等价形式 x² − 2x − 3 = 0。也要留意伪装的二次方程：x(x + 5) = 24、(x + 1)² = 9 以及 3/x = x + 2，都可以用一点代数变形化归为标准形式。
- Highest power 2: if you see x², or x multiplied by x, you are dealing with a quadratic. 最高次数为 2：只要看到 x²，或 x 与 x 相乘，就是二次方程。
- Standard form first: always rearrange to "= 0" before choosing a method. 先化标准形式：选择方法之前，一定要整理成"等于 0"的形式。
- Simplify if you can: dividing out a common factor makes factorising much easier. 能化简就化简：除以公因式会让因式分解容易很多。
2. The Zero Product Principle | 零乘积原理
The factorising method depends completely on the zero product principle. If two numbers or expressions multiply to give zero, then at least one of them must be zero. This single idea lets us split a factorised quadratic into two simple linear equations, each of which is solved in one line.
因式分解法完全依赖于零乘积原理。如果两个数或两个式子的乘积为零，那么其中至少有一个必须为零。正是这一条简单的想法，让我们能够把因式分解后的二次方程拆成两个简单的一次方程，每个都只需一行就能解出。
If p × q = 0, then p = 0 or q = 0
ro. For example, x² − 3x + 2 = 0 is already prepared for factorising, but x² − 3x = −2 needs rearranging to x² − 3x + 2 = 0 first. A common exam trap is to factorise an equation such as x² − 3x = −2 and then wrongly write x = 0 or x = 3.
该原理只有在乘积等于零时才成立。所以，在因式分解之前，你必须先把方程整理成一边为零的形式。例如，x² − 3x + 2 = 0 已经可以直接因式分解，但 x² − 3x = −2 需要先移项为 x² − 3x + 2 = 0。常见的考试陷阱是对方程 x² − 3x = −2 直接因式分解，然后错误地写出 x = 0 或 x = 3。
- Correct: x² − 3x + 2 = 0 → (x − 1)(x − 2) = 0 → x = 1 or x = 2. 正确做法。
- Incorrect: x² − 3x = −2 → x(x − 3) = −2 → x = −2 or x − 3 = −2. 错误做法：右边不为零时不能拆分。
3. Factorising x² + bx + c | 因式分解 x² + bx + c
² + bx + c = (x + m)(x + n). In other words, you are looking for a pair of numbers whose product is the constant term and whose sum is the coefficient of x.
当 x² 的系数为 1 时，二次式形如 x² + bx + c。因式分解时，需要找两个数 m 和 n，使得 m × n = c 且 m + n = b。于是 x² + bx + c = (x + m)(x + n)。换句话说，你要找的是一对数字：它们的乘积等于常数项，它们的和等于 x 的系数。
x² + b x + c = (x + m)(x + n), where m × n = c and m + n = b
only 2 + 3 = 5, so x² + 5x + 6 = (x + 2)(x + 3). Signs are decided by b and c together: if c is positive, m and n have the same sign (both positive when b is positive, both negative when b is negative); if c is negative, m and n have opposite signs.
要系统地逐一尝试。对于 x² + 5x + 6，需要两个数相乘得 6、相加得 5；6 的数对只有 1 × 6 和 2 × 3，其中只有 2 + 3 = 5，所以 x² + 5x + 6 = (x + 2)(x + 3)。符号由 b 与 c 共同决定：若 c 为正，则 m 与 n 同号（b 为正时同为正，b 为负时同为负）；若 c 为负，则 m 与 n 异号。
| Quadratic 二次式 | Factorised 因式分解 | Roots 根 |
| x² + 5x + 6 | (x + 2)(x + 3) | x = −2, x = −3 |
| x² − 7x + 12 | (x − 3)(x − 4) | x = 3, x = 4 |
| x² + 2x − 15 | (x + 5)(x − 3) | x = −5, x = 3 |
| x² − 4x − 12 | (x − 6)(x + 2) | x = 6, x = −2 |
ce how each root is the negative of the number inside its bracket - this is the single most common place students lose marks.
务必在脑中展开括号进行检查：(x + 5)(x − 3) 得到 x² − 3x + 5x − 15 = x² + 2x − 15，与原式一致。注意每个根都是括号内数字的相反数——这是学生最常丢分的地方。
4. Factorising ax² + bx + c when a ≠ 1 | 当 a ≠ 1 时的因式分解
When the coefficient of x² is not 1, the quadratic has the form ax² + bx + c. The most reliable exam technique is the "ac method": multiply a and c, find two numbers that multiply to ac and add to b, split the middle term, then factorise in pairs.
当 x² 的系数不是 1 时，二次式形如 ax² + bx + c。最可靠的考试技巧是"ac 法"：先计算 a 与 c 的乘积，找出两个数使它们的乘积等于 ac、和等于 b，再把中间项拆开，然后分组分解。
(x + 3) + 1(x + 3) = (2x + 1)(x + 3). Setting each bracket to zero gives x = −½ or x = −3.
以 2x² + 7x + 3 为例。这里 a × c = 2 × 3 = 6，b = 7，所以两个数是 6 和 1。拆开中间项：2x² + 6x + x + 3。分组分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。令每个括号等于零，得到 x = −½ 或 x = −3。
x − 2) = (3x − 4)(x − 2), giving x = 4/3 or x = 2. You can also use trial and error with the factors of a and c, but the ac method is faster and less error-prone under time pressure.
再看一个例子：3x² − 10x + 8。此时 a × c = 24，b = −10，所以两个数是 −6 和 −4（因为 −6 × −4 = 24 且 −6 + −4 = −10）。于是 3x² − 6x − 4x + 8 = 3x(x − 2) − 4(x − 2) = (3x − 4)(x − 2)，得到 x = 4/3 或 x = 2。你也可以用 a 与 c 的因数去试凑，但 ac 法更快，在时间压力下也更不容易出错。
- Step 1: write a, b, c and compute ac. 第一步：写出 a、b、c 并计算 ac。
- Step 2: find the number pair for ac that adds to b. 第二步：找出乘积为 ac、和为 b 的数对。
- Step 3: split the middle term into those two parts. 第三步：把中间项拆成这两部分。
- Step 4: factorise in pairs and check by expanding. 第四步：分组分解，并展开检验。
- Step 5: use the zero product principle to finish. 第五步：用零乘积原理求出根。
5. Special Cases: Difference of Two Squares and Perfect Squares | 特殊情形：平方差与完全平方
ics with no x term, this solves equations such as x² − 9 = 0 instantly, giving (x + 3)(x − 3) = 0, so x = 3 or x = −3.
有两种模式值得牢记，因为它们能节省大量时间。第一种是平方差公式：a² − b² = (a + b)(a − b)。把它用在没有 x 项的二次方程上，可以立刻解出 x² − 9 = 0，得到 (x + 3)(x − 3) = 0，所以 x = 3 或 x = −3。
a² − b² = (a + b)(a − b)
one repeated root, x = 3. Recognising these patterns is not just a shortcut - the discriminant of a perfect square is always zero, and examiners often ask you to state how many roots there are.
第二种模式是完全平方：a² + 2ab + b² = (a + b)²，a² − 2ab + b² = (a − b)²。像 x² − 6x + 9 = 0 这样的方程可化为 (x − 3)² = 0，它只有一个重根 x = 3。识别这些模式不只是捷径——完全平方的判别式永远为零，而考官常常要求你说明方程有几个根。
a² ± 2a b + b² = (a ± b)²
| Equation 方程 | Pattern 模式 | Solution 解 |
| x² − 25 = 0 | Difference of squares 平方差 | x = 5, x = −5 |
| 4x² − 49 = 0 | Difference of squares 平方差 | x = 7/2, x = −7/2 |
| x² + 10x + 25 = 0 | Perfect square 完全平方 | x = −5 (repeated 重根) |
| 9x² − 12x + 4 = 0 | Perfect square 完全平方 | x = 2/3 (repeated 重根) |
6. The Quadratic Formula | 求根公式
rd form, whatever the values of a, b and c. You should memorise it exactly, including the ± sign and the denominator 2a.
并非每个二次方程都能用整齐的整数因式分解，有些甚至完全无法分解。求根公式可以解任何标准形式的二次方程，无论 a、b、c 取什么值。你必须一字不差地记住它，包括 ± 号和分母 2a。
x = ( −b ± √b² − 4a c ) ÷ ( 2a )
the formula, then simplify. For x² − 4x + 1 = 0 we have a = 1, b = −4, c = 1, so x = (4 ± √16 − 4) ÷ 2 = (4 ± √12) ÷ 2 = (4 ± 2√3) ÷ 2 = 2 ± √3. When a root is a surd, leave it in exact form unless the question says otherwise.
使用公式的步骤是：先把方程写成标准形式，仔细写出 a、b、c 的值，代入公式，然后化简。对于 x² − 4x + 1 = 0，有 a = 1、b = −4、c = 1，所以 x = (4 ± √16 − 4) ÷ 2 = (4 ± √12) ÷ 2 = (4 ± 2√3) ÷ 2 = 2 ± √3。当根是无理数时，除非题目另有要求，否则应保留精确形式。
- Substitute in brackets: use (−4)² rather than −4² to avoid sign errors. 代入时加括号：写 (−4)² 而不是 −4²，以免出现符号错误。
- Watch the negative b: if b is negative, −b becomes positive. 注意负的 b：如果 b 是负数，−b 就变成正数。
- Simplify the surd: √12 = 2√3, and always cancel common factors. 化简根式：√12 = 2√3，并注意约去公因式。
- Round only at the end: keep full accuracy until the final answer. 只在最后一步取近似值。
7. The Discriminant: How Many Roots? | 判别式：方程有几个根
The expression under the square root, b² − 4ac, is called the discriminant, often written as Δ. It tells you how many real roots a quadratic has before you do any solving, which is exactly what many exam questions ask for.
根号下的表达式 b² − 4ac 称为判别式，通常记作 Δ。它能在你动手求解之前就告诉你二次方程有多少个实根，而这正是许多考题所要求的。
Δ = b² − 4a c
| Discriminant 判别式 | Number of real roots 实根个数 | Graph 图像 |
| Δ > 0 | Two distinct real roots 两个不同实根 | Crosses the x-axis twice 与 x 轴相交两次 |
| 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 | Does not meet the x-axis 与 x 轴不相交 |
2² − 4 × 1 × 5 = 4 − 20 = −16, so it has no real roots. A typical exam question might say "show that x² + kx + 9 = 0 has equal roots" n Δ is not a perfect square, the roots are irrational surds and factorising will not work.
例如，x² + 2x + 5 = 0 的 Δ = 2² − 4 × 1 × 5 = 4 − 20 = −16，所以它没有实根。典型的考题可能是"证明 x² + kx + 9 = 0 有等根"并求 k：令 Δ = k² − 36 = 0，得到 k = 6 或 k = −6。注意当 Δ 不是完全平方数时，根是无理数，因式分解法就行不通了。
8. Worked Examples: Comparing the Two Methods | 例题：两种方法对照
The best way to see the difference between the methods is to solve the same equation both ways. Consider 2(x²
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