📚 Solving Quadratic Equations for IGCSE Maths | IGCSE 数学：解二次方程
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, graphs and real-world problems, and examiners often test both the method and the interpretation of the solutions. This article covers the standard form, three main solving techniques, the discriminant, graphical interpretation, word problems and the common mistakes that cost marks.
二次方程是 IGCSE 数学中最重要的主题之一。它们出现在代数、图像和实际问题中，考官经常同时考查解题方法和对方程解的解释。本文涵盖标准形式、三种主要解法、判别式、图像解释、应用题以及最容易丢分的常见错误。
1. Standard Form of a Quadratic Equation | 二次方程的标准形式
A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of x is 2, and that is exactly what makes the equation quadratic rather than linear.
二次方程是任何可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 是常数且 a ≠ 0。x 的最高次数是 2，正是这一点使方程成为二次方程而不是一次方程。
ax² + bx + c = 0, a ≠ 0
he sign belongs to the coefficient: b is −5, not 5, and this matters a great deal when you use the quadratic formula.
如果 a = 0，方程就变成一次方程，而不是二次方程。例如 3x² − 5x + 2 = 0 中 a = 3、b = −5、c = 2。注意符号属于系数：b 是 −5 而不是 5，这一点在使用求根公式时非常关键。
n descending powers of x. For example, x² + 4 = 5x should be rewritten as x² − 5x + 4 = 0.
求解前，一定要先把方程整理成一边等于零的形式。各项必须按 x 的降幂排列。例如 x² + 4 = 5x 应改写为 x² − 5x + 4 = 0。
| Equation | 方程 | a | b | c |
| 3x² − 5x + 2 = 0 | 3 | −5 | 2 |
| x² − 5x + 6 = 0 | 1 | −5 | 6 |
| 2x² + 7x + 3 = 0 | 2 | 7 | 3 |
| x² + 6x + 2 = 0 | 1 | 6 | 2 |
first be multiplied by x to give x² − 5x + 6 = 0. Always tidy up before choosing a method.
有时二次方程隐藏在括号或分数之中。展开 (x + 1)(x − 3) = 0 得到 x² − 2x − 3 = 0；而像 x + 6 ÷ x = 5 这样的方程必须先两边乘以 x，得到 x² − 5x + 6 = 0。选择解法之前，务必先整理成形。
2. Solving by Factorising | 因式分解法
two numbers that multiply to +6 and add to −5: they are −2 and −3. So (x − 2)(x − 3) = 0.
当二次表达式可以干净地分解为两个一次括号时，因式分解法通常是最快的方法。对于 x² − 5x + 6 = 0，我们寻找两个数，它们相乘得 +6，相加得 −5：这两个数是 −2 和 −3。因此 (x − 2)(x − 3) = 0。
and x = 3. Always check each solution by substituting it back into the original equation.
根据零乘积性质，如果 (x − 2)(x − 3) = 0，那么 x − 2 = 0 或 x − 3 = 0。解为 x = 2 和 x = 3。每个解都要代回原方程检验。
x² − 5x + 6 = (x − 2)(x − 3) = 0 → x = 2 or x = 3
nd two numbers that multiply to 6 and add to 7, namely 6 and 1. Then 2x² + 6x + x + 3 = 0, so 2x(x + 3) + 1(x + 3) = 0 and finally (2x + 1)(x + 3) = 0.
当 x² 的系数不是 1 时，例如 2x² + 7x + 3 = 0，因式分解可能需要使用拆分中间项的方法。先算 a × c：2 × 3 = 6，再找两个数，相乘得 6、相加得 7，即 6 和 1。于是 2x² + 6x + x + 3 = 0，即 2x(x + 3) + 1(x + 3) = 0，最终 (2x + 1)(x + 3) = 0。
2x² + 7x + 3 = (2x + 1)(x + 3) = 0 → x = −1/2 or x = −3
= 0 → x = 3 or x = −3. A common factor gives x² − 5x = 0 → x(x − 5) = 0 → x = 0 or x = 5.
考试中经常出现两种特殊结构。平方差形式：x² − 9 = 0 → (x + 3)(x − 3) = 0 → x = 3 或 x = −3。提公因式形式：x² − 5x = 0 → x(x − 5) = 0 → x = 0 或 x = 5。
Never divide both sides of an equation by x to "simplify" instead, because a quadratic normally has two solutions and you must not lose one of them.
千万不要为了"化简"而把方程两边同时除以 x。在 x² − 5x = 0 中除以 x 会丢掉根 x = 0。应该因式分解，因为二次方程通常有两个解，不能丢掉其中一个。
3. Completing the Square | 配方法
method is especially useful when the quadratic does not factorise neatly and when finding the vertex of a parabola.
配方法把二次方程改写为 a(x − h)² + k = 0 的形式。当二次方程不容易因式分解以及需要求抛物线的顶点时，这种方法特别有用。
+ 3)² − 9. So the equation becomes (x + 3)² − 9 + 2 = 0, which simplifies to (x + 3)² = 7.
对于 x² + 6x + 2 = 0，先把 x² + 6x 写成 (x + 3)² − 9。于是方程变为 (x + 3)² − 9 + 2 = 0，简化为 (x + 3)² = 7。
Taking square roots gives x + 3 = ±√7, so x = −3 ± √7. The exact solutions are x = −3 + √7 and x = −3 − √7, which are approximately 0.354 and −6.354.
开平方得到 x + 3 = ±√7，因此 x = −3 ± √7。精确解为 x = −3 + √7 和 x = −3 − √7，约为 0.354 和 −6.354。
x² + 6x + 2 = (x + 3)² − 9 + 2 = 0 → (x + 3)² = 7 → x = −3 ± √7
b ÷ 2)² − (b ÷ 2)². Half of 6 is 3, and 3² = 9, which is exactly what we subtracted above.
首项系数为 1 时的通用法则很简单：把 x 的系数取一半放进括号，再减去它的平方。用符号表示即 x² + bx = (x + b ÷ 2)² − (b ÷ 2)²。6 的一半是 3，3² = 9，这正是上面减去的数。
3 = 0, which leads to 2(x + 2)² − 8 + 3 = 0, so 2(x + 2)² = 5 and (x + 2)² = 5 ÷ 2 = 2.5.
当 a ≠ 1 时，先把系数提到括号外。例如 2x² + 8x + 3 = 0。在括号内配方：2[(x + 2)² − 4] + 3 = 0，得到 2(x + 2)² − 8 + 3 = 0，于是 2(x + 2)² = 5，(x + 2)² = 5 ÷ 2 = 2.5。
Taking square roots gives x + 2 = ±√2.5, so x = −2 ± √2 form also tells you immediately that the minimum point of y = 2x² + 8x + 3 is at (−2, −5).
开平方得 x + 2 = ±√2.5，所以 x = −2 ± √2.5，即 x ≈ −0.419 或 x ≈ −3.581（保留三位小数）。配方式还直接告诉我们 y = 2x² + 8x + 3 的最小值点在 (−2, −5)。
4. The Quadratic Formula | 求根公式
any quadratic equation, whether or not it factorises. If ax² + bx + c = 0 and a ≠ 0, then:
求根公式可以解任何二次方程，不论它是否能因式分解。如果 ax² + bx + c = 0 且 a ≠ 0，那么：
x = [ −b ± √b² − 4ac ] ÷ 2a
Write the formula down first, then substitute carefully. A useful habit is to work out b² − 4ac separately before putting it under the square root, because most exam errors in this topic are arithmetic rather than algebraic.
先写出公式，再仔细代入。一个好习惯是先把 b² − 4ac 单独算出来，再放进根号里，因为这个考点中大多数考场错误其实是算术错误，而不是代数错误。
3. Both roots are exact and can be checked by factorising: 3x² − 5x + 2 = (3x − 2)(x − 1).
以 3x² − 5x + 2 = 0 为例，a = 3、b = −5、c = 2。则 b² − 4ac = 25 − 24 = 1，于是 x = (5 ± 1) ÷ 6。得到 x = 6 ÷ 6 = 1 和 x = 4 ÷ 6 = 2 ÷ 3。两个根都是精确值，可用因式分解检验：3x² − 5x + 2 = (3x − 2)(x − 1)。
Now take x² + 4x + 1 = 0. Here b² − 4ac = 16 − 4 = 12, so x = (−4 ± √12) ÷ 2 = (−4 ± 2√3) ÷ 2 = −2 ± √3 y x ≈ −0.268 or x ≈ −3.732. Simplify surds where possible, and only round at the very end.
再看 x² + 4x + 1 = 0。此时 b² − 4ac = 16 − 4 = 12，所以 x = (−4 ± √12) ÷ 2 = (−4 ± 2√3) ÷ 2 = −2 ± √3。数值上 x ≈ −0.268 或 x ≈ −3.732。能化简的根式要化简，并且只在最后一步才取近似值。
Unless the question says otherwise, give non-exact answers to 2 decimal places or 3 significant figures, and show the substitution in your working. In calculator papers you may use the equation solver, but you must still write the formula and the values of a, b and c to earn method marks.
除非题目另有说明，非精确答案要保留两位小数或三位有效数字，并写出代入过程。在使用计算器的试卷中可以用解方程功能，但仍必须写出公式以及 a、b、c 的取值，才能拿到方法分。
| Method | 方法 | Best used when | 最适用情形 |
| Factorising | 因式分解 | Coefficients are small integers; brackets are obvious | 系数为小整数，括号容易看出 |
| Completing the square | 配方 | Question asks for vertex, minimum value or exact surd form | 题目要求顶点、最小值或精确根式形式 |
| Quadratic formula | 求根公式 | Nothing factorises, or the question says "using the formula" | 无法分解，或题目指定"用公式" |
5. The Discriminant | 判别式
The expression under the square root, b² − 4ac, is called the discriminant and is usually written as Δ. It tells you how many real roots a quadratic equation has without you having to solve it.
根号下的表达式 b² − 4ac 称为判别式，通常写作 Δ。它能在不解方程的情况下告诉你二次方程有多少个实数根。
Δ = b² − 4ac
- Δ > 0: two distinct real roots; the parabola crosses the x-axis twice.
Δ > 0：两个不同的实根；抛物线与 x 轴相交两次。 - Δ = 0: one repeated real root; the parabola touches the x-axis at the vertex.
Δ = 0：一个重根；抛物线在顶点处与 x 轴相切。 - Δ < 0: no real roots; the parabola never meets the x-axis.
Δ < 0：没有实根；抛物线与 x 轴永不相交。
2x² + 3x + 5 = 0 has Δ = 9 − 40 = −31 < 0, so it has no real solutions at IGCSE level.
例如 x² − 6x + 9 = 0 的 Δ = 36 − 36 = 0，所以有一个重根 x = 3，事实上 (x − 3)² = 0。方程 2x² + 3x + 5 = 0 的 Δ = 9 − 40 = −31 < 0，因此在 IGCSE 范围内没有实数解。
Discriminant questions are often phrased as "show that" or "find the value of k" −6. If a question says the equation has two distinct real roots, write Δ > 0 and solve the inequality.
判别式题目常以"证明"或"求 k 的值"的形式出现。例如 x² + kx + 9 = 0 有两个相等实根时 Δ = 0，故 k² − 36 = 0，k = 6 或 k = −6。如果题目说方程有两个不同实根，就写 Δ > 0 并解不等式。
tersections of a line and a curve. To find where y = 2x + 5 meets y = x² + 1, set x² + 1 =
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