📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are one of the most frequently tested topics in IGCSE Mathematics. They appear in almost every exam paper, either as direct solving questions or embedded within graphs, word problems, and even coordinate geometry. This article provides a complete, exam-focused guide to solving quadratic equations, covering factorisation, the quadratic formula, completing the square, the discriminant, and common pitfalls.
二次方程是 IGCSE 数学中考查频率最高的知识点之一。几乎每份试卷都会出现二次方程：既可能直接要求解方程，也可能融入函数图像、应用题乃至坐标几何之中。本文将提供一份完整且紧扣考纲的解题指南，涵盖因式分解法、求根公式法、配方法、判别式以及常见易错点。
1. What Is a Quadratic Equation | 什么是二次方程
A quadratic equation is an equation that can be written in the standard form
二次方程是指可以写成以下标准形式的方程：
ax² + bx + c = 0, where a ≠ 0
The highest power of the variable x is 2, which is why the equation is called "quadratic" (from the Latin word for "square"). Here a, b, and c are constants, and a cannot be zero because otherwise the equation would become linear.
方程中变量的最高次数为 2，因此称为"二次"方程。这里的 a、b、c 是常数，且 a 不能为零——如果 a = 0，方程就退化为一次方程了。
before solving it. For instance, 4x² = 9 − 2x must first be rewritten as 4x² + 2x − 9 = 0.
例如，3x² + 5x − 2 = 0 是一个二次方程，其中 a = 3，b = 5，c = −2。在 IGCSE 考试中，你常常需要先将方程整理成标准形式再求解。例如，4x² = 9 − 2x 必须先改写为 4x² + 2x − 9 = 0。
2. Solving by Factorisation | 因式分解法
Factorisation is the fastest method when the quadratic has simple integer roots. The key principle is the zero product property: if the product of two expressions is zero, then at least one of them must be zero.
当二次方程具有简单的整数根时，因式分解法是最快捷的方法。其核心原理是零乘积性质：如果两个因式的乘积为零，那么至少其中一个因式必须为零。
Step 1: Rearrange the equation into the form ax² + bx + c = 0.
第一步：将方程整理成 ax² + bx + c = 0 的形式。
Step 2: Factorise the left-hand side into two brackets.
第二步：将等号左边分解为两个括号相乘的形式。
Step 3: Set each bracket equal to zero and solve for x.
第三步：令每个括号等于零，分别解出 x。
Worked Example 1: Solve x² + 7x + 12 = 0.
例 1：解方程 x² + 7x + 12 = 0。
12 and add to give 7. These numbers are 3 and 4, so x² + 7x + 12 = (x + 3)(x + 4). Hence:
我们寻找两个数，它们相乘等于 12，相加等于 7。这两个数是 3 和 4，因此 x² + 7x + 12 = (x + 3)(x + 4)。于是：
(x + 3)(x + 4) = 0 → x = −3 or x = −4
Worked Example 2: Solve 2x² + 5x − 3 = 0.
例 2：解方程 2x² + 5x − 3 = 0。
factor pairs of −3. We find (2x − 1)(x + 3) = 2x² + 6x − x − 3 = 2x² + 5x − 3. Therefore:
这里 x² 的系数不是 1，因此我们需要寻找形如 (2x + m)(x + n) 的两个括号。由于 2x² 必须由 2x × x 得到，我们尝试 −3 的各组因子，可得 (2x − 1)(x + 3) = 2x² + 6x − x − 3 = 2x² + 5x − 3。因此：
(2x − 1)(x + 3) = 0 → x = ½ or x = −3
difference of two squares: x² − 9 = (x − 3)(x + 3). This pattern is often tested directly.
还要记住平方差公式的特例：x² − 9 = (x − 3)(x + 3)。这一模式经常被直接考查。
3. Solving by the Quadratic Formula | 求根公式法
When a quadratic cannot be factorised easily, the quadratic formula always works. You are expected to memorise it for IGCSE exams.
当二次方程不易因式分解时，求根公式法永远有效。IGCSE 考试要求你熟记这个公式。
x = (−b ± √b² − 4ac) / (2a)
means that you must calculate two values: one with a plus sign and one with a minus sign.
这个公式给出 ax² + bx + c = 0 的解。符号 ± 表示你需要计算两个值：一个取加号，一个取减号。
Worked Example: Solve 2x² + 3x − 5 = 0 using the formula.
例题：用求根公式解方程 2x² + 3x − 5 = 0。
Here a = 2, b = 3, c = −5. Substitute into the formula:
这里 a = 2，b = 3，c = −5。代入公式：
x = (−3 ± √(3² − 4 × 2 × (−5))) / (2 × 2)
x = (−3 ± √9 + 40) / 4 = (−3 ± √49) / 4 = (−3 ± 7) / 4
/4 = −2.5. Always check your answers by substituting them back into the original equation.
因此 x = (−3 + 7)/4 = 1，或 x = (−3 − 7)/4 = −2.5。务必把答案代回原方程进行检验。
A common exam tip: when the question says "give your answers correct to 2 decimal places", it is a hint that factorisation will not work and you should use the formula. Do not force factorisation on a non-factorisable quadratic.
一个常见的考试技巧：当题目要求"答案精确到两位小数"时，这是在暗示该方程无法因式分解，应使用求根公式。不要硬用因式分解法去解一个不可分解的二次方程。
4. Solving by Completing the Square | 配方法
his method is especially useful for analysing graphs, because it directly reveals the vertex of the parabola.
配方法将二次方程转化为 (x + p)² + q = 0 的形式。这种方法在分析函数图像时特别有用，因为它能直接揭示抛物线的顶点。
Worked Example: Solve x² + 6x + 2 = 0 by completing the square.
例题：用配方法解方程 x² + 6x + 2 = 0。
= 3, and use (x + 3)² = x² + 6x + 9. Therefore x² + 6x = (x + 3)² − 9. Substituting back:
取 x 的系数的一半，即 6 ÷ 2 = 3，利用 (x + 3)² = x² + 6x + 9。因此 x² + 6x = (x + 3)² − 9。代回原方程：
(x + 3)² − 9 + 2 = 0 → (x + 3)² = 7
Now take the square root of both sides, remembering the ± sign:
现在对两边开平方根，切记加上 ± 号：
x + 3 = ±√7 → x = −3 ± √7
p)² + q also tells you that the turning point of the graph y = (x + p)² + q is at (−p, q).
当 x² 的系数不为 1 时，同样的方法仍然适用，但需要先把整个方程除以该系数。配方后的形式 (x + p)² + q 还能告诉你，函数 y = (x + p)² + q 的图像顶点位于 (−p, q)。
5. The Discriminant | 判别式
The expression b² − 4ac inside the quadratic formula is called the discriminant. It tells you how many real roots a quadratic equation has, without actually solving it.
求根公式中的 b² − 4ac 被称为判别式。它能在不真正解方程的情况下，告诉我们二次方程有多少个实数根。
- If b² − 4ac > 0, the equation has two distinct real roots.
- If b² − 4ac = 0, the equation has exactly one repeated real root.
- If b² − 4ac < 0, the equation has no real roots.
- 若 b² − 4ac > 0，方程有两个不相等的实数根。
- 若 b² − 4ac = 0，方程有两个相等的实数根（即一个重根）。
- 若 b² − 4ac < 0，方程没有实数根。
Worked Example: Determine the number of real roots of 3x² − 2x + 5 = 0.
例题：判断方程 3x² − 2x + 5 = 0 的实数根个数。
b² − 4ac = (−2)² − 4 × 3 × 5 = 4 − 60 = −56 < 0. Therefore the equation has no real roots.
这里 a = 3，b = −2，c = 5，所以 b² − 4ac = (−2)² − 4 × 3 × 5 = 4 − 60 = −56 < 0。因此方程没有实数根。
In IGCSE, you may also be asked a reverse problem: "Find the value of k such that the equation has equal roots." Simply set b² − 4ac = 0 and solve for k.
在 IGCSE 中，你还可能遇到逆向问题："求 k 的值，使方程有相等的根。"只需令 b² − 4ac = 0 并解出 k 即可。
6. Quadratic Graphs and Roots | 二次函数图像与根
of a gives a U-shaped parabola that opens upwards, while a negative value of a gives an inverted U-shape.
方程 ax² + bx + c = 0 的实数根恰好对应函数 y = ax² + bx + c 图像与 x 轴的交点。当 a > 0 时，抛物线开口朝上呈 U 形；当 a < 0 时，抛物线开口朝下呈倒 U 形。
When you sketch a quadratic graph in an exam, mark these key features:
在考试中画二次函数草图时，务必标出以下关键要素：
- The roots (x-intercepts), found by setting y = 0.
- The y-intercept, found by setting x = 0; it is always equal to c.
- e turning point, found by completing the square or using the line of symmetry x = −b/(2a).
- 根（与 x 轴的交点）：令 y = 0 求得。
- 与 y 轴的交点：令 x = 0 求得，其纵坐标恒等于 c。
- 顶点：通过配方法求得，或用对称轴 x = −b/(2a) 求出。
t at (0, 3), and a vertex at (2, −1). These three features are enough to sketch the parabola accurately.
例如，y = x² − 4x + 3 = (x − 1)(x − 3) 的根为 x = 1 和 x = 3，与 y 轴的交点为 (0, 3)，顶点为 (2, −1)。这三个特征足以让你准确地画出抛物线草图。
7. Solving Word Problems | 解应用题
IGCSE problems often hide a quadratic equation inside a real-world context, such as area, geometry, or projectile motion. The key is to translate the words into algebra and then solve.
IGCSE 题目经常将二次方程隐藏在现实情境中，例如面积、几何或抛体运动。关键是先把文字转化为代数式，再解方程。
Worked Example: A rectangle has length (x + 4) cm and width (x − 1) cm. Its area is 28 cm². Find x.
例题：一个长方形的长为 (x + 4) cm，宽为 (x − 1) cm，面积为 28 cm²。求 x 的值。
Write the area equation and expand:
列出面积方程并展开：
(x + 4)(x − 1) = 28 → x² + 3x − 4 = 28 → x² + 3x − 32 = 0
ession does not factorise nicely, so use the quadratic formula with a = 1, b = 3, c = −32:
该式子不易因式分解，因此使用求根公式，其中 a = 1，b = 3，c = −32：
x = (−3 ± √9 + 128) / 2 = (−3 ± √137) / 2
be negative, so x ≈ 4.35. Always reject negative or impossible solutions in word problems.
因此 x ≈ 4.35 或 x ≈ −7.35。由于长度不能为负，所以 x ≈ 4.35。在应用题中务必舍去负数解或不合理的解。
A common exam type is the "product of two consecutive integers" + x − 56 = 0 = (x + 8)(x − 7), giving x = 7 (or x = −8 if negative integers are allowed).
另一类常见题型是"两个连续整数之积"问题：若 x(x + 1) = 56，则 x² + x − 56 = 0 = (x + 8)(x − 7)，得 x = 7（若允许负整数，则 x = −8 也可）。
8. Equations That Reduce to Quadratics | 可化为二次方程的方程
example, x⁴ − 5x² + 4 = 0 is a quartic, but if we let u = x², it becomes u² − 5u + 4 = 0.
有些方程本身不是二次形式，但可以通过换元或去分母化为二次方程。例如，x⁴ − 5x² + 4 = 0 是一个四次方程，但若令 u = x²，它就变成 u² − 5u + 4 = 0。
ack, x² = 1 gives x = ±1, and x² = 4 gives x = ±2, so the full solution set is x = ±1, ±2.
解 u² − 5u + 4 = 0 得 (u − 1)(u − 4) = 0，因此 u = 1 或 u = 4。代回原变量：x² = 1 得 x = ±1，x² = 4 得 x = ±2，所以完整的解集为 x = ±1，±2。
Similarly, equations with fractions, such as, can be cleared by multiplying both sides by the common denominator. For example,
类似地，含有分数的方程可以通过两边同乘公分母来消去分母。例如：
3/(x − 1) + 2 = x
th the quadratic formula. Remember to check that any solution does not make the original denominator zero.
两边同乘 (x − 1)：3 + 2(x − 1) = x(x − 1)，化简得 3 + 2x − 2 = x² − x，即 x² − 3x − 1 = 0。然后用求根公式求解。切记检验所得解是否会使原分母为零。
9. Common Mistakes and How to Avoid Them | 常见错误及其避免方法
Many marks are lost in IGCSE exams not from a lack of knowledge, but from careless algebra. Here are the most common pitfalls.
在 IGCSE 考试中，很多分数并非由于知识欠缺而丢失，而是因为粗心的代数运算。以下是最常见的陷阱。
- Forgetting the ± sign when taking square roots. If x² = 16, then x = ±4, not just x = 4.
- ch as x² = 5x. This loses the root x = 0. Instead, rearrange to x² − 5x = 0 and factorise.
- Sign errors when substituting negative values into the quadratic formula, especially with −4ac when c is negative.
- Forgetting to rearrange the equation into standard form before applying the formula or factorisation.
- Writing "x = 2 or 3" instead of "x = 2 or x = 3" — always write the variable each time.
- 开平方根时忘记 ± 号。若 x² = 16，则 x = ±4，而不是仅 x = 4。
- 为化简方程 x² = 5x 而直接除以 x，这会丢失根 x = 0。应移项为 x² − 5x = 0 再因式分解。
- 将负数代入求根公式时出现符号错误，尤其是 c 为负数时的 −4ac。
- 在使用求根公式或因式分解前，忘记将方程整理成标准形式。
- 将解写成 "x = 2 or 3" 而不是 "x = 2 or x = 3"——每次都要写出变量。
A good habit is to substitute your answers back into the original equation. If both sides balance, your solution is almost certainly correct.
一个好习惯是把答案代回原方程验证。若左右两边相等，你的解就基本确定是正确的。
10. Exam Tips and Strategy | 考试技巧与策略
When you see a quadratic equation in the exam, quickly decide which method to use. Start by checking whether it can be factorised; if it cannot, go straight to the quadratic formula. Only use completing the square if the question explicitly asks for it, since it takes longer.
在考试中看到二次方程时，请快速决定使用哪种方法。先检查是否能因式分解；若不能，直接用求根公式。只有当题目明确要求时再用配方法，因为它耗时较长。
Manage your time wisely. A standard "solve" question is worth 2 to 4 marks. Show every step, because IGCSE mark schemes award method marks even if your final answer is slightly wrong. In particular, write down the substitution into the quadratic formula before calculating.
合理分配时间。一道标准的"求解"题通常为 2 到 4 分。务必写出每一步，因为 IGCSE 评分标准会为正确的方法步骤给分，即使最终答案稍有偏差。尤其是向求根公式中代入数值的过程一定要写出来。
Finally, connect the algebra to the graph. If a question gives you a sketch showing a parabola crossing the x-axis at two points, the solutions are simply the x-coordinates of those intersection points. Reading a graph is often faster than solving algebraically.
最后，将代数与图像联系起来。如果题目给出一个抛物线草图并显示它与 x 轴相交于两点，那么方程的解就是这两个交点各自的 x 坐标。读图往往比代数求解更快。
With consistent practice, solving quadratic equations becomes a reliable source of marks. Master the four methods — factorisation, formula, completing the square, and the discriminant — and you will be well prepared for any quadratic question in your IGCSE exam.
只要坚持练习，解二次方程就会成为稳定的得分点。掌握四种方法——因式分解、求根公式、配方法和判别式，你就能从容应对 IGCSE 考试中的任何二次方程问题。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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