📚 Solving Quadratic Equations | 解一元二次方程
Quadratic equations are one of the most important topics in the IGCSE Mathematics syllabus. They appear in almost every paper, often as a standalone problem or as part of coordinate geometry, functions, and word problems. This guide will take you through the three main methods of solving a quadratic equation, explain the discriminant, and help you avoid common pitfalls in exams.
一元二次方程是 IGCSE 数学大纲中最重要的考点之一。几乎每份试卷都会出现，既可能作为独立题目，也可能结合坐标几何、函数和应用题一起考查。本指南将带你掌握解一元二次方程的三种主要方法，理解判别式的含义，并帮助你在考试中避开常见陷阱。
1. Standard Form and Key Definitions | 标准形式与关键定义
A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. The value of a cannot be zero, because if a = 0, the equation becomes linear, not quadratic.
一元二次方程是可以写成 ax² + bx + c = 0 形式的任何方程，其中 a、b、c 是实数，且 a ≠ 0。a 不能为零，因为如果 a = 0，方程就变成一次方程，而不是二次方程。
The highest power of the variable x is 2, which is why the equation is called “quadratic” (from the Latin word “quadratus”, meaning square). The term ax² is called the quadratic term, bx is the linear term, and c is the constant term.
变量 x 的最高次数是 2，因此这类方程被称为“二次方程”（quadratic 源自拉丁语 quadratus，意为“平方”）。ax² 称为二次项，bx 称为一次项，c 称为常数项。
When you are asked to “solve” a quadratic equation, you are finding the values of x that make the equation true. These values are called the roots or solutions of the equation. A quadratic equation can have two distinct real roots, one repeated real root, or no real roots at all.
当你被要求“解”一个一元二次方程时，实际上是在寻找使方程成立的 x 值。这些值被称为方程的根或解。一元二次方程可以有两个不同的实数根、一个重根，或者没有实数根。
ax² + bx + c = 0, a ≠ 0
The coefficient b or c may be zero. For example, x² − 9 = 0 and 2x² + 5x = 0 are both quadratic equations. However, the equation must contain an x² term; otherwise it is not quadratic.
系数 b 或 c 可能为零。例如，x² − 9 = 0 和 2x² + 5x = 0 都是二次方程。但方程中必须包含 x² 项，否则就不是二次方程。
2. Solving by Factorisation | 因式分解法
Factorisation is often the fastest method when the quadratic expression factorises neatly. The idea is to rewrite ax² + bx + c as a product of two linear expressions, then use the fact that if the product of two factors is zero, at least one factor must be zero.
当二次表达式能够整齐地分解时，因式分解法通常是最快的方法。其思路是将 ax² + bx + c 改写成两个一次表达式的乘积，然后利用“若两个因式的积为零，则至少有一个因式为零”这一性质。
Consider the equation x² + 5x + 6 = 0. We need two numbers that multiply to give 6 and add to give 5. The numbers 2 and 3 satisfy both conditions, so we can factorise as (x + 2)(x + 3) = 0.
考虑方程 x² + 5x + 6 = 0。我们需要找到两个数，它们相乘得 6，相加得 5。数字 2 和 3 同时满足这两个条件，因此可以分解为 (x + 2)(x + 3) = 0。
Since the product is zero, either x + 2 = 0 or x + 3 = 0. Solving these two linear equations gives x = −2 or x = −3. Always check your answer by substituting each root back into the original equation.
因为乘积为零，所以要么 x + 2 = 0，要么 x + 3 = 0。解这两个一次方程得到 x = −2 或 x = −3。始终记得将每个根代回原方程进行验证。
When the coefficient a is not 1, you must use a different strategy. For example, solve 2x² + 7x + 3 = 0. Multiply a and c: 2 × 3 = 6. Now find two numbers that multiply to 6 and add to 7. These are 6 and 1. Rewrite the middle term: 2x² + 6x + x + 3 = 0. Then factor by grouping: 2x(x + 3) + 1(x + 3) = 0, giving (2x + 1)(x + 3) = 0. Hence x = −½ or x = −3.
当系数 a 不等于 1 时，需要使用不同的策略。例如，解 2x² + 7x + 3 = 0。将 a 和 c 相乘：2 × 3 = 6。现在找到两个数，相乘得 6，相加得 7。这两个数是 6 和 1。改写中间项：2x² + 6x + x + 3 = 0。然后分组分解：2x(x + 3) + 1(x + 3) = 0，得到 (2x + 1)(x + 3) = 0。因此 x = −½ 或 x = −3。
Factorisation is only usable when the roots are rational numbers. If the discriminant (which we will meet later) is not a perfect square, you cannot factorise over integers and must use another method.
因式分解法仅在根为有理数时才能使用。如果判别式（稍后讲解）不是完全平方数，就无法在整数范围内分解，必须使用其他方法。
3. Solving by Completing the Square | 配方法
Completing the square transforms a quadratic into the form p(x + q)² + r = 0. This method always works and is especially useful for finding the turning point of a parabola or when the quadratic formula is not allowed by the examiner.
配方法将二次方程转化为 p(x + q)² + r = 0 的形式。这种方法总是有效，尤其适用于求抛物线的顶点，或当考试不允许使用求根公式时。
For a simple quadratic like x² + 6x − 7 = 0, take half of the coefficient of x, which is 3, and square it to get 9. Rewrite: x² + 6x + 9 − 9 − 7 = 0, so (x + 3)² − 16 = 0. Then (x + 3)² = 16, giving x + 3 = ±4, so x = 1 or x = −7.
对于像 x² + 6x − 7 = 0 这样的简单二次方程，取 x 系数的一半，即 3，再平方得到 9。改写：x² + 6x + 9 − 9 − 7 = 0，即 (x + 3)² − 16 = 0。然后 (x + 3)² = 16，得到 x + 3 = ±4，所以 x = 1 或 x = −7。
When a is not 1, first factor a out of the x² and x terms. For example, solve 2x² + 8x − 10 = 0. Divide everything by 2 first: x² + 4x − 5 = 0. Then complete the square: (x + 2)² − 4 − 5 = 0, so (x + 2)² = 9, giving x = 1 or x = −5.
当 a 不等于 1 时，先将 a 从 x² 项和 x 项中提出。例如，解 2x² + 8x − 10 = 0。先将所有项除以 2：x² + 4x − 5 = 0。然后配方：(x + 2)² − 4 − 5 = 0，即 (x + 2)² = 9，得到 x = 1 或 x = −5。
Completing the square also reveals the vertex of the parabola y = ax² + bx + c. If y = (x + 3)² − 16, the turning point is at (−3, −16), and the line of symmetry is x = −3.
配方还可以揭示抛物线 y = ax² + bx + c 的顶点。如果 y = (x + 3)² − 16，则顶点为 (−3, −16)，对称轴为 x = −3。
Always be careful with the sign when completing the square. For x² − 8x, half of −8 is −4, and (−4)² = 16, so x² − 8x = (x − 4)² − 16.
配方时务必注意符号。对于 x² − 8x，−8 的一半是 −4，(−4)² = 16，所以 x² − 8x = (x − 4)² − 16。
4. The Quadratic Formula | 求根公式
The quadratic formula is the most reliable method for solving any quadratic equation. It works for all cases, including when roots are irrational or when the equation cannot be factorised.
求根公式是解任意一元二次方程最可靠的方法。它适用于所有情况，包括根为无理数或方程无法因式分解的情形。
x = (−b ± √(b² − 4ac)) / 2a
To use the formula, identify a, b and c from the equation in standard form, then substitute into the formula. For example, solve 2x² − 4x − 3 = 0. Here a = 2, b = −4, c = −3.
使用公式时，先从标准形式的方程中确定 a、b、c，然后代入公式。例如，解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。
Substitute: x = (4 ± √(16 − 4 × 2 × (−3))) / (2 × 2) = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4. Since √40 = 2√10, we get x = (4 ± 2√10) / 4 = 1 ± (√10) / 2.
代入：x = (4 ± √(16 − 4 × 2 × (−3))) / (2 × 2) = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4。因为 √40 = 2√10，所以 x = (4 ± 2√10) / 4 = 1 ± (√10) / 2。
Notice that b is squared, so a negative b becomes positive. Many students lose marks by forgetting to square the negative sign. Also remember that the entire numerator is divided by 2a, not just the part under the square root.
注意 b 要平方，所以负的 b 平方后变成正数。很多学生因为忘记给负号平方而失分。还要记住，整个分子都要除以 2a，而不仅仅是根号下的部分。
Always write the formula down before substituting. This shows clear working to the examiner and reduces careless errors. Give your final answer to 3 significant figures unless the question says otherwise.
代入之前先写出公式。这样既能向考官展示清晰的解题过程，也能减少粗心错误。除非题目另有要求，否则最终答案保留 3 位有效数字。
5. The Discriminant | 判别式
The expression b² − 4ac inside the square root is called the discriminant. It tells us how many real roots a quadratic equation has without solving it fully.
根号内的表达式 b² − 4ac 被称为判别式。它能够在不必完全解方程的情况下，告诉我们一元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 (also called a double root). If b² − 4ac < 0, there are no real roots.
如果 b² − 4ac > 0，方程有两个不同的实数根。如果 b² − 4ac = 0，方程恰好有一个重复的实数根（也称为重根）。如果 b² − 4ac < 0，则没有实数根。
Graphically, a positive discriminant means the parabola crosses the x-axis at two points; a zero discriminant means it touches the x-axis at exactly one point; and a negative discriminant means it never touches the x-axis.
从图像上看，判别式为正意味着抛物线与 x 轴有两个交点；判别式为零意味着抛物线与 x 轴相切于一点；判别式为负意味着抛物线始终不与 x 轴相交。
For example, find the value of k such that x² + kx + 9 = 0 has one repeated root. The discriminant must be zero: k² − 4 × 1 × 9 = 0, so k² = 36, giving k = 6 or k = −6.
例如，求 k 的值，使 x² + kx + 9 = 0 有一个重根。判别式必须为零：k² − 4 × 1 × 9 = 0，所以 k² = 36，得到 k = 6 或 k = −6。
The discriminant is also useful for determining whether an equation can be solved by factorisation. If the discriminant is a perfect square, the roots are rational and factorisation is possible.
判别式还可用于判断方程是否可以用因式分解法求解。如果判别式是完全平方数，则根为有理数，可以分解。
6. Writing a Quadratic from Its Roots | 由根构造二次方程
Another common IGCSE question asks you to form a quadratic equation given its roots. If the roots are α and β, then the equation can be written as (x − α)(x − β) = 0.
另一类常见的 IGCSE 题目是给定根，要求构造一个二次方程。如果根是 α 和 β，则可以写成 (x − α)(x − β) = 0。
For roots 3 and −2, the equation is (x − 3)(x + 2) = 0, which expands to x² − x − 6 = 0. Note that this is not the only possible equation; multiplying by any non-zero constant gives an equivalent family of equations.
对于根 3 和 −2，方程为 (x − 3)(x + 2) = 0，展开得到 x² − x − 6 = 0。注意这并不是唯一的方程；乘以任意非零常数可以得到一个等价的方程族。
If the roots are given as fractions or surds, the factor form is the safest starting point. For roots ½ and −¾, write (2x − 1)(4x + 3) = 0 and then expand.
如果根是分数或根式，因式形式是最稳妥的起点。对于根 ½ 和 −¾，写成 (2x − 1)(4x + 3) = 0，然后展开。
Sometimes the question states that the roots differ by a certain amount or that one root is twice the other. In such cases, represent the roots algebraically, substitute into the sum and product relationships, and solve for the unknown coefficient.
有时题目会说明两根相差某个数值，或者一个根是另一个根的两倍。这种情况下，用代数表示根，代入两根之和与两根之积的关系，并求解未知系数。
7. Word Problems and Applications | 应用题与实际应用
Quadratic equations frequently appear in word problems involving areas, projectile motion, number relationships, and geometry. The key is to translate the English sentence into algebra carefully.
二次方程经常出现在涉及面积、抛体运动、数字关系和几何的应用题中。关键是将题目语言准确地转化为代数式。
Consider the following classic problem: The area of a rectangle is 84 cm². Its length is 5 cm longer than its width. Find the dimensions. Let the width be x cm. Then the length is (x + 5) cm, so x(x + 5) = 84.
看一个经典问题：一个矩形的面积是 84 cm²。它的长比宽长 5 cm。求各边尺寸。设宽为 x cm，则长为 (x + 5) cm，所以 x(x + 5) = 84。
Expanding gives x² + 5x − 84 = 0. Factorise: (x + 12)(x − 7) = 0. Hence x = −12 or x = 7. Since a width cannot be negative, we discard −12. The width is 7 cm and the length is 12 cm.
展开得到 x² + 5x − 84 = 0。因式分解：(x + 12)(x − 7) = 0。因此 x = −12 或 x = 7。因为宽度不能为负，所以舍去 −12。宽为 7 cm，长为 12 cm。
Always check whether negative solutions are meaningful in context. If the variable represents a length, time or count, reject negative values. If the question asks for a possible value, state the unit in your final answer.
始终检查负数解在具体情境中是否有意义。如果变量表示长度、时间或数量，应舍去负值。如果题目要求给出具体值，最终答案要带上单位。
For projectile motion problems of the form h = ut − 5t², set h to the given height and solve for t. Remember that t must be non-negative when it represents time.
对于 h = ut − 5t² 形式的抛体运动问题，将 h 设为给定高度并求解 t。记住，当 t 表示时间时，t 必须为非负数。
8. Solving by the Sum and Product of Roots | 利用两根之和与积解题
For a quadratic equation ax² + bx + c = 0, the sum of the roots is −b/a and the product of the roots is c/a. These relationships are surprisingly powerful in extended-response questions.
对于一元二次方程 ax² + bx + c = 0，两根之和为 −b/a，两根之积为 c/a。这两个关系式在扩展题中非常有用。
For example, in the equation 2x² + 6x − 8 = 0, the sum of roots is −6/2 = −3 and the product is −8/2 = −4. If one root is 1, the other root must be −4, since 1 + (−4) = −3 and 1 × (−4) = −4.
例如，在方程 2x² + 6x − 8 = 0 中，两根之和为 −6/2 = −3，两根之积为 −8/2 = −4。如果一个根是 1，则另一个根必然是 −4，因为 1 + (−4) = −3 且 1 × (−4) = −4。
This method allows you to find the second root instantly when one root is given. It also helps when you need to construct a new quadratic with roots that are transformations of the original roots, such as doubled roots or reciprocal roots.
这种方法在已知一根时可以立即求出另一根。当需要构造一个根为原根变换（如加倍根或倒数根）的新二次方程时，它也非常有帮助。
If a new equation has roots 2α and 2β, the sum is 2(α + β) and the product is 4αβ. Substitute the known sum and product of the original equation to build the new one.
如果新方程的根为 2α 和 2β，则两根之和为 2(α + β)，两根之积为 4αβ。将原方程已知的两根之和与积代入，即可构造新方程。
9. Common Mistakes and Exam Tips | 常见错误与考试技巧
One of the most frequent errors is forgetting to rearrange the equation into the form ax² + bx + c = 0 before identifying a, b and c. If the equation is given as 2x² − 3 = 5x, you must first subtract 5x from both sides.
最常见的错误之一是忘记先将方程整理成 ax² + bx + c = 0 的形式，就直接确定 a、b、c。如果给出的方程是 2x² − 3 = 5x，必须先两边同时减去 5x。
Another common mistake is dividing by x. If you have x² = 3x, dividing both sides by x gives x = 3, but you lose the solution x = 0. Instead, rearrange to x² − 3x = 0 and factorise x(x − 3) = 0.
另一个常见错误是两边同时除以 x。如果你有 x² = 3x，两边除以 x 得到 x = 3，但会丢失 x = 0 这个解。正确做法是整理为 x² − 3x = 0，然后分解为 x(x − 3) = 0。
When using the quadratic formula, check your values of a, b and c once more before substituting. A single sign error inside the discriminant changes the entire answer. Use brackets to keep terms separate.
使用求根公式时，代入前再检查一遍 a、b、c 的值。判别式中一个符号错误就会改变整个答案。用括号将各项隔开，保持清晰。
In questions that ask for answers correct to 2 decimal places, use the quadratic formula and your calculator carefully. Do not round intermediate values; only round the final answer.
在要求答案精确到小数点后两位的题目中，请仔细使用求根公式和计算器。不要四舍五入中间值，只对最终答案进行四舍五入。
Read the instruction at the top of the question paper. Some papers allow calculators, while others do not. If a non-calculator paper, factorisation and completing the square are the only feasible methods.
仔细阅读试卷顶部的说明。有些试卷允许使用计算器，有些则不允许。在非计算器试卷上，因式分解法和配方法是最可行的两种方法。
Finally, always check your answers by substituting back into the original equation. This takes only a few seconds and can save valuable marks. If both roots satisfy the equation, you can be confident in your working.
最后，始终通过代回原方程来验证答案。这只需几秒钟，却能挽回宝贵的分数。如果两个根都能使原方程成立，那么你的解答过程就是正确的。
10. Practice Exam Questions | 真题练习
Working through exam-style questions is essential for mastering quadratic equations. Below are three representative problems that cover the methods you have just learned.
做真题风格的练习对掌握一元二次方程至关重要。以下是三道有代表性的题目，覆盖了刚才学到的各种方法。
Question 1: Solve 3x² − 5x − 2 = 0 by factorisation. Solution: find two numbers that multiply to 3 × (−2) = −6 and add to −5. These are −6 and 1. Rewrite: 3x² − 6x + x − 2 = 0, then 3x(x − 2) + 1(x − 2) = 0, so (3x + 1)(x − 2) = 0. Hence x = −⅓ or x = 2.
第 1 题：用因式分解法解 3x² − 5x − 2 = 0。解法：找到两个数，相乘得 3 × (−2) = −6，相加得 −5。这两个数是 −6 和 1。改写：3x² − 6x + x − 2 = 0，然后 3x(x − 2) + 1(x − 2) = 0，所以 (3x + 1)(x − 2) = 0。因此 x = −⅓ 或 x = 2。
Question 2: Solve x² + 6x + 2 = 0, giving your answers correct to 3 significant figures. Using the formula: x = (−6 ± √(36 − 8)) / 2 = (−6 ± √28) / 2. So x = (−6 + 5.2915…) / 2 = −0.354, or x = (−6 − 5.2915…) / 2 = −5.646. The answers are x = −0.354 and x = −5.65.
第 2 题：解 x² + 6x + 2 = 0，答案保留 3 位有效数字。使用公式：x = (−6 ± √(36 − 8)) / 2 = (−6 ± √28) / 2。所以 x = (−6 + 5.2915…) / 2 = −0.354，或 x = (−6 − 5.2915…) / 2 = −5.646。答案为 x = −0.354 和 x = −5.65。
Question 3: The sum of the squares of two consecutive positive integers is 145. Find the integers. Let n be the smaller integer. Then n² + (n + 1)² = 145. Expanding: n² + n² + 2n + 1 = 145, so 2n² + 2n − 144 = 0, which simplifies to n² + n − 72 = 0. Factorise: (n + 9)(n − 8) = 0. Hence n = 8, and the integers are 8 and 9.
第 3 题：两个连续正整数的平方和为 145。求这两个整数。设较小的整数为 n。则 n² + (n + 1)² = 145。展开：n² + n² + 2n + 1 = 145，即 2n² + 2n − 144 = 0，化简为 n² + n − 72 = 0。因式分解：(n + 9)(n − 8) = 0。因此 n = 8，这两个整数为 8 和 9。
Quadratic equations are a gateway to higher-level mathematics, including functions, graphs and integration. Mastering the three solving methods and the discriminant will prepare you for both Paper 2 and Paper 4 of the IGCSE Mathematics examination, as well as for AS and A Level study.
一元二次方程是通往高等数学的桥梁，包括函数、图像和积分。掌握三种解法和判别式，将为你在 IGCSE 数学 Paper 2 和 Paper 4 中取得好成绩打下坚实基础，也为 AS 和 A Level 的学习做好准备。
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