📚 Solving Quadratic Equations for IGCSE Mathematics | IGCSE 数学：二次方程求解
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from algebra and graphs to real-life modelling. This revision guide explains the standard form, the main solving methods, the discriminant, graphical interpretation, and common exam pitfalls.
二次方程贯穿 IGCSE 数学大纲，从代数、图像到实际问题建模都会出现。本复习指南将讲解标准形式、主要解法、判别式、图像意义以及考试中常见的失分点。
You will meet quadratic equations in the algebra, equations and graphs strands of both the Core and Extended papers. Extended candidates must also handle completing the square, the quadratic formula, the discriminant and quadratic inequalities, so it is worth mastering all the methods rather than relying on just one.
在 Core 和 Extended 试卷的代数、方程与图像部分都会遇到二次方程。Extended 考生还必须掌握配方法、求根公式、判别式和二次不等式，因此最好把各种方法都练熟，而不是只依赖其中一种。
1. What Is a Quadratic Equation? | 什么是二次方程
A quadratic equation is a polynomial equation of degree 2. The highest power of the variable is x².
二次方程是一个次数为 2 的多项式方程，其变量的最高次幂是 x²。
The general form is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.
一般形式为 ax² + bx + c = 0，其中 a、b、c 是常数，且 a ≠ 0。
inear, not quadratic. For example, x² − 3x + 2 = 0 is quadratic, but 2x + 1 = 0 is linear.
如果 a = 0，方程就变成一次方程，而不是二次方程。例如 x² − 3x + 2 = 0 是二次方程，而 2x + 1 = 0 是一次方程。
t into y² − 5y + 4 = 0, which you can solve normally and then convert back to values of x.
二次方程也会以"伪装"形式出现。像 x⁴ − 5x² + 4 = 0 这样的方程本身不是二次方程，但令 y = x² 后就变成 y² − 5y + 4 = 0，可以按常规方法求解，再把 y 换回 x。
2. Standard Form and Key Terms | 标准形式与关键术语
erm, bx is the linear term, and c is the constant term. The coefficient a is called the leading coefficient.
在 ax² + bx + c = 0 中，ax² 是二次项，bx 是一次项，c 是常数项。系数 a 称为首项系数。
A root, or solution, is a value of x that makes the equation true. Quadratic equations can have two real roots, one repeated real root, or no real roots.
根（或解）是使方程成立的 x 值。二次方程可以有两个实数根、一个重复实数根，或没有实数根。
ax² + bx + c = 0, a ≠ 0
5x − 12 = 0, and 2x(x − 3) = 5 must first be expanded and rearranged as 2x² − 6x − 5 = 0.
在选择题型解法之前，务必先将方程整理成这个标准形式。例如 3x² = 12 − 5x 要先整理为 3x² + 5x − 12 = 0；2x(x − 3) = 5 要先展开并整理为 2x² − 6x − 5 = 0。
A useful habit is to write down a, b and c explicitly once the equation is in standard form, including their signs. Most marks lost in this topic come from a missing minus sign rather than from a lack of understanding.
一个很实用的习惯是：方程整理成标准形式后，把 a、b、c 连同符号明确写出来。这个板块丢分大多源于漏掉负号，而不是理解上的问题。
3. Solving by Factorisation | 因式分解法
Factorisation is usually the fastest method when the quadratic expression factorises neatly. Follow these steps: set one side to 0, factorise, set each bracket equal to 0, and solve.
当二次式可以容易地因式分解时，因式分解法通常是最快的方法。步骤如下：将一边写成 0，进行因式分解，令每个括号等于 0，然后解方程。
If (x − p)(x − q) = 0, then x = p or x = q
Example 1: Solve x² + 7x + 12 = 0.
例 1：解 x² + 7x + 12 = 0。
x² + 7x + 12 = (x + 3)(x + 4) = 0
Therefore x + 3 = 0 or x + 4 = 0, giving x = −3 or x = −4.
因此 x + 3 = 0 或 x + 4 = 0，得到 x = −3 或 x = −4。
Example 2: Solve 2x² + 5x − 3 = 0.
例 2：解 2x² + 5x − 3 = 0。
Here a = 2, so the brackets must start as (2x …)(x …). The product of the two number terms must be −3, and the cross terms must add to +5x.
这里 a = 2，所以括号应以 (2x …)(x …) 开头。两个常数项的乘积必须是 −3，交叉项相加后必须得到 +5x。
2x² + 5x − 3 = (2x − 1)(x + 3) = 0
½ or x = −3. You can always check by substituting a root back into the original equation.
所以 2x − 1 = 0 或 x + 3 = 0，得到 x = ½ 或 x = −3。你随时可以把根代回原方程来检验。
+ 5), and any common factor should be taken out first, for example 3x² − 12x = 3x(x − 4).
有两个特殊模式能节省时间。平方差如 x² − 25 可分解为 (x − 5)(x + 5)；有公因式时应先提取，例如 3x² − 12x = 3x(x − 4)。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q. This method works even when the expression does not factorise, and it also gives the vertex of the graph directly.
配方法把二次式改写成 (x + p)² + q 的形式。即使式子无法因式分解，这种方法也照样适用，同时还能直接给出图像的顶点。
x² + bx + c = (x + b/2)² + c − (b/2)²
To complete the square, halve the coefficient of x, write (x + b/2)², then subtract the square of that half-coefficient to keep the expression unchanged.
配方时，先取 x 的系数的一半，写出 (x + b/2)²，再减去这个半系数的平方，以保持式子与原式相等。
Example 3: Solve x² + 6x + 1 = 0 by completing the square.
例 3：用配方法解 x² + 6x + 1 = 0。
(x + 3)² − 9 + 1 = 0 → (x + 3)² = 8
x + 3 = ±√8 → x = −3 ± 2√2
So x ≈ 0.17 or x ≈ −5.83 to 2 decimal places.
所以 x ≈ 0.17 或 x ≈ −5.83（保留两位小数）。
− 2)² − 3, which tells you instantly that the minimum value is −3 and it occurs at x = 2.
当 a ≠ 1 时，先把系数 a 从前两项提出。例如 2x² − 8x + 5 = 2(x² − 4x) + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3，由此立刻可知最小值为 −3，且在 x = 2 处取得。
5. The Quadratic Formula | 求根公式
The quadratic formula solves any quadratic equation, whether or not it factorises, as long as you substitute accurately.
无论方程能否因式分解，只要代入准确，求根公式都能求解任何二次方程。
x = (−b ± √b² − 4ac) ÷ 2a
The formula is given on the IGCSE formula sheet, but you must still know when to use it and how to handle the signs. Write down a, b and c first, then substitute inside brackets to avoid calculator errors.
该公式会出现在 IGCSE 公式表中，但你仍需知道何时使用它以及如何处理符号。先写出 a、b、c，再用括号代入，以避免计算器出错。
Example 4: Solve 2x² + 5x − 3 = 0 using the formula.
例 4：用求根公式解 2x² + 5x − 3 = 0。
Here a = 2, b = 5, c = −3, so b² − 4ac = 25 + 24 = 49.
这里 a = 2，b = 5，c = −3，所以 b² − 4ac = 25 + 24 = 49。
x = (−5 ± √49) ÷ 4 = (−5 ± 7) ÷ 4
This gives x = 2 ÷ 4 = ½ or x = −12 ÷ 4 = −3, matching the factorised answer in Example 2.
由此得到 x = 2 ÷ 4 = ½ 或 x = −12 ÷ 4 = −3，与例 2 的因式分解答案一致。
In Extended papers you are often told to give answers correct to 3 significant figures. Do not round part-way through the calculation; keep full accuracy on your calculator and round only in the final answer.
在 Extended 试卷中，常要求答案保留三位有效数字。不要在计算中途四舍五入，应让计算器保持完整精度，只在最后一步取近似值。
6. The Discriminant and the Number of Roots | 判别式与根的个数
The expression b² − 4ac is called the discriminant, usually written as Δ. It tells you how many real roots the equation has before you solve it.
表达式 b² − 4ac 称为判别式，通常记作 Δ。它能在求解之前告诉你方程有多少个实数根。
| Discriminant Δ | 判别式 | Nature of roots | 根的性质 |
|---|---|
| Δ > 0 | Two distinct real roots 两个不同的实数根 |
| Δ = 0 | One repeated real root 一个重复实数根 |
| Δ < 0 | No real roots 没有实数根 |
+ 2x + 5 = 0, Δ = 4 − 20 = −16, which is negative, so the equation has no real solutions.
对 x² − 4x + 4 = 0，Δ = 16 − 16 = 0，所以有一个重复根 x = 2。对 x² + 2x + 5 = 0，Δ = 4 − 20 = −16，为负数，所以方程没有实数解。
A common exam question gives an unknown constant, for instance "find k so that x² + kx + 9 = 0 has equal roots". Setting Δ = k² − 36 = 0 gives k = 6 or k = −6. Remember there are usually two answers.
常见考题会给出一个未知常数，例如"求 k 使 x² + kx + 9 = 0 有等根"。令 Δ = k² − 36 = 0，得 k = 6 或 k = −6。记住这类题通常有两个答案。
7. Graphical Interpretation | 图像意义
oots of ax² + bx + c = 0 are exactly the x-coordinates where the curve crosses the x-axis.
y = ax² + bx + c 的图像是一条抛物线。ax² + bx + c = 0 的实数根正好是曲线与 x 轴交点的横坐标。
If the curve crosses the x-axis twice there are two real roots; if it just touches the axis there is one repeated root; if it stays completely above or below the axis there are no real roots.
如果曲线与 x 轴相交两次，就有两个实数根；如果只是相切，则有一个重复根；如果完全在 x 轴上方或下方，则没有实数根。
mber of x-intercepts equals the number of real solutions, and the sign of Δ decides which case applies.
这与判别式直接相关。因为 y = 0 就是 x 轴，所以 x 轴交点的个数等于实数解的个数，而 Δ 的符号决定了属于哪种情况。
You can also solve a quadratic graphically by reading off the intersections, though the IGCSE mark scheme normally expects an algebraic check unless the question says otherwise.
你也可以通过读图求交点来图解二次方程，但除非题目另有说明，IGCSE 评分标准通常还是要求代数验证。
8. Key Features of Quadratic Graphs | 二次函数图像的关键特征
Three features are examined regularly: the shape, the y-intercept and the turning point. If a > 0 the parabola opens upwards and has a minimum; if a < 0 it opens downwards and has a maximum.
有三个特征经常考：开口方向、y 轴截距和顶点。若 a > 0，抛物线开口向上，有最小值；若 a < 0，开口向下，有最大值。
mply gives y = c. The x-intercepts come from solving the equation, usually by factorising.
令 x = 0 即可得到 y 轴截距，结果就是 y = c。x 轴截距则通过解方程得到，通常用因式分解。
Axis of symmetry: x = −b ÷ 2a
the roots are x = 1 and x = 3, so the axis of symmetry is x = 2 and the vertex is (2, −1).
对称轴经过顶点；当有两个根时，它正好位于两根的正中间。对 y = x² − 4x + 3，根为 x = 1 和 x = 3，所以对称轴是 x = 2，顶点是 (2, −1)。
(x − 2)² − 1 shows the vertex immediately. Choose whichever method the question hints at.
配方法不用解方程也能给出同样的信息：y = (x − 2)² − 1 一眼就能看出顶点。根据题目提示选择合适的方法即可。
9. Quadratics in Real-Life Problems | 二次方程的实际应用
Many IGCSE questions ask you to form a quadratic equation from a word problem, solve it, and then interpret the answer in context.
许多 IGCSE 题目要求你根据文字题建立二次方程、求解，然后结合情境解释答案。
number puzzles, and projectile motion where height depends on time, such as h = 20t − 5t².
常见情境包括面积问题、连续整数、数字谜题，以及高度随时间变化的抛体运动，例如 h = 20t − 5t²。
Example 5: A rectangle is 3 cm longer than it is wide and has an area of 40 cm². Find its width.
例 5：一个矩形比宽长 3 cm，面积为 40 cm²，求它的宽。
Let the width be x cm, so the length is x + 3 cm. Then x(x + 3) = 40.
设宽为 x cm，则长为 x + 3 cm。于是 x(x + 3) = 40。
x² + 3x − 40 = 0 → (x + 8)(x − 5) = 0
−8 or x = 5. Since a length cannot be negative, the width is 5 cm and the length is 8 cm.
解得 x = −8 或 x = 5。由于长度不能为负，所以宽为 5 cm，长为 8 cm。
This rejection step is worth a mark. Whenever a problem describes a physical quantity, state clearly why the negative root is discarded.
这个"舍去"步骤是有分的。只要题目描述的是实际量，就要明确说明为什么舍去负根。
10. Quadratic and Linear Simultaneous Equations | 二次与一次联立方程
c equation, for example y = x + 1 and y = x² − 1, and asks for the points of intersection.
Extended 常见题型是给出一个一次方程和一个二次方程，例如 y = x + 1 与 y = x² − 1，要求求出交点。
The reliable method is substitution. Rearrange the linear equation if necessary, substitute into the quadratic, collect everything on one side and solve.
最可靠的方法是代入法。必要时先整理一次方程，代入二次方程，把所有项移到一边后求解。
x + 1 = x² − 1 → x² − x − 2 = 0 → (x − 2)(x + 1) = 0
So x = 2 or x = −1, giving y = 3 or y = 0. The intersection points are (2, 3) and (−1, 0).
所以 x = 2 或 x = −1，对应 y = 3 或 y = 0。交点坐标为 (2, 3) 与 (−1, 0)。
Always give both coordinates of each intersection point, and remember that the number of intersection points equals the number of real solutions of the resulting quadratic.
每个交点都要写出两个坐标；同时要记住，交点个数等于所得二次方程的实数解个数。
11. Quadratic Inequalities | 二次不等式
A quadratic inequality such as x² − x − 6 > 0 is solved by finding the critical values first, then deciding which regions satisfy the inequality.
解 x² − x − 6 > 0 这样的二次不等式，先求出临界值，再判断哪些区间满足不等式。
x² − x − 6 = (x − 3)(x + 2) = 0 → x = 3 or x = −2
Because the coefficient of x² is positive, the parabola opens upwards, so the expression is positive outside the roots and negative between them.
由于 x² 的系数为正，抛物线开口向上，所以式子在两根之外为正，在两根之间为负。
x² − x − 6 > 0 → x < −2 or x > 3
x² − x − 6 < 0 → −2 < x < 3
e inequality involves ≤ or ≥, the critical values themselves are included in the solution.
画一个粗略的抛物线草图是避免选错区间最稳妥的办法。如果不等式含 ≤ 或 ≥，则临界值本身也包含在解集中。
12. Common Mistakes and Exam Tips | 常见失分点与应试建议
The most frequent error is losing a sign when writing down b or c, especially when b is negative. Write the formula with brackets before substituting.
最常见的错误是写 b 或 c 时丢符号，尤其是 b 为负数时。代入前先把公式写成带括号的形式。
0; instead rearrange to x² − 5x = 0 and factorise as x(x − 5) = 0, giving x = 0 or x = 5.
解 x² = 5x 这类方程时，千万不要两边同除以 x。除以 x 会丢掉 x = 0 这个根；应整理为 x² − 5x = 0，分解为 x(x − 5) = 0，得到 x = 0 或 x = 5。
Other repeat offenders: forgetting to rearrange into standard form, giving only one root when there are two, rounding too early, and misreading "correct to 3 significant figures" as 3 decimal places.
其他高频错误包括：忘记整理成标准形式、有两个根却只写一个、过早四舍五入，以及把"保留三位有效数字"误读成三位小数。
- Check every answer by substituting it back into the original equation. 把每个答案代回原方程检验。
- Learn one method thoroughly and use the others as backup. 先彻底掌握一种方法，其余作为备选。
- Learn the discriminant cases, because they are quick marks. 熟记判别式的三种情况，这是一种"快速得分点"。
- Show the substitution step; method marks survive arithmetic slips. 把代入过程写出来，方法分不会因计算失误而全丢。
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