📚 Solving Quadratic Equations | 二次方程求解
Quadratic equations are one of the fundamental topics in IGCSE Mathematics, appearing in algebra, graphs, and problem-solving. Understanding how to solve them efficiently is essential for success in the Edexcel exam. This article covers all major methods: factorising, using the quadratic formula, completing the square, and graphical interpretation, together with the discriminant, worked examples and the exam mistakes that cost the most marks.
二次方程是 IGCSE 数学的基础课题之一，涉及代数、图像和实际问题解决。掌握高效的求解方法对于在 Edexcel 考试中取得成功至关重要。本文涵盖了所有主要方法：因式分解法、二次公式法、配方法以及图像解读，并补充判别式、典型例题和最易失分的考试错误。
1. Introduction to Quadratic Equations | 二次方程入门
umbers, and a is not equal to zero. For example, 2x² + 3x - 5 = 0 is a quadratic equation.
二次方程是可以写成标准形式 a x² + b x + c = 0 的任何方程，其中 a、b 和 c 是实数，且 a 不等于零。例如，2x² + 3x - 5 = 0 就是一个二次方程。
The highest power of the variable x is 2, which gives the equation its name 'quadratic' (from the Latin 'quadratus' meaning square). In Edexcel IGCSE, you are expected to solve quadratics by factorising, by applying the quadratic formula, by completing the square, and by using graphs.
变量 x 的最高次幂是 2，因此得名 '二次'（来源于拉丁语 'quadratus'，意为平方）。在 Edexcel IGCSE 中，你需要通过因式分解、应用二次公式、配方法以及使用图像来解二次方程。
Before choosing a method, always rearrange the equation so that one side is exactly zero. Terms hidden behind brackets, fractions or a 'show that' instruction must all be expanded and collected first, otherwise every later step will be wrong.
在选择方法之前，务必先把方程整理成一边恰好为零。隐藏在括号、分数或 '证明' 指令背后的项必须先展开并合并同类项，否则后续每一步都会出错。
- Standard form / 标准形式: a x² + b x + c = 0, a ≠ 0
- Degree / 次数: the highest power of x is 2
- Roots / 根: the values of x that make the equation true
- Solutions / 解: a quadratic has at most two real roots
2. Solving by Factorising (a = 1) | 因式分解法 (a = 1 的情况)
need numbers with product 12 and sum 7: they are 3 and 4. So we write (x + 3)(x + 4) = 0.
当 x² 的系数为 1 时，因式分解需要找到两个数，它们的乘积等于 c，且和等于 b。对于方程 x² + 7x + 12 = 0，我们需要乘积为 12、和为 7 的两个数：它们是 3 和 4。因此，我们写成 (x + 3)(x + 4) = 0。
= 0 gives x = -4. Always check solutions by substituting back into the original equation.
利用零乘积性质，令每个括号等于零：x + 3 = 0 得到 x = -3；x + 4 = 0 得到 x = -4。始终将解代回原方程进行检查。
(x + 3)(x + 4) = 0 → x = -3 or x = -4
gives 16 - 28 + 12 = 0. Both roots satisfy the equation, so the factorisation is correct.
检验：代入 x = -3 得 9 - 21 + 12 = 0，代入 x = -4 得 16 - 28 + 12 = 0。两个根都满足方程，说明因式分解正确。
(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12, which matches the original expression exactly.
你也可以把括号乘回去检验因式分解：(x + 3)(x + 4) = x² + 4x + 3x + 12 = x² + 7x + 12，与原式完全一致。
3. Factorising with a Negative Constant | 常数项为负的因式分解
sum -3 gives -5 and 2, factorising to (x - 5)(x + 2) = 0, with solutions x = 5 and x = -2.
注意：如果 c 为负，则两个数异号。对于 x² - 3x - 10 = 0，乘积为 -10、和为 -3，得到 -5 和 2，因式分解为 (x - 5)(x + 2) = 0，解为 x = 5 和 x = -2。
s -3 and -5, so the equation becomes (x - 3)(x - 5) = 0 and the roots are x = 3 and x = 5.
当 b 为负、c 为正时，两个数必须都是负数。对于 x² - 8x + 15 = 0，乘积为 15、和为 -8，得到 -3 和 -5，于是方程变为 (x - 3)(x - 5) = 0，根为 x = 3 和 x = 5。
square such as x² - 6x + 9 = 0 factorises as (x - 3)² = 0, giving the repeated root x = 3.
考试中有两种特殊情况极为常见。平方差形式如 x² - 9 = 0 可分解为 (x - 3)(x + 3) = 0，得到 x = 3 和 x = -3。完全平方形式如 x² - 6x + 9 = 0 可分解为 (x - 3)² = 0，得到重根 x = 3。
0, so x = 0 or x = 4. Never divide both sides by x, because that destroys the root x = 0.
如果每一项都含有公因式 x，应先提取。对于 2x² - 8x = 0，写成 2x(x - 4) = 0，因此 x = 0 或 x = 4。绝不要两边同时除以 x，因为这样会丢掉根 x = 0。
4. Solving by Factorising (a ≠ 1) | 因式分解法 (a ≠ 1)
and c (2 × -3 = -6) and look for two numbers with product -6 and sum 5. They are 6 and -1.
当 a 不等于 1 时，例如 2x² + 5x - 3 = 0，我们使用系统方法：先计算 a 与 c 的乘积 (2 × -3 = -6)，再寻找乘积为 -6、和为 5 的两个数。它们是 6 和 -1。
se in pairs: 2x(x + 3) - 1(x + 3) = 0, which gives the common bracket (2x - 1)(x + 3) = 0.
用这两个数拆分中间项：2x² + 6x - x - 3 = 0。然后两两分组分解：2x(x + 3) - 1(x + 3) = 0，得到公因式括号 (2x - 1)(x + 3) = 0。
2x² + 5x - 3 = 0 → (2x - 1)(x + 3) = 0 → x = ½ or x = -3
Checking the first root: 2(½)² + 5(½) - 3 = 0.5 + 2.5 - 3 = 0, which confirms the answer.
令每个括号等于零，得到 x = ½ 和 x = -3。检验第一个根：2(½)² + 5(½) - 3 = 0.5 + 2.5 - 3 = 0，答案正确。
An alternative and often faster route is the 'guess and check' method for simple coefficients, but the splitting method above always works and is the safest choice when a is larger than 2 or when the numbers are awkward.
另一种有时更快的方法是针对简单系数的“试凑法”，但上述拆分法总是可行，当 a 大于 2 或数字较为刁钻时，它是最稳妥的选择。
5. The Quadratic Formula | 二次公式法
tic formula always works, provided the equation is already in the form a x² + b x + c = 0.
并不是每个二次方程都能用整数因式分解。当因式分解行不通时，只要方程已经写成 a x² + b x + c = 0 的形式，二次公式就一定有效。
x = (-b ± √b² - 4ac) ÷ 2a
two values: one using + and one using -. Write down a, b and c before substituting, and use brackets when the coefficients are negative, because sign errors are the most common cause of lost marks here.
符号 ± 意味着你必须算出两个值：一个用 +、一个用 -。代入前先写出 a、b 和 c，系数为负时务必加括号，因为符号错误是这里最常见的失分原因。
Example: solve x² - 4x - 1 = 0. Here a = 1, b = -4, c = -1, so b² - 4ac = 16 + 4 = 20.
例：解 x² - 4x - 1 = 0。这里 a = 1，b = -4，c = -1，所以 b² - 4ac = 16 + 4 = 20。
x = (4 ± √20) ÷ 2 = (4 ± 2√5) ÷ 2 = 2 ± √5
So x = 2 + √5 ≈ 4.24 or x = 2 - √5 nswer in surd form unless the question asks for 3 significant figures or 2 decimal places.
因此 x = 2 + √5 ≈ 4.24 或 x = 2 - √5 ≈ -0.24。除非题目要求 3 位有效数字或 2 位小数，否则应保留根式形式。
- Write a, b, c clearly / 明确写出 a、b、c
- Square the whole of b, including its sign / b 要连符号一起平方
- Simplify the surd before dividing / 先化简根式再相除
- Give both roots / 两个根都要写出来
6. The Discriminant | 判别式
The expression under the square root, b² - 4ac, is called the discriminant and is often written as Δ. It tells you how many real roots a quadratic has, without solving 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几个实根。
| Discriminant / 判别式 | Number of real roots / 实根个数 | Graph / 图像 |
| b² - 4ac > 0 | Two distinct roots / 两个不同实根 | Crosses the x-axis twice / 与 x 轴交于两点 |
| b² - 4ac = 0 | One repeated root / 一个重根 | Touches the x-axis once / 与 x 轴相切 |
| b² - 4ac < 0 | No real roots / 无实根 | Does not meet the x-axis / 不与 x 轴相交 |
lt; 0, so the equation has no real solutions and the curve lies entirely above the x-axis.
例如，在 x² - 4x - 1 = 0 中判别式为 20 > 0，因此有两个不同实根。在 x² + 2x + 5 = 0 中判别式为 4 - 20 = -16 < 0，因此方程无实数解，曲线完全位于 x 轴上方。
Exam questions frequently reverse the logic: 'find the value of k for which x² + kx + 9 = 0 has equal roots'. Setting b² - 4ac = 0 gives k² - 36 = 0, so k = 6 or k = -6.
考试常反向设问：“求使 x² + kx + 9 = 0 有相等实根的 k 值”。令 b² - 4ac = 0 得 k² - 36 = 0，因此 k = 6 或 k = -6。
7. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q, which makes both the roots and the vertex of the graph easy to read off.
配方法把二次式改写为 (x + p)² + q 的形式，使方程的根与图像的顶点都变得容易读出。
t inside the bracket, then subtract its square. So x² + b x + c = (x + b/2)² - (b/2)² + c.
对于 x² + b x + c，规则是：取 b 的一半放入括号，再减去它的平方。即 x² + b x + c = (x + b/2)² - (b/2)² + c。
y completing the square. Half of 6 is 3, so x² + 6x + 1 = (x + 3)² - 9 + 1 = (x + 3)² - 8.
例：用配方法解 x² + 6x + 1 = 0。6 的一半是 3，所以 x² + 6x + 1 = (x + 3)² - 9 + 1 = (x + 3)² - 8。
(x + 3)² = 8 → x + 3 = ±2√2 → x = -3 ± 2√2
the point (-3, -8), and the minimum value of the expression is -8. This is why completing the square is also the standard tool for finding maximum and minimum values.
同一形式还能立刻看出曲线 y = x² + 6x + 1 的顶点在点 (-3, -8)，该表达式的最小值为 -8。这正是配方法也是求最大值、最小值标准工具的原因。
8. Graphical Interpretation | 图像解读
. Its roots are exactly the x-coordinates of the points where the curve crosses the x-axis, which is why 'solve' and 'find the x-intercepts' mean the same thing.
y = a x² + b x + c 的图像是一条抛物线。它的根正是曲线与 x 轴交点的横坐标，这就是“解方程”和“求 x 轴截距”含义相同的原因。
If a > 0 the parabola opens upwards with a minimum point; if a < 0 it opens downwards with a maximum point. A graph question usually asks you to read the roots to 1 decimal place, so mark where the curve cuts the axis and read across carefully.
若 a > 0，抛物线开口向上，有最低点；若 a < 0，开口向下，有最高点。图像题通常要求你把根读到 1 位小数，所以要标出曲线与轴的交点并仔细对应读数。
of y = x² - 2x - 3, read where y = 0. To solve x² = 2x + 3, rewrite it as x² - 2x - 3 = 0 and read the same points.
常见的考试陷阱是要求你借助给定的图像解一个非标准形式的方程。例如，用 y = x² - 2x - 3 的图像解 x² - 2x - 3 = 0，就是读 y = 0 的位置。而要解 x² = 2x + 3，只需改写为 x² - 2x - 3 = 0，再读同样的点。
the line y = c. Drawing that horizontal line on the grid is normally worth an extra mark.
对于像 y = x² - 2x - 1 这样的相关曲线，通过加减变换得到目标方程，再与直线 y = c 求交点。在网格上画出这条水平线通常还能多得一分。
9. Forming and Solving Word Problems | 列方程解应用题
Many quadratic questions begin with a real-life context: areas of rectangles, consecutive integers, or motion under gravity. The routine is always the same: define the unknown, write an equation, rearrange to zero, then solve.
许多二次方程题目以实际情境开头：矩形面积、连续整数或重力作用下的运动。解题步骤始终相同：设未知数、列方程、整理为零、再求解。
Example: a rectangle is 3 cm longer than it is wide and has an area of 40 cm². Let the width be x, so the length is x + 3, and x(x +
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