📚 Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 一元二次方程：因式分解、配方法与求根公式
In IGCSE Mathematics, quadratic equations appear across algebra, graphs and problem-solving. A quadratic equation involves a squared term and can often have two solutions. This revision guide covers factorising, completing the square, the quadratic formula, the discriminant and graph interpretation, so that you can recognise a quadratic, choose the fastest method, and connect the algebra to the curve.
在 IGCSE 数学中，一元二次方程广泛出现在代数、图像和应用题中。一元二次方程含有二次项，通常有两个解。本复习指南涵盖因式分解法、配方法、求根公式、判别式以及图像解读，帮助你识别一元二次方程、选择最快的方法，并把代数式与抛物线图像对应起来。
1. Recognising a Quadratic Equation | 认识一元二次方程
A quadratic equation is an equation in which the highest power of the variable is 2. In IGCSE Mathematics, it is usually written with x as the variable and can have two solutions, one solution or no real solutions.
一元二次方程是指变量的最高次数为 2 的方程。在 IGCSE 数学中，它通常以 x 为变量，可能有两个解、一个解或没有实数解。
because they contain an x² term. The squared term is what makes them quadratic, not the number of terms you can see at first glance.
例如，x² + 5x + 6 = 0、2x² − 3x = 1 和 x² − 9 = 0 都是一元二次方程，因为它们含有 x² 项。判断是否为二次方程的关键在于是否存在二次项，而不是表面上的项数。
c, because the x in the denominator makes the expression fractional rather than a polynomial of degree 2.
像 x³ + 2x = 1 这样的方程不是二次方程，因为它的最高次数是 3。同样，1 ÷ x + x = 3 也不是二次方程，因为分母中的 x 使式子成为分式而不是二次多项式。
write it as x² + 5x + 6 = 0. The equals sign is what turns an expression into something you can solve.
你还需要区分"整式"与"方程"。x² + 5x + 6 是二次整式，可以因式分解；但只有写成 x² + 5x + 6 = 0 以后，它才有"解"。等号是让整式变成可求解方程的关键。
2. Standard Form and Key Terminology | 标准形式与关键术语
are numbers and a ≠ 0. If a were 0, the equation would become linear instead of quadratic.
任何一元二次方程都可以整理成标准形式 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。如果 a = 0，方程就退化为一次方程，不再是二次方程。
rmula. Subtract 1 from both sides to obtain 2x² − 3x − 1 = 0, so a = 2, b = −3 and c = −1.
例如，使用求根公式之前，必须先把 2x² − 3x = 1 整理好。两边同时减去 1，得到 2x² − 3x − 1 = 0，于是 a = 2，b = −3，c = −1。
Here is a quick reference for identifying the coefficients correctly. Remember that the sign in front of a term belongs to that coefficient.
下面这张表可以帮助你正确识别各项系数。请注意：某一项前面的符号属于它自己的系数。
| Equation | 方程 | a | b | c |
| x² + 5x + 6 = 0 | 1 | 5 | 6 |
| 2x² − 3x − 1 = 0 | 2 | −3 | −1 |
| x² − 9 = 0 | 1 | 0 | −9 |
| 3x² + x = 0 | 3 | 1 | 0 |
The answers to a quadratic equation are called roots, solutions or zeros. In an IGCSE exam these words are used interchangeably, so do not be confused when a question asks for "the roots of the equation" or "the values of x for which the curve crosses the axis".
一元二次方程的结果称为"根"、"解"或"零点"。在 IGCSE 考试中这些词可以互换使用，因此当题目问"方程的根"或"曲线与坐标轴交点的 x 值"时，你不必困惑，它们是同一件事。
3. Solving by Factorising | 因式分解法求解
Factorising is the quickest method when the quadratic factorises into whole-number brackets. It relies on the zero product property: if the product of two factors is zero, then at least one of the factors must be zero.
当二次式能分解为整数括号时，因式分解是最快的方法。它的依据是"零乘积性质"：若两个因式的乘积为零，则其中至少有一个因式为零。
If pq = 0, then p = 0 or q = 0.
x² + 5x + 6 = (x + 2)(x + 3) = 0, and setting each bracket to zero gives x = −2 or x = −3.
以 x² + 5x + 6 = 0 为例。我们需要找两个数，它们的乘积是 +6，和是 +5，这两个数是 +2 和 +3。因此 x² + 5x + 6 = (x + 2)(x + 3) = 0，令每个括号等于零，得到 x = −2 或 x = −3。
Three special patterns appear constantly in exams. Recognising them saves a great deal of time.
考试中有三种特殊模式反复出现，能识别它们可以节省大量时间。
- Common factor | 提取公因式: x² − 5x = x(x − 5) = 0, so x = 0 or x = 5.
- Difference of two squares | 平方差: x² − 9 = (x − 3)(x + 3) = 0, so x = 3 or x = −3.
- Perfect square | 完全平方: x² + 6x + 9 = (x + 3)² = 0, so x = −3 (a repeated root).
rovement across the brackets: 2x² + 7x + 3 = (2x + 1)(x + 3) = 0, giving x = −½ or x = −3.
当 x² 的系数不是 1 时，例如 2x² + 7x + 3 = 0，你可能需要拆分中间项或尝试不同的括号组合：2x² + 7x + 3 = (2x + 1)(x + 3) = 0，得到 x = −½ 或 x = −3。
4. Completing the Square: The Technique | 配方法：技术要点
Completing the square rewrites a quadratic so that the variable appears only once, inside a squared bracket. This form is essential for finding the vertex of a curve and for solving equations that do not factorise.
配方法把二次式改写为"变量只出现一次、且位于一个平方括号内"的形式。这种形式对于求抛物线顶点以及求解无法因式分解的方程都非常重要。
For a monic quadratic x² + bx, the rule is straightforward: take half of b, square it, add it and subtract it.
对于首项系数为 1 的二次式 x² + bx，规则很直接：取 b 的一半，平方，然后加上再减去这个数。
x² + bx = (x + b ÷ 2)² − (b ÷ 2)²
= (x + 3)² − 8. Check by expanding: (x + 3)² − 8 = x² + 6x + 9 − 8 = x² + 6x + 1. Correct.
例如：x² + 6x + 1。6 的一半是 3，3² = 9，所以 x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8。展开验证：(x + 3)² − 8 = x² + 6x + 9 − 8 = x² + 6x + 1，结果正确。
x² + 8x + 5: first write 2(x² + 4x) + 5, then complete inside the bracket as 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3.
当 a ≠ 0 时，先把系数 a 从前两项中提出，再配方；注意减去的常数项要乘回 a。以 2x² + 8x + 5 为例：先写成 2(x² + 4x) + 5，再在括号内配方得 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3。
The completed-square form tells you the minimum point of the curve immediately, which is why examiners ask for it even when the equation cannot be solved nicely.
配方式能立刻告诉你曲线的最低点，这也是即使方程无法漂亮求解、考官仍要求配方的原因。
5. Solving by Completing the Square | 用配方法求根
e root of both sides and remember the ± sign. This gives two roots whenever q is positive.
当方程写成 (x + p)² = q 之后，两边开平方，并牢记 ± 号。只要 q 为正，就能得到两个根。
(x + p)² = q → x = −p ± √q
² + 6x + 1 = 0. From the previous section, (x + 3)² − 8 = 0, so (x + 3)² = 8 and x + 3 = ±√8. Since √8 = 2√2, the exact answers are x = −3 + 2√2 or x = −3 − 2√2, which are approximately −0.172 and −5.828.
解 x² + 6x + 1 = 0。由上一节可知 (x + 3)² − 8 = 0，所以 (x + 3)² = 8，x + 3 = ±√8。因为 √8 = 2√2，精确答案为 x = −3 + 2√2 或 x = −3 − 2√2，约等于 −0.172 和 −5.828。
lutions, because a square cannot be negative. This is the same conclusion the discriminant will give you, but you reach it without any formula.
如果配方式为 (x + p)² = q 且 q < 0，则没有实数解，因为平方不可能为负数。这与判别式给出的结论一致，但你无需使用任何公式就能得到它。
IGCSE questions usually ask for exact surd answers when the equation does not factorise, and for decimals only when the question states "give your answers correct to 2 decimal places".
在 IGCSE 中，当方程无法因式分解时，通常要求给出精确的根式答案；只有当题目明确写"保留两位小数"时才写小数。
6. The Quadratic Formula | 求根公式
t factorises. It is derived by completing the square on the general form ax² + bx + c = 0.
求根公式可以求解任何一元二次方程，无论它能否因式分解。它正是对一般式 ax² + bx + c = 0 进行配方推导出来的。
x = (−b ± √b² − 4ac) ÷ (2a)
1 = 0, where a = 2, b = −3 and c = −1. Then b² − 4ac = (−3)² − 4 × 2 × (−1) = 9 + 8 = 17.
把它用于 2x² − 3x − 1 = 0，此时 a = 2，b = −3，c = −1。于是 b² − 4ac = (−3)² − 4 × 2 × (−1) = 9 + 8 = 17。
x = (3 ± √17) ÷ 4 ≈ 1.781 or −0.281
. Always write the formula with brackets around the substituted values before simplifying.
代入负的 b 值是最常见的出错来源。因为 b = −3，所以 −b 变为 +3。建议先带括号把数值代入公式，再化简，可以避免符号错误。
oduct rules: the sum of the roots is −b ÷ a and the product of the roots is c ÷ a. These are useful for checking your answers quickly.
在拓展内容中，ax² + bx + c = 0 的两个根还满足"和与积"的关系：两根之和为 −b ÷ a，两根之积为 c ÷ a。它们可以帮你快速检验答案。
7. 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 the equation has before you solve it.
根号下的表达式 b² − 4ac 称为判别式，通常记为 Δ。它能在你求解之前就告诉你方程有多少个实数根。
Δ = b² − 4ac
| Value of Δ | Δ 的取值 | Nature of roots | 根的性质 | Graph | 图像 |
| Δ > 0 | Two distinct real roots | 两个不同实根 | Crosses the x-axis twice | 与 x 轴有两个交点 |
| Δ = 0 | One repeated real root | 一个重根 | Touches the x-axis | 与 x 轴相切 |
| Δ < 0 | No real roots | 无实根 | Does not meet the x-axis | 与 x 轴无交点 |
e are no real roots. For 2x² − 3x − 1 = 0, Δ = 17 > 0, so there are two distinct roots.
用具体数字检验这三种情况。对于 x² + 4x + 4 = 0，Δ = 16 − 16 = 0，故 x = −2 为重根。对于 x² + x + 1 = 0，Δ = 1 − 4 = −3 < 0，故无实根。对于 2x² − 3x − 1 = 0，Δ = 17 > 0，故有两个不同实根。
for which the equation has equal roots. Set Δ = 0 and solve the resulting equation for k.
考题常给出含有未知常数 k 的二次方程，问 k 取何值时方程有等根。做法是令 Δ = 0，再解出关于 k 的方程。
8. Graphs of Quadratic Functions | 二次函数图像
. When a > 0 the curve opens upwards and has a minimum point; when a < 0 the curve opens downwards and has a maximum point.
y = ax² + bx + c 的图像是一条抛物线。当 a > 0 时开口向上，有最低点；当 a < 0 时开口向下，有最高点。
tly the x-coordinates where the parabola crosses the x-axis. This is why the discriminant and the number of x-intercepts always agree.
方程 ax² + bx + c = 0 的解，正好是抛物线与 x 轴交点的横坐标。这就是判别式与 x 轴交点个数总是一致的原因。
, the y-intercept is (0, 1), and the curve is symmetric about the vertical line through its vertex.
令 x = 0 可得 y 轴截距 y = c。因此对于 y = x² + 6x + 1，y 轴截距为 (0, 1)，且曲线关于过顶点的竖直直线对称。
Graph questions may also ask you to interpret the intersection of a line and a curve. Setting the line equal to the quadratic produces a new quadratic equation, and its discriminant tells you how many intersection points there are.
图像题还可能要求你解读直线与曲线的交点。把直线方程与二次函数相等，会得到一个新的二次方程，其判别式就告诉你交点有多少个。
9. Vertex, Axis of Symmetry and Completed Square Form | 顶点、对称轴与配方式
x lies on this line, so its x-coordinate is −b ÷ 2a; substitute back to find the y-coordinate.
y = ax² + bx + c 的对称轴是竖直线 x = −b ÷ 2a。顶点在这条直线上，因此顶点的横坐标为 −b ÷ 2a；再代回原式即可求出纵坐标。
(−p, q) and the axis of symmetry is x = −p. No differentiation or substitution is needed.
在配方式 y = a(x + p)² + q 中，顶点直接就是 (−p, q)，对称轴为 x = −p，不需要求导或代入计算。
alue of the function is −8. Because a = 1 > 0, this is a minimum rather than a maximum.
对于 y = (x + 3)² − 8，顶点为 (−3, −8)，对称轴为 x = −3，函数的最小值为 −8。由于 a = 1 > 0，这是最小值而不是最大值。
Typical exam wording includes "write down the coordinates of the turning point", "state the equation of the line of symmetry" and "find the minimum value of the function". All three are answered in one step from the completed-square form.
常见题干包括"写出转折点坐标""写出对称轴方程""求函数的最小值"。这三种问法都能由配方式一步得出答案。
10. Modelling and Problem Solving | 应用题与建模
Quadratic equations appear in word problems about areas, consecutive numbers, speeds and projectiles. The method is always the same: define a variable, translate the words into an equation, rearrange to standard form, solve, then reject any answer that is
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