📚 Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与求根公式
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from pure algebra to geometry, measurement and modelling. They are equations of degree two, and they can describe projectile paths, areas of rectangles, and many optimisation problems. In this revision guide, you will learn how to recognise quadratic equations, solve them using three main methods, and interpret their graphs and discriminants.
二次方程贯穿 IGCSE 数学大纲，从纯代数到几何、测量和建模都有出现。它们是二次方程，可以描述抛体轨迹、矩形面积以及许多最优化问题。在本复习指南中，你将学会识别二次方程，用三种主要方法求解，并理解图像与判别式。
1. Recognising a Quadratic Equation | 识别二次方程
a, b and c are constants and a ≠ 0. If a = 0, the equation becomes linear, not quadratic.
如果方程中变量的最高次数为 2，则该方程称为二次方程。它总可以写成标准形式 ax² + bx + c = 0，其中 a、b、c 是常数且 a ≠ 0。如果 a = 0，方程就变成一次方程，而不是二次方程。
ax² + bx + c = 0, a ≠ 0
ighest power of x is 2. However, 2x + 5 = 0 is linear because the highest power of x is 1.
例如，x² + 3x − 4 = 0 是二次方程，因为 x 的最高次数为 2。然而，2x + 5 = 0 是一次方程，因为 x 的最高次数为 1。
ch as x(x + 2) = 15. You must expand and rearrange them into standard form before solving.
二次方程也可能以变形形式出现，例如 x(x + 2) = 15。你必须先展开并整理成标准形式，然后再求解。
2. Solving by Factorising | 因式分解法求解
Factorising is often the fastest method when the quadratic expression has simple integer roots. First write the equation in standard form. Then factorise the left side into two brackets. Finally set each factor equal to zero and solve.
当二次式具有简单整数根时，因式分解通常是最快的方法。首先将方程写成标准形式，然后将左边因式分解为两个括号，最后令每个因式等于零并求解。
ion becomes (x + 2)(x + 3) = 0. Therefore x + 2 = 0 or x + 3 = 0, giving x = −2 or x = −3.
示例：求解 x² + 5x + 6 = 0。该式可因式分解为 (x + 2)(x + 3)，因此方程变为 (x + 2)(x + 3) = 0。于是 x + 2 = 0 或 x + 3 = 0，得到 x = −2 或 x = −3。
Always check your factorisation by expanding the brackets. A correct factorisation must return the original quadratic expression.
一定要通过展开括号来检查因式分解是否正确。正确的因式分解必须还原为原来的二次式。
If the quadratic has a coefficient of x² that is not 1, such as 2x² + 5x − 3, factorising requires more care. You can use the ac method or trial and error to find the correct pair of brackets.
如果二次式中 x² 的系数不为 1，例如 2x² + 5x − 3，因式分解就需要更加小心。你可以使用 ac 方法或试错法来找到正确的括号对。
3. The Zero Product Property | 零乘积性质
then at least one of the factors must be zero. In symbols, if AB = 0, then A = 0 or B = 0.
因式分解法依赖于零乘积性质：如果两个因式的乘积为零，那么至少有一个因式为零。用符号表示，如果 AB = 0，则 A = 0 或 B = 0。
AB = 0 → A = 0 or B = 0
This is why after factorising a quadratic equation, you set each bracket equal to zero separately. It works because the product is zero only when one of the factors is zero.
这就是为什么将二次方程因式分解后，要分别令每个括号等于零。它的成立是因为只有当其中一个因式为零时，乘积才为零。
. The zero product property only applies when the product equals zero, not six or any other non-zero number.
一个常见错误是写出 (x + 2)(x + 3) = 6，然后得出 x + 2 = 6 或 x + 3 = 6。零乘积性质只适用于乘积等于零的情况，而不适用于等于 6 或任何其他非零数的情况。
Before applying the zero product property, you must rearrange the equation so that one side is exactly zero. Otherwise the method fails.
在应用零乘积性质之前，你必须对方程进行整理，使一边恰好为零。否则该方法会失效。
4. Completing the Square | 配方法
Completing the square is a powerful method that works for every quadratic equation. It is especially useful when the quadratic cannot be factorised easily. The key idea is to rewrite the expression x² + bx as part of a perfect square.
配方法是一种适用于每个二次方程的强大方法。当二次式不容易因式分解时，它尤其有用。关键思想是将表达式 x² + bx 改写为一个完全平方的一部分。
uare for x² + bx, add and subtract (b ÷ 2)². This gives x² + bx = (x + b ÷ 2)² − (b ÷ 2)².
要为 x² + bx 配方，需加上并减去 (b ÷ 2)²。这会得到 x² + bx = (x + b ÷ 2)² − (b ÷ 2)²。
alf of it is 3, and 3² = 9. Add 9 to both sides: x² + 6x + 9 = 4. This gives (x + 3)² = 4.
示例：用配方法求解 x² + 6x + 5 = 0。首先改写为 x² + 6x = −5。x 的系数为 6，它的一半是 3，而 3² = 9。两边同时加 9：x² + 6x + 9 = 4。由此得到 (x + 3)² = 4。
x² + 6x + 5 = 0 → x² + 6x = −5 → x² + 6x + 9 = 4 → (x + 3)² = 4
the square root of both sides: x + 3 = ±2. Therefore x = −3 ± 2, giving x = −1 or x = −5.
接下来对两边取平方根：x + 3 = ±2。因此 x = −3 ± 2，得到 x = −1 或 x = −5。
If the coefficient of x² is not 1, divide the whole equation by that coefficient before completing the square.
如果 x² 的系数不为 1，在配方之前要先用该系数除以整个方程。
5. The Quadratic Formula | 求根公式
= 0 directly. It is derived by completing the square on the general form. The formula is:
求根公式可直接给出任何二次方程 ax² + bx + c = 0 的解。它是对一般形式进行配方而推导出来的。公式为：
x = (−b ± √b² − 4ac) ÷ 2a
You should use the quadratic formula when factorising is difficult or when the coefficients are not integers. It always works, provided you substitute the values of a, b and c carefully.
当因式分解困难或系数不是整数时，你应当使用求根公式。只要你仔细代入 a、b、c 的值，它总是可行。
3, c = −2. Substitute into the formula: x = (−3 ± √(3² − 4 × 2 × (−2))) ÷ (2 × 2) = (−3 ± √9 + 16) ÷ 4 = (−3 ± √25) ÷ 4 = (−3 ± 5) ÷ 4.
示例：求解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。代入公式：x = (−3 ± √(3² − 4 × 2 × (−2))) ÷ (2 × 2) = (−3 ± √9 + 16) ÷ 4 = (−3 ± √25) ÷ 4 = (−3 ± 5) ÷ 4。
This gives the two solutions x = (−3 + 5) ÷ 4 = 1 ÷ 2 and x = (−3 − 5) ÷ 4 = −2.
由此得到两个解：x = (−3 + 5) ÷ 4 = 1 ÷ 2 和 x = (−3 − 5) ÷ 4 = −2。
Always write the formula exactly as shown and use brackets around negative values. A sign error in b or c is one of the most common causes of lost marks.
务必按照所示方式写出公式，并在负值处使用括号。b 或 c 的符号错误是失分最常见的原因之一。
6. The Discriminant | 判别式
It tells you how many real solutions the quadratic equation has without solving it fully.
判别式是求根公式中平方根下的部分：Δ = b² − 4ac。它无需完全求解方程，就能告诉你二次方程有多少个实数解。
Δ = b² − 4ac
d real root. If Δ < 0, it has no real roots, because the square root of a negative number is not real.
如果 Δ > 0，方程有两个不同的实数根。如果 Δ = 0，方程有一个重根。如果 Δ < 0，方程没有实数根，因为负数的平方根不是实数。
² − 4x + 4 = 0 has Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0, so it has one repeated root x = 2.
例如，x² − 4x + 4 = 0 的判别式为 Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0，所以它有一个重根 x = 2。
In graphical terms, the discriminant determines how many times the parabola crosses the x-axis. A positive discriminant means two intersections, zero means one touching point, and negative means no intersection.
从图像上看，判别式决定抛物线穿过 x 轴的次数。判别式为正表示有两个交点，为零表示有一个切点，为负表示没有交点。
7. Graphical Interpretation | 图像解释
solutions of ax² + bx + c = 0 are the x-coordinates where the parabola crosses the x-axis.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，开口向上；如果 a < 0，开口向下。方程 ax² + bx + c = 0 的解就是抛物线与 x 轴交点的横坐标。
dinate of the vertex can be found by substituting this x value into the original equation.
抛物线的顶点位于对称轴上，对称轴方程为 x = −b ÷ 2a。顶点的纵坐标可以通过将这个 x 值代入原方程求得。
x = −b ÷ 2a
The y-intercept of the parabola is simply the constant term c, because when x = 0, y = c.
抛物线的 y 轴截距就是常数项 c，因为当 x = 0 时，y = c。
Understanding the graph helps you check whether your algebraic solutions are reasonable. For example, if the discriminant is negative, the parabola is entirely above or below the x-axis, so there are no real x-intercepts.
理解图像有助于检查你的代数解是否合理。例如，如果判别式为负，抛物线完全位于 x 轴上方或下方，因此没有实数 x 轴截距。
8. Solving Word Problems | 解应用题
Many IGCSE questions ask you to form a quadratic equation from a real-life situation. The key steps are: define the unknown, write an equation from the given information, solve it, and then check that the solution makes sense in context.
许多 IGCSE 题目要求你根据实际情境建立二次方程。关键步骤是：设未知数，根据已知信息写出方程，求解方程，然后检查解在实际情境中是否有意义。
et the width be x cm. Then the length is x + 3 cm, and the area equation is x(x + 3) = 40.
示例：一个矩形的长比宽多 3 cm，面积为 40 cm²。设宽为 x cm，则长为 x + 3 cm，面积方程为 x(x + 3) = 40。
5) = 0, so x = −8 or x = 5. Since width cannot be negative, x = 5 cm. The length is 8 cm.
展开并整理：x² + 3x = 40，因此 x² + 3x − 40 = 0。因式分解得 (x + 8)(x − 5) = 0，所以 x = −8 或 x = 5。由于宽不能为负，x = 5 cm。长为 8 cm。
Always reject solutions that do not fit the physical situation, such as negative lengths, times or areas. Also check that your answers satisfy the original conditions.
一定要舍弃不符合物理情境的解，例如负的长度、时间或面积。还要检查你的答案是否满足原始条件。
9. Common Mistakes | 常见错误
equation is not quadratic, and using the quadratic formula would require division by zero.
一个常见错误是忘记标准形式中 a 不能为零。如果 a = 0，方程就不是二次方程，使用求根公式时会出现除以零的情况。
Another common mistake is making sign errors when factorising. For example, x² − 5x + 6 factorises as
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