📚 Quadratic Equations | IGCSE 数学：二次方程
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from basic factorising to graph sketching and real-life modelling. Mastering them gives you reliable marks in both the Core and Extended papers.
二次方程贯穿 IGCSE 数学课程，从基础因式分解到图像绘制和实际建模。掌握二次方程能让你在 Core 和 Extended 试卷中稳定得分。
1. What Is a Quadratic Equation? | 什么是二次方程？
x, is 2. Its standard form is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.
二次方程是次数为 2 的方程。也就是说，变量（通常是 x）的最高次数为 2。其标准形式为 ax² + bx + c = 0，其中 a、b、c 是常数，且 a ≠ 0。
adratic because the highest power of x is 2. However, 4x − 7 = 0 is linear, not quadratic.
例如：3x² − 5x + 2 = 0 是二次方程，因为 x 的最高次数为 2。而 4x − 7 = 0 是一次方程，不是二次方程。
The condition a ≠ 0 is essential. If a were equal to 0, the x² term would disappear and the equation would become linear.
条件 a ≠ 0 非常重要。如果 a 等于 0，x² 项就会消失，方程就会变成一次方程。
2. Standard Form and Coefficients | 标准形式与系数
The number a is called the quadratic coefficient, b the linear coefficient and c the constant term.
求解前，一定要把方程整理为 ax² + bx + c = 0 的形式。a 称为二次项系数，b 称为一次项系数，c 称为常数项。
c = 1. For −x² + 4x = 5, first move the 5: −x² + 4x − 5 = 0, so a = −1, b = 4 and c = −5.
对于 2x² − 3x + 1 = 0，我们有 a = 2，b = −3，c = 1。对于 −x² + 4x = 5，先把 5 移项：−x² + 4x − 5 = 0，因此 a = −1，b = 4，c = −5。
Identifying a, b and c correctly is the first step in most solution methods, especially when using the quadratic formula.
正确识别 a、b 和 c 是大多数解题方法的第一步，尤其是在使用求根公式时。
| Equation | 方程 | a | b | c |
|---|---|---|---|
| x² − 5x + 6 = 0 | 1 | −5 | 6 |
| 2x² − 3 = 0 | 2 | 0 | −3 |
| −x² + 4x = 5 | −1 | 4 | −5 |
3. Solving by Factorising | 因式分解法
Factorising is the fastest method when the quadratic can be written as a product of two brackets. After factorising, set each bracket equal to zero and solve.
当二次式可以写成两个括号的乘积时，因式分解法是最快的方法。因式分解后，令每个括号等于零并求解。
0. We need two numbers that multiply to 6 and add to −5: −2 and −3. So (x − 2)(x − 3) = 0.
例如，解 x² − 5x + 6 = 0。我们需要两个数，乘积为 6，和为 −5：−2 和 −3。所以 (x − 2)(x − 3) = 0。
= 0 or x − 3 = 0, giving x = 2 or x = 3. Always write both solutions in your final answer.
然后 x − 2 = 0 或 x − 3 = 0，得到 x = 2 或 x = 3。最终答案一定要写出两个解。
- Step 1: Write the equation in standard form. | 第一步：把方程写成标准形式。
- Step 2: Factorise the quadratic expression. | 第二步：因式分解二次表达式。
- Step 3: Set each factor equal to zero and solve. | 第三步：令每个因式等于零并求解。
- Step 4: Check by substituting your values back into the original equation. | 第四步：把解代回原方程检验。
4. Solving by Completing the Square | 配方法
Completing the square is useful when factorising is difficult or when you need to find the turning point of a graph. The key identity is shown below.
当因式分解比较困难，或者需要求图像的顶点时，配方法非常有用。核心恒等式如下。
x² + bx = (x + b ÷ 2)² − (b ÷ 2)²
xample, solve x² + 6x + 2 = 0. First rewrite x² + 6x as (x + 3)² − 9 because (6 ÷ 2)² = 9.
例如，解 x² + 6x + 2 = 0。首先把 x² + 6x 改写为 (x + 3)² − 9，因为 (6 ÷ 2)² = 9。
So the equation becomes (x + 3)² − 9 + 2 = 0, which simplifies to (x + 3)² = 7.
因此方程变为 (x + 3)² − 9 + 2 = 0，化简为 (x + 3)² = 7。
Taking the square root of both sides gives x + 3 = ±√7, so x = −3 ± √7.
两边开平方得到 x + 3 = ±√7，所以 x = −3 ± √7。
x = −3 + √7 or x = −3 − √7
5. The Quadratic Formula | 公式法
If a quadratic cannot be factorised easily, use the quadratic formula. It works for every quadratic equation in standard form.
如果二次方程不容易因式分解，可以使用求根公式。它适用于所有标准形式的二次方程。
x = (−b ± √b² − 4ac) ÷ (2a)
Example: Solve 2x² + 3x − 2 = 0. Here a = 2, b = 3 and c = −2.
例如：解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。
Substitute into the formula: x = (−3 ± √(3² − 4 × 2 × (−2))) ÷ (2 × 2).
代入公式：x = (−3 ± √(3² − 4 × 2 × (−2))) ÷ (2 × 2)。
This simplifies to x = (−3 ± √9 + 16) ÷ 4 = (−3 ± √25) ÷ 4 = (−3 ± 5) ÷ 4.
化简为 x = (−3 ± √9 + 16) ÷ 4 = (−3 ± √25) ÷ 4 = (−3 ± 5) ÷ 4。
So x = (−3 + 5) ÷ 4 = 2 ÷ 4 = ½ or x = (−3 − 5) ÷ 4 = −8 ÷ 4 = −2.
所以 x = (−3 + 5) ÷ 4 = 2 ÷ 4 = ½ 或 x = (−3 − 5) ÷ 4 = −8 ÷ 4 = −2。
6. Discriminant and Nature of Roots | 判别式与根的性质
mula: D = b² − 4ac. It tells us the number and type of roots without solving the equation.
判别式是求根公式中平方根号内的表达式：D = b² − 4ac。它能告诉我们方程根的数量和类型，而无需实际求解。
- If D > 0, there are two distinct real roots. | 如果 D > 0，方程有两个不相等的实数根。
- If D = 0, there is one repeated real root. | 如果 D = 0，方程有一个重根。
- If D < 0, there are no real roots. | 如果 D < 0，方程没有实数根。
x² + 4x + 4 = 0, D = 4² − 4 × 1 × 4 = 16 − 16 = 0, so there is exactly one repeated root.
例如：对于 x² + 4x + 4 = 0，D = 4² − 4 × 1 × 4 = 16 − 16 = 0，所以恰好有一个重根。
Example: For x² + x + 1 = 0, D = 1² − 4 × 1 × 1 = 1 − 4 = −3, so there are no real roots.
例如：对于 x² + x + 1 = 0，D = 1² − 4 × 1 × 1 = 1 − 4 = −3，所以没有实数根。
7. Graphs of Quadratic Functions | 二次函数图像
. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上，有一个最低点。如果 a < 0，抛物线开口向下，有一个最高点。
is always (0, c). The x-intercepts, if they exist, are the real roots of ax² + bx + c = 0.
y 轴截距始终是 (0, c)。如果存在 x 轴截距，它们就是 ax² + bx + c = 0 的实数根。
The axis of symmetry passes through the turning point. Its equation is given below.
对称轴经过顶点。它的方程如下。
Axis of symmetry: x = −b ÷ (2a)
x² − 4x + 3, the axis is x = −(−4) ÷ (2 × 1) = 2, so the turning point has x-coordinate 2.
对于 y = x² − 4x + 3，对称轴为 x = −(−4) ÷ (2 × 1) = 2，因此顶点的 x 坐标为 2。
8. Turning Point and Completed Square Form | 顶点与配方式
+ k gives the vertex (h, k) directly. The graph is a parabola shifted h units horizontally and k units vertically from the origin.
配方式 y = a(x − h)² + k 直接给出顶点 (h, k)。图像是由原点平移 h 个单位水平距离和 k 个单位竖直距离得到的抛物线。
For y = (x − 2)² + 3, the vertex is (2, 3) and the graph opens upwards because a = 1 &
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