📚 Mastering Quadratic Equations for IGCSE Mathematics | 掌握 IGCSE 数学二次方程
Quadratic equations appear throughout the IGCSE Mathematics syllabus, from algebraic manipulation to graph sketching and real-life modelling. A strong grasp of the standard form, factorising, completing the square, and the quadratic formula will give you reliable methods for any exam question.
二次方程贯穿 IGCSE 数学大纲，从代数运算到图像绘制和现实建模都会考查。熟练掌握标准形式、因式分解、配方法和求根公式，能让你在考试中应对任何相关题目。
1. What Is a Quadratic Equation? | 什么是二次方程？
he equation can always be rearranged into the standard form ax² + bx + c = 0, where a ≠ 0.
二次方程是次数为 2 的多项式方程。也就是说，变量 x 的最高次数是 2，方程总可以整理成标准形式 ax² + bx + c = 0，其中 a ≠ 0。
are both quadratic equations. The presence of the x² term is essential; without it the equation would be linear or constant.
例如，x² − 3x + 2 = 0 和 2x² + 5x = 0 都是二次方程。x² 项的存在是关键；如果没有 x² 项，方程就变成一次方程或常数方程。
2. Standard Form and Key Terms | 标准形式与关键术语
The standard form of a quadratic equation is written as:
二次方程的标准形式写作：
ax² + bx + c = 0
Here a, b and c are constants, and a cannot be zero. The term ax² is called the quadratic term, bx is the linear term, and c is the constant term.
其中 a、b 和 c 是常数，且 a 不能为零。ax² 项称为二次项，bx 项称为一次项，c 称为常数项。
ide equal to zero. For example, x² + 4 = 4x should first be rearranged to x² − 4x + 4 = 0.
解题前，应始终合并同类项并将方程一边设为零。例如，x² + 4 = 4x 应首先整理为 x² − 4x + 4 = 0。
3. Solving by Factorising | 因式分解法求解
rising is usually the fastest method. The key idea is that if pq = 0, then p = 0 or q = 0.
当二次式可以写成两个括号的乘积时，因式分解通常是最快的方法。核心思想是：如果 pq = 0，那么 p = 0 或 q = 0。
We look for two numbers that multiply to 6 and add to −5, which are −2 and −3. Therefore:
例题：解 x² − 5x + 6 = 0。我们寻找两个数，它们相乘为 6，相加为 −5，这两个数是 −2 和 −3。因此：
x² − 5x + 6 = (x − 2)(x − 3) = 0
or x = 3. Always check your answers by substituting them back into the original equation.
令每个括号分别为零，得到 x = 2 或 x = 3。始终将答案代回原方程进行检验。
4. Solving by Completing the Square | 配方法求解
Completing the square is useful when factorising is not straightforward, and it also helps with finding the turning point of a quadratic graph. The method transforms ax² + bx + c into the form a(x + p)² + q.
当因式分解不容易进行时，配方法非常有用，它也有助于求二次图像顶点。该方法将 ax² + bx + c 转化为 a(x + p)² + q 的形式。
Example: Solve x² + 6x + 2 = 0 by completing the square. First rewrite:
例题：用配方法解 x² + 6x + 2 = 0。首先改写：
x² + 6x = (x + 3)² − 9
So x² + 6x + 2 = (x + 3)² − 7 = 0. Then (x + 3)² = 7, giving x = −3 ± √7.
因此 x² + 6x + 2 = (x + 3)² − 7 = 0。于是 (x + 3)² = 7，求得 x = −3 ± √7。
5. Solving with the Quadratic Formula | 公式法求解
The quadratic formula works for every quadratic equation, even when factorising is difficult or impossible. It is derived from completing the square and states:
求根公式适用于所有二次方程，即使因式分解困难或无法进行时也可使用。它由配方法推导而来，公式为：
x = (−b ± √b² − 4ac) ÷ 2a
Example: Solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, c = −2. Substitute into the formula:
例题：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)
x = (−3 ± √9 + 16) ÷ 4 = (−3 ± 5) ÷ 4
−2. The formula is especially reliable if you take care with negative signs and brackets.
得到 x = 1/2 或 x = −2。只要注意负号和括号，公式法非常可靠。
6. The Discriminant and Nature of Roots | 判别式与根的性质
The discriminant is the expression inside the square root in the quadratic formula:
判别式是求根公式中平方根内的表达式：
Δ = b² − 4ac
real roots. If Δ = 0, there is one repeated real root. If Δ < 0, there are no real roots.
它告诉你方程有多少个实数根。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。
× 1 × 5 = 16 − 20 = −4, so there are no real roots. This is a common IGCSE question style.
例题：对于 x² + 4x + 5 = 0，Δ = 4² − 4 × 1 × 5 = 16 − 20 = −4，因此没有实数根。这是 IGCSE 常见题型。
7. Quadratic Graphs and Turning Points | 二次图像与顶点
. 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，有最低点；如果 a < 0，抛物线开口向下，有最高点。
The turning point has x-coordinate:
顶点的 x 坐标为：
x = −b ÷ 2a
Substitute this x-value into the equation to find the y-coordinate. Completing the square can also show the turning point directly from the form a(x + p)² + q as (−p, q).
将该 x 值代入方程可求出 y 坐标。配方法也可直接由 a(x + p)² + q 看出顶点为 (−p, q)。
8. Applications and Word Problems | 应用题与文字题
Quadratic equations model many real-life situations, including area, projectile motion, and profit. In word problems, define the unknown as x, translate the conditions into an equation, solve it, and reject any solution that does not fit the context.
二次方程可以模拟许多现实情境，包括面积、抛体运动和利润。解文字题时，先设未知数为 x，将条件转化为方程，解方程，并舍去不符合题意的解。
ves (x + 12)(x − 11) = 0, so x = 11, since x must be positive. The integers are 11 and 12.
例题：两个连续正整数的积为 132。设这两个整数为 x 和 x + 1。则 x(x + 1) = 132，所以 x² + x − 132 = 0。因式分解得 (x + 12)(x − 11) = 0，因为 x 必须是正数，所以 x = 11。这两个整数是 11 和 12。
9. Common Mistakes to Avoid | 常见错误提醒
ero before factorising. Another is dividing both sides by x and losing the solution x = 0.
许多分数因可避免的错误而丢失。一个常见错误是在因式分解前忘记将方程设为零。另一个错误是两边同除以 x，从而丢失 x = 0 这个解。
- Do not divide by a variable unless you are certain it is not zero. | 除非你确定变量不为零，否则不要除以变量。
- Do not confuse (−b ± √b² − 4ac) ÷ 2a with −b ± √b² − 4ac ÷ 2a. | 不要把 (−b ± √b² − 4ac) ÷ 2a 与 −b ± √b² − 4ac ÷ 2a 混淆。
- Always check solutions in the original equation. | 总要把解代回原方程检验。
10. Exam Tips and Practice Strategy | 考试技巧与练习策略
In the exam, first identify the method that is most efficient: factorising if the expression is simple, the formula if coefficients are messy, and completing the square if the question asks for the turning point.
考试中，首先要识别最有效的方法：表达式简单时用因式分解，系数较繁琐时用公式，要求顶点时用配方法。
Practise mixed questions under timed conditions, and write down every step clearly. This helps you gain method marks even if the final answer is wrong.
在限时条件下练习混合题型，并清楚写出每一步。这样即使最终答案错误，也能获得步骤分。
| Method 方法 | Best Used When 最佳使用条件 |
| Factorising 因式分解 | The expression factorises easily. 表达式容易因式分解。 |
| Completing the square 配方法 | You need the turning point or to derive the formula. 需要求顶点或推导公式。 |
| Quadratic formula 求根公式 | Any quadratic, especially messy coefficients. 任何二次方程，尤其是系数复杂时。 |
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