📚 Mastering Quadratic Equations for Cambridge IGCSE Mathematics | 剑桥 IGCSE 数学：二次方程完全攻略
A quadratic equation is one of the most important topics in the Cambridge IGCSE Mathematics syllabus. It appears in both the Core and Extended papers and is often linked to graphs, algebra, and real-life problems. This article will guide you through the key methods for solving quadratic equations, interpreting the discriminant, sketching graphs, and tackling exam-style questions with confidence.
二次方程是剑桥 IGCSE 数学大纲中最重要的主题之一。它同时出现在核心卷和扩展卷中，并且经常与图像、代数和实际问题相结合。本文将引导你掌握解二次方程的关键方法、理解判别式、绘制函数图像，并自信地应对考试题型。
1. Recognising Quadratic Equations | 识别二次方程
A quadratic equation in one variable is any equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. If a = 0, the equation is no longer quadratic but linear, so the condition on a is essential.
一元二次方程是任何可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b 和 c 是常数，并且 a ≠ 0。如果 a = 0，方程就不再是二次方程，而是一次方程，因此 a 不为零这个条件非常关键。
For example, 3x² - 5x + 2 = 0 is quadratic with a = 3, b = -5 and c = 2. Equations such as x² = 9 or 2x(x - 4) = 0 can be rearranged into the standard form, so they are also quadratic.
例如，3x² - 5x + 2 = 0 是一个二次方程，其中 a = 3、b = -5、c = 2。像 x² = 9 或 2x(x - 4) = 0 这样的方程也可以整理成标准形式，因此它们也是二次方程。
2. Solving by Factorisation | 因式分解法求解
Factorisation is often the fastest method when the quadratic expression can be factorised easily. The key principle is the zero product property: if pq = 0, then either p = 0 or q = 0 or both.
如果二次表达式容易因式分解，因式分解法通常是最快的方法。其核心原理是零乘积性质：如果 pq = 0，那么 p = 0 或 q = 0 或两者都为零。
To solve x² - 5x + 6 = 0, we factorise the left-hand side as (x - 2)(x - 3) = 0. Setting each factor equal to zero gives x - 2 = 0 or x - 3 = 0, so the solutions are x = 2 and x = 3.
要解 x² - 5x + 6 = 0，我们把左边因式分解为 (x - 2)(x - 3) = 0。令每个因式等于零，得到 x - 2 = 0 或 x - 3 = 0，因此解为 x = 2 和 x = 3。
When the coefficient of x² is not 1, such as in 2x² - 7x + 3 = 0, we can split the middle term: 2x² - 6x - x + 3 = 0, then factor by grouping to get (2x - 1)(x - 3) = 0. This gives x = 1/2 or x = 3.
当 x² 的系数不是 1 时，例如 2x² - 7x + 3 = 0，我们可以拆分中间项：2x² - 6x - x + 3 = 0，然后分组分解得到 (2x - 1)(x - 3) = 0。这样解为 x = 1/2 或 x = 3。
3. Solving by Completing the Square | 配方法求解
Completing the square is a reliable algebraic method that works for any quadratic equation. The idea is to rewrite ax² + bx + c in the form a(x + p)² + q, which can then be solved using square roots.
配方法是一种适用于任何二次方程的可靠代数方法。其思路是把 ax² + bx + c 改写成 a(x + p)² + q 的形式，然后再用平方根来求解。
For the equation x² + 6x + 1 = 0, we first move the constant: x² + 6x = -1. Half of the coefficient of x is 3, and 3² = 9, so we add 9 to both sides: x² + 6x + 9 = 8. This becomes (x + 3)² = 8. Taking square roots gives x + 3 = ±√8, so x = -3 ± 2√2.
对于方程 x² + 6x + 1 = 0，我们先把常数项移到右边：x² + 6x = -1。一次项系数的一半是 3，而 3² = 9，所以我们在两边同时加上 9：x² + 6x + 9 = 8。这变成 (x + 3)² = 8。两边开平方得到 x + 3 = ±√8，所以 x = -3 ± 2√2。
When the coefficient of x² is not 1, factor it out first. For 2x² - 8x + 5 = 0, divide every term by 2 to get x² - 4x + 5/2 = 0, then complete the square: x² - 4x = -5/2, add (-2)² = 4 to both sides, giving (x - 2)² = 3/2. Thus x = 2 ± √3/2.
当 x² 的系数不是 1 时，要先把它提取出来。例如 2x² - 8x + 5 = 0，每项除以 2 得到 x² - 4x + 5/2 = 0，然后配方：x² - 4x = -5/2，两边加上 (-2)² = 4，得到 (x - 2)² = 3/2。因此 x = 2 ± √3/2。
4. The Quadratic Formula | 求根公式
The quadratic formula is the most powerful tool for solving any quadratic equation ax² + bx + c = 0. It is derived from completing the square and is given by the formula below.
求根公式是解任何二次方程 ax² + bx + c = 0 的最有力工具。它由配方法推导而来，公式如下。
x = (-b ± √b² - 4ac) / (2a)
To use the formula, identify a, b and c from the standard form, then substitute them carefully. For 2x² + 3x - 2 = 0, we have a = 2, b = 3 and c = -2. Substitution gives x = (-3 ± √3² - 4 × 2 × -2) / (2 × 2), which simplifies to x = (-3 ± √25) / 4, so x = (-3 ± 5) / 4. The two roots are x = 1/2 and x = -2.
使用公式时，先从标准形式中找出 a、b 和 c，然后仔细代入。对于 2x² + 3x - 2 = 0，我们有 a = 2、b = 3、c = -2。代入后得到 x = (-3 ± √3² - 4 × 2 × -2) / (2 × 2)，化简为 x = (-3 ± √25) / 4，所以 x = (-3 ± 5) / 4。两个根分别是 x = 1/2 和 x = -2。
Always check that the equation is equal to zero before identifying a, b and c. Also, remember to place the entire numerator over 2a, not just the square root part.
在确定 a、b 和 c 之前，一定要先确认方程等于零。另外，要记住整个分子部分都要除以 2a，而不仅仅是平方根部分。
5. The Discriminant and Nature of Roots | 判别式与根的性质
The expression under the square root in the quadratic formula, b² - 4ac, is called the discriminant. It tells us the nature of the roots without solving the equation fully.
求根公式中平方根号下的表达式 b² - 4ac 被称为判别式。它可以在不全解方程的情况下告诉我们根的性质。
If the discriminant is positive, the equation has two distinct real roots. If it is zero, there is exactly one real root (a repeated root). If it is negative, there are no real roots, but two complex roots for Extension level study.
如果判别式为正，方程有两个不同的实根。如果判别式为零，方程恰好有一个实根（重根）。如果判别式为负，方程没有实根，但在扩展层次学习中会有两个复数根。
For example, x² - 6x + 9 = 0 has discriminant (-6)² - 4 × 1 × 9 = 36 - 36 = 0, so it has one repeated root, x = 3. The equation x² + x + 1 = 0 has discriminant 1 - 4 = -3, so it has no real roots.
例如，x² - 6x + 9 = 0 的判别式为 (-6)² - 4 × 1 × 9 = 36 - 36 = 0，所以它有一个重根 x = 3。方程 x² + x + 1 = 0 的判别式为 1 - 4 = -3，所以它没有实根。
6. Sketching Quadratic Graphs | 二次函数图像绘制
The graph of a quadratic function y = ax² + bx + c is a parabola. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.
二次函数 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。
The vertex (turning point) can be found using x = -b / (2a). Substituting this value into the function gives the corresponding y-coordinate. The graph crosses the y-axis at c, since y = c when x = 0. The x-intercepts, if they exist, are the roots of the equation ax² + bx + c = 0.
顶点（转折点）可以用 x = -b / (2a) 来求。把这个值代回函数就可以得到对应的 y 坐标。图像在 c 处与 y 轴相交，因为当 x = 0 时 y = c。如果存在 x 轴截距，它们就是方程 ax² + bx + c = 0 的根。
For y = x² - 4x + 3, the roots are x = 1 and x = 3, the y-intercept is 3, and the vertex is at x = -(-4) / (2 × 1) = 2, giving the point (2, -1).
对于 y = x² - 4x + 3，它的根是 x = 1 和 x = 3，y 轴截距是 3，顶点在 x = -(-4) / (2 × 1) = 2 处，对应点为 (2, -1)。
7. Simultaneous Equations Involving Quadratics | 含二次式的联立方程组
A common Cambridge IGCSE question asks you to solve a linear and a quadratic equation simultaneously. The standard method is substitution: replace one variable in the quadratic equation using the linear equation.
剑桥 IGCSE 常见题型是求解一个一次方程和一个二次方程的联立方程。标准方法是代入法：利用一次方程把二次方程中的一个变量替换掉。
For the system y = x² + 2x - 3 and y = 2x + 1, set the two expressions for y equal: x² + 2x - 3 = 2x + 1. Subtract 2x + 1 from both sides to get x² - 4 = 0. This factorises as (x - 2)(x + 2) = 0, so x = 2 or x = -2. Substitute these into y = 2x + 1 to get y = 5 and y = -3. The solutions are (2, 5) and (-2, -3).
对于方程组 y = x² + 2x - 3 和 y = 2x + 1，令两个 y 的表达式相等：x² + 2x - 3 = 2x + 1。两边同时减去 2x + 1，得到 x² - 4 = 0。分解为 (x - 2)(x + 2) = 0，所以 x = 2 或 x = -2。把这些代入 y = 2x + 1，得到 y = 5 和 y = -3。解为 (2, 5) 和 (-2, -3)。
Always give simultaneous solutions as coordinate pairs, and check both values satisfy both original equations.
联立方程的解一定要写成坐标对，并且要检查两个值是否都满足原方程组中的两个方程。
8. Applications and Word Problems | 实际应用与文字题
Quadratic equations often model real-life situations such as projectile motion, areas of shapes, and product problems. Translating the words into an equation is the first and most important step.
二次方程经常用来模拟实际生活中的情况，例如抛体运动、图形面积和乘积问题。把文字转化为方程是第一步，也是最重要的一步。
A typical area problem: a rectangle has length (x + 3) cm and width x cm. If the area is 40 cm², form the equation x(x + 3) = 40, which expands to x² + 3x - 40 = 0. Factorising gives (x + 8)(x - 5) = 0. Since length cannot be negative, x = 5 cm is the valid solution.
一个典型的面积问题：一个矩形的长为 (x + 3) cm，宽为 x cm。如果面积是 40 cm²，那么可列出方程 x(x + 3) = 40，展开得到 x² + 3x - 40 = 0。因式分解为 (x + 8)(x - 5) = 0。由于长度不能为负数，所以 x = 5 cm 是有效解。
In projectile problems, the height h metres after t seconds might be given by h = -5t² + 20t + 1. To find when the object hits the ground, set h = 0 and solve the resulting quadratic equation.
在抛体问题中，t 秒后的高度 h 米可能由 h = -5t² + 20t + 1 给出。要求物体何时落地，就令 h = 0 并解所得的二次方程。
9. Common Mistakes and Exam Tips | 常见错误与应试技巧
One common mistake is dividing both sides of an equation by x, which can cause a root to be lost. For x² = 4x, this would give only x = 4, but the correct method is x² - 4x = 0, then x(x - 4) = 0, so x = 0 or x = 4.
一个常见错误是方程两边同时除以 x，这会导致丢失一个根。例如对于 x² = 4x，直接除以 x 只会得到 x = 4，但正确方法是 x² - 4x = 0，然后 x(x - 4) = 0，所以 x = 0 或 x = 4。
Another frequent error is mishandling the signs in the quadratic formula, especially when b or c is negative. Always use brackets when substituting negative values to avoid mistakes.
另一个常见错误是在求根公式中弄错符号，尤其是当 b 或 cPublished by TutorHao | Mathematics Revision Series | aleveler.com
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