📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are one of the most frequently tested topics in IGCSE Mathematics. Whether you are taking the Core or Extended paper, you will meet questions on expanding brackets, factorising, solving equations, and sketching graphs of quadratic functions. This revision guide explains every key skill in a clear, step-by-step way, with worked examples and exam-style advice.
二次方程是 IGCSE 数学中考查频率最高的内容之一。无论你参加 Core 还是 Extended 试卷，你都会遇到展开括号、因式分解、解方程以及绘制二次函数图像的题目。本复习指南以清晰、循序渐进的方式讲解每一个关键技能，并配有例题和贴近考试的解题建议。
1. Standard Form of a Quadratic Equation | 二次方程的标准形式
A quadratic equation is any equation that can be rearranged into the standard form below:
任何能通过移项整理成下面标准形式的方程，都叫做二次方程：
ax² + bx + c = 0, where a ≠ 0
Here a, b and c are constants and x is the variable. The value of a cannot be zero, because if a = 0 the x² term disappears and the equation becomes linear.
这里 a、b、c 是常数，x 是变量。a 的值不能为 0，因为若 a = 0，x² 项就消失了，方程就变成了一次方程。
For example, 3x² − 5x + 2 = 0 is a quadratic equation with a = 3, b = −5 and c = 2. However, 2x − 7 = 0 is linear, not quadratic, because it contains no x² term.
例如，3x² − 5x + 2 = 0 是一个二次方程，其中 a = 3，b = −5，c = 2。而 2x − 7 = 0 是一次方程，不是二次方程，因为它不含 x² 项。
2. Expanding and Simplifying | 展开与化简
Before solving quadratics, you must be confident at expanding products of brackets. The general rule is to multiply every term in the first bracket by every term in the second bracket:
在解二次方程之前，你必须熟练展开括号乘积。通用法则是：用第一个括号中的每一项去乘第二个括号中的每一项：
(a + b)(c + d) = ac + ad + bc + bd
Three special results appear very often in IGCSE papers and are well worth memorising:
三个特殊结论在 IGCSE 试卷中出现频率极高，非常值得牢记：
- Perfect square (完全平方式): (a + b)² = a² + 2ab + b²
- Perfect square (完全平方式): (a − b)² = a² − 2ab + b²
- Difference of two squares (平方差): (a + b)(a − b) = a² − b²
For example, expand (2x − 3)(x + 4). Multiply each term in the first bracket by each term in the second: 2x × x = 2x², 2x × 4 = 8x, (−3) × x = −3x and (−3) × 4 = −12. Collecting like terms gives 2x² + 5x − 12.
例如，展开 (2x − 3)(x + 4)。用第一个括号中的每一项乘第二个括号中的每一项：2x × x = 2x²，2x × 4 = 8x，(−3) × x = −3x，(−3) × 4 = −12。合并同类项后得到 2x² + 5x − 12。
3. Factorising Quadratics | 因式分解二次式
Factorising is the reverse process of expanding. For a quadratic of the form x² + bx + c, we look for two numbers m and n such that m + n = b and m × n = c. Then x² + bx + c = (x + m)(x + n).
因式分解是展开的逆过程。对于形如 x² + bx + c 的二次式，我们需要找到两个数 m 和 n，使得 m + n = b 且 m × n = c。于是 x² + bx + c = (x + m)(x + n)。
Example: factorise x² + 7x + 12. We need two numbers that add to 7 and multiply to 12; they are 3 and 4. Therefore x² + 7x + 12 = (x + 3)(x + 4).
例如：分解因式 x² + 7x + 12。我们需要两个和为 7、积为 12 的数，它们是 3 和 4。因此 x² + 7x + 12 = (x + 3)(x + 4)。
When the coefficient of x² is not 1, such as 2x² + 5x + 3, use the ‘ac’ method: multiply a and c to get 6, then find two numbers that multiply to 6 and add to 5; these are 2 and 3. Split the middle term: 2x² + 2x + 3x + 3, then factor in pairs: 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1).
当 x² 的系数不是 1 时，例如 2x² + 5x + 3，可使用 ‘ac’ 法：将 a 与 c 相乘得 6，然后找两个积为 6、和为 5 的数，即 2 和 3。拆开中间项：2x² + 2x + 3x + 3，再分组提取公因式：2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1)。
4. Solving by Factorisation | 用因式分解法求解
The key principle is that if the product of two factors is zero, then at least one factor must be zero:
关键原理是：若两个因式的乘积为零，则至少有一个因式必须为零：
If p × q = 0, then p = 0 or q = 0
Solve x² − 5x + 6 = 0. First factorise: (x − 2)(x − 3) = 0. Then set each factor to zero: x − 2 = 0 gives x = 2, and x − 3 = 0 gives x = 3. The solution set is x = 2 or x = 3.
解方程 x² − 5x + 6 = 0。首先分解因式：(x − 2)(x − 3) = 0。然后令每个因式为零：x − 2 = 0 得 x = 2，x − 3 = 0 得 x = 3。解集为 x = 2 或 x = 3。
Always check your answers by substituting them back into the original equation. For x = 2: 4 − 10 + 6 = 0, correct. For x = 3: 9 − 15 + 6 = 0, correct.
务必把答案代回原方程检验。当 x = 2 时：4 − 10 + 6 = 0，成立；当 x = 3 时：9 − 15 + 6 = 0，也成立。
5. Solving by Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q. For an expression of the form x² + bx, add and subtract (b/2)²:
配方法将二次式改写为 (x + p)² + q 的形式。对于形如 x² + bx 的表达式，加上并减去 (b/2)²：
x² + bx = (x + b/2)² − (b/2)²
Solve x² + 6x − 7 = 0 by completing the square. Since b = 6, half of b is 3. Write x² + 6x = (x + 3)² − 9. The equation becomes (x + 3)² − 9 − 7 = 0, that is (x + 3)² = 16. Taking square roots gives x + 3 = ±4, so x = 1 or x = −7.
用配方法解 x² + 6x − 7 = 0。因为 b = 6，b 的一半是 3。将 x² + 6x 写成 (x + 3)² − 9。原方程变为 (x + 3)² − 9 − 7 = 0，即 (x + 3)² = 16。两边开平方得 x + 3 = ±4，所以 x = 1 或 x = −7。
This method is especially useful because the completed-square form directly reveals the turning point of the graph, as you will see in Section 9.
这种方法特别有用，因为配方后的形式可以直接揭示图像的顶点坐标，你将在第 9 节中看到这一点。
6. The Quadratic Formula | 求根公式
Every quadratic equation ax² + bx + c = 0 can be solved using the quadratic formula. This formula works even when factorisation is difficult or impossible.
每一个二次方程 ax² + bx + c = 0 都可以用求根公式来解。即使因式分解很困难或无法进行，这个公式也依然有效。
x = (−b ± √(b² − 4ac)) / (2a)
Solve 2x² + 3x − 2 = 0 using the formula. Here a = 2, b = 3 and c = −2. Substitute into the formula:
用求根公式解 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
Therefore x = (−3 + 5) / 4 = 1/2 or x = (−3 − 5) / 4 = −2. The solutions are x = 1/2 and x = −2.
因此 x = (−3 + 5) / 4 = 1/2，或 x = (−3 − 5) / 4 = −2。解为 x = 1/2 和 x = −2。
Note that the ± sign means there are two possible solutions: one using the plus sign and one using the minus sign.
注意 ± 符号表示有两个可能的解：一个取加号，一个取减号。
7. The Discriminant | 判别式
The expression b² − 4ac inside the square root is called the discriminant, often written as Δ. Its sign tells us how many real roots the equation has.
根号里面的表达式 b² − 4ac 叫做判别式，常用 Δ 表示。它的正负号告诉我们方程有多少个实数根。
| Discriminant | Number of real roots | Graph behaviour |
| Δ > 0 | Two distinct real roots | The curve crosses the x-axis twice |
| Δ = 0 | One repeated real root | The curve touches the x-axis once |
| Δ < 0 | No real roots | The curve never touches the x-axis |
For example, the equation 2x² + 3x − 2 = 0 has discriminant 25, which is positive, so it has two distinct real roots, x = 1/2 and x = −2.
例如，方程 2x² + 3x − 2 = 0 的判别式为 25，是正数，因此它有两个不同的实数根，即 x = 1/2 和 x = −2。
If the discriminant is negative, you should state that there are no real roots; do not try to take the square root of a negative number on a non-calculator IGCSE paper.
如果判别式为负，你应该说明方程没有实数根；在 IGCSE 不使用计算器的试卷中，不要在负数下开平方根。
8. Graphs of Quadratic Functions | 二次函数图像
The graph of y = ax² + bx + c is a smooth curve called a parabola. The sign of a controls the shape:
函数 y = ax² + bx + c 的图像是一条平滑曲线，称为抛物线。a 的符号决定其开口方向：
- If a > 0, the parabola opens upwards like a ‘U’ shape and has a minimum turning point.
- 如果 a > 0，抛物线开口向上，像字母 ‘U’，有一个最低点（极小值点）。
- If a < 0, the parabola opens downwards like an 'n' shape and has a maximum turning point.
- 如果 a < 0，抛物线开口向下，像字母 'n'，有一个最高点（极大值点）。
The curve always crosses the y-axis at the point (0, c), because substituting x = 0 leaves y = c. The x-intercepts, if they exist, are the real roots of ax² + bx + c = 0.
抛物线总是与 y 轴相交于点 (0, c)，因为代入 x = 0 后只剩下 y = c。与 x 轴的交点（若存在）就是方程 ax² + bx + c = 0 的实数根。
For example, y = x² − 4x + 3 has a = 1 > 0, so it is a ‘U’ shape. It crosses the y-axis at (0, 3). Solving x² − 4x + 3 = 0 gives (x − 1)(x − 3) = 0, so the x-intercepts are at x = 1 and x = 3.
例如，y = x² − 4x + 3 中 a = 1 > 0，所以图像是 ‘U’ 形。它与 y 轴交于 (0, 3)。解 x² − 4x + 3 = 0 得 (x − 1)(x − 3) = 0，因此与 x 轴的交点是 x = 1 和 x = 3。
9. Roots, Axis of Symmetry and Turning Point | 根、对称轴与顶点
Every parabola is symmetrical about a vertical line called the axis of symmetry. This line passes through the midpoint of the two roots and also through the turning point.
每条抛物线都关于一条竖直线对称，这条线叫做对称轴。对称轴穿过两个根的中点，也穿过顶点。
Axis of symmetry: x = −b / (2a)
The turning point lies on this axis, so its x-coordinate is x = −b/(2a). Substitute this value into y = ax² + bx + c to find the y-coordinate.
顶点就在对称轴上，所以其横坐标为 x = −b/(2a)。把这个值代入 y = ax² + bx + c 即可求得纵坐标。
Alternatively, complete the square to write the function in the form y = (x − p)² + q. The turning point is then (p, q), and because (x − p)² is never negative, the value y = q is the minimum (if the parabola opens upwards).
另一种方法是配方法，把函数写成 y = (x − p)² + q 的形式。此时顶点就是 (p, q)。由于 (x − p)² 永远不小于 0，y = q 就是最小值（当抛物线开口向上时）。
Example: y = x² − 4x + 5 = (x − 2)² + 1. The turning point is (2, 1), and the axis of symmetry is x = 2. The minimum value of y is 1.
例如：y = x² − 4x + 5 = (x − 2)² + 1。顶点是 (2, 1)，对称轴为 x = 2。y 的最小值为 1。
10. Real-World Applications | 实际应用
Quadratic equations model many real-life situations involving area, motion and optimisation. A classic IGCSE problem involves projectile motion: an object is thrown upwards and its height h metres after t seconds is given by h = 20t − 5t².
二次方程可用来建立许多实际情境的模型，例如面积、运动和优化问题。一个经典的 IGCSE 应用题是抛体运动：物体被向上抛出，t 秒后的高度 h 米满足 h = 20t − 5t²。
Question: when does the object hit the ground? The ground corresponds to h = 0. Solve 20t − 5t² = 0. Factorise: 5t(4 − t) = 0, so t = 0 or t = 4. The object starts on the ground at t = 0 and lands again after 4 seconds.
问题：物体何时落地？地面对应 h = 0。解 20t − 5t² = 0。分解因式：5t(4 − t) = 0，所以 t = 0 或 t = 4。物体在 t = 0 时从地面出发，4 秒后再次落地。
Quadratic equations also appear in area problems. For instance, a rectangle has length x + 3 and width x, with area 28 cm². Then x(x + 3) = 28, so x² + 3x − 28 = 0. Factorising gives (x + 7)(x − 4) = 0, so x = 4 (since x = −7 is impossible for a length). The rectangle is 4 cm by 7 cm.
二次方程也常出现在面积问题中。例如，一个长方形长 x + 3、宽 x，面积为 28 cm²。则 x(x + 3) = 28，即 x² + 3x − 28 = 0。因式分解得 (x + 7)(x − 4) = 0，所以 x = 4（因为长度不能为负，舍去 x = −7）。这个长方形是 4 cm × 7 cm。
When a problem produces a negative root that does not fit the context, always reject it and state why.
当问题产生不符合实际情况的负根时，一定要舍去，并说明理由。
11. Exam Tips and Common Mistakes | 考试技巧与常见错误
Here are the most important points to remember in your IGCSE exam:
以下是在 IGCSE 考试中最需要记住的要点：
- Always rearrange the equation into
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