📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are among the most frequently tested topics in IGCSE Mathematics. They connect algebra, graphs and problem solving, and they appear again at A-Level. A solid understanding of quadratics is essential for any student aiming for a top grade.
二次方程是 IGCSE 数学中考查最频繁的知识点之一。它把代数、图像和应用题联系在一起，在 A-Level 中也会再次出现。扎实掌握二次方程，是任何想拿高分的学生必备的基础。
1. What Is a Quadratic Equation? | 什么是二次方程
are constants and a ≠ 0. The highest power of the variable is 2, which is why it is called a "quadratic" or "second-degree" equation.
二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。变量的最高次数是 2，因此称为“二次”或“二阶”方程。
ax² + bx + c = 0
a is called the leading coefficient, b is the linear coefficient, and c is the constant term.
例如，2x² − 5x + 3 = 0 是二次方程，而 2x − 5 = 0 是一次（线性）方程，2x³ − 5 = 0 是三次方程。系数 a 称为二次项系数，b 称为一次项系数，c 称为常数项。
- A quadratic expression has degree 2. | 二次表达式的最高次数为 2。
- The coefficient a cannot be zero; if a = 0, the equation becomes linear. | 系数 a 不能为零；若 a = 0，方程就变成了线性方程。
- A quadratic equation usually has two solutions, called roots. | 二次方程通常有两个解，称为根。
2. Expanding and Factorising | 展开与因式分解
on of (x + p)(x + q) gives x² + (p + q)x + pq. For example, (x + 3)(x + 4) = x² + 7x + 12.
在解二次方程之前，你必须要熟练展开括号。(x + p)(x + q) 的展开结果是 x² + (p + q)x + pq。例如，(x + 3)(x + 4) = x² + 7x + 12。
(x + 3)(x + 4) = x² + 7x + 12
multiply to 12 and add to 7; those numbers are 3 and 4, so x² + 7x + 12 = (x + 3)(x + 4).
因式分解是展开的逆过程。要把 x² + 7x + 12 因式分解，我们需要找两个数，它们相乘等于 12、相加等于 7；这两个数是 3 和 4，因此 x² + 7x + 12 = (x + 3)(x + 4)。
- x² − 9 = (x − 3)(x + 3) — difference of two squares. | 平方差公式。
- x² + 6x + 9 = (x + 3)² — perfect square. | 完全平方公式。
- 2x² + 5x + 3 = (2x + 3)(x + 1) — factorising with a leading coefficient. | 二次项系数不为 1 时的十字相乘法。
3. Solving by Factorising | 用因式分解法求解
is zero, then at least one of them must be zero. That is, if AB = 0, then A = 0 or B = 0.
最简洁的方法是使用零因子定律：如果两个因式的乘积为零，那么至少其中一个因式为零。也就是说，若 AB = 0，则 A = 0 或 B = 0。
get (x − 2)(x − 3) = 0. By the null factor law, x − 2 = 0 or x − 3 = 0, so x = 2 or x = 3.
例：解 x² − 5x + 6 = 0。因式分解得 (x − 2)(x − 3) = 0。根据零因子定律，x − 2 = 0 或 x − 3 = 0，因此 x = 2 或 x = 3。
- Rearrange the equation into the form ax² + bx + c = 0. | 先把方程整理成 ax² + bx + c = 0 的形式。
- Factorise the left-hand side completely. | 将左边完全因式分解。
- Set each factor equal to zero and solve the resulting linear equations. | 令每个因式分别等于零，并解所得的一次方程。
- Check your solutions by substituting back into the original equation. | 代回原方程检验答案。
(x − 5)(x + 5) = 0, giving x = 5 or x = −5. This is the difference of two squares method.
特例：x² − 25 = 0 可写成 (x − 5)(x + 5) = 0，得 x = 5 或 x = −5。这就是平方差公式法。
4. The Quadratic Formula | 求根公式
When factorising is difficult or impossible, we use the quadratic formula, which works for every quadratic equation. This formula is derived from the method of completing the square and is valid for all real values of a, b and c, provided a ≠ 0.
当因式分解困难甚至无法进行时，我们使用求根公式，它对所有二次方程都适用。该公式由配方法推导而来，只要 a ≠ 0，对任意实数 a、b、c 都成立。
x = (−b ± √b² − 4ac) / 2a
e: Solve 2x² − 4x − 3 = 0. Here a = 2, b = −4 and c = −3. Substituting these values gives:
例：解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。代入得：
x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4
Since √40 = 2√10, the two solutions are x = (4 + 2√10) / 4 and x = (4 − 2√10) / 4. You can simplify these to x = (2 ± √10) / 2 if you wish.
因为 √40 = 2√10，所以两个解为 x = (4 + 2√10) / 4 和 x = (4 − 2√10) / 4。如需化简，可得 x = (2 ± √10) / 2。
- Be careful with negative signs when substituting b. | 代入 b 时注意负号。
- Make sure the equation is in standard form before reading a, b and c. | 读取 a、b、c 之前，确保方程是标准形式。
- The ± symbol gives two solutions: one with + and one with −. | ± 符号给出两个解：一个用 +，一个用 −。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form (x + p)² + q. This is useful not only for solving equations but also for finding the vertex of a parabola.
配方法将二次式改写为 (x + p)² + q 的形式。这不仅可用于解方程，也有助于找到抛物线的顶点。
ce (x + 3)² = x² + 6x + 9, we must subtract 9 to keep the expression unchanged. Therefore:
以 x² + 6x + 2 为例。取 6 的一半是 3，写成 (x + 3)²。因为 (x + 3)² = x² + 6x + 9，所以必须减去 9 才能保持式子不变。因此：
x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7
To solve (x + 3)² − 7 = 0, we write (x + 3)² = 7, then x + 3 = ±√7, so x = −3 ± √7.
要解 (x + 3)² − 7 = 0，先写 (x + 3)² = 7，然后 x + 3 = ±√7，所以 x = −3 ± √7。
- For the general quadratic ax² + bx + c, first divide by a if a ≠ 1. | 对于一般的 ax² + bx + c，若 a ≠ 1，先除以 a。
- Half the coefficient of x, then square it. | 取 x 系数的一半，然后平方。
- Write the final answer in the form a(x + p)² + q. | 最终答案写成 a(x + p)² + q 的形式。
6. The Discriminant | 判别式
sign: Δ = b² − 4ac. It tells us the number and type of roots without solving the equation.
判别式是求根公式中根号内的部分：Δ = b² − 4ac。它无需解方程就能告诉我们根的个数与类型。
Δ = b² − 4ac
| Discriminant Δ | Number of Real Roots | 实数根的个数 |
| Δ > 0 | Two distinct real roots | 两个不相等的实数根 |
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