Mastering Quadratic Equations and Graphs for IGCSE Mathematics | 掌握 IGCSE 数学二次方程与图像

📚 Mastering Quadratic Equations and Graphs for IGCSE Mathematics | 掌握 IGCSE 数学二次方程与图像

Quadratic equations are a central topic in IGCSE Mathematics, bridging algebra, coordinate geometry and real-world problem solving. A confident grasp of factorising, using the quadratic formula, and interpreting graphs will help you tackle a wide range of exam questions.

二次方程是 IGCSE 数学的核心主题，连接了代数、坐标几何和现实问题解决。熟练掌握因式分解、二次公式以及图像解读，能帮助你应对各种考试题型。

1. What is a Quadratic Expression? | 什么是二次表达式？

A quadratic expression is any algebraic expression of the form ax² + bx + c, where a, b and c are constants and a ≠ 0.

二次表达式是形如 ax² + bx + c 的代数式，其中 a、b、c 为常数且 a ≠ 0。

If a = 0, the expression becomes linear bx + c, so the condition a ≠ 0 is essential. In IGCSE exams, a quadratic expression is usually written with descending powers of x, but it may appear in different orders.

如果 a = 0，表达式就变成一次式 bx + c，因此 a ≠ 0 这个条件至关重要。在 IGCSE 考试中，二次表达式通常按 x 的降幂书写，但也可能以不同顺序出现。

Common examples include x² + 5x + 6, 2x² − 3x + 1 and −x² + 4x. The coefficient a controls the shape and direction of the related parabola.

常见例子包括 x² + 5x + 6、2x² − 3x + 1 和 −x² + 4x。系数 a 控制相应抛物线的形状和开口方向。

2. Expanding and Factorising Quadratics | 二次式的展开与因式分解

Expanding two binomials such as (x + p)(x + q) gives x² + (p+q)x + pq. This is the foundation of many IGCSE algebraic manipulations.

展开两个二项式如 (x + p)(x + q) 得到 x² + (p+q)x + pq。这是许多 IGCSE 代数运算的基础。

(x + p)(x + q) = x² + (p + q)x + pq

Factorising reverses this process: to factorise x² + 7x + 10, find two numbers with sum 7 and product 10, namely 2 and 5. Therefore x² + 7x + 10 = (x + 2)(x + 5).

因式分解是这个过程的逆运算：要因式分解 x² + 7x + 10，需要找到两个数，使它们的和为 7、积为 10，即 2 和 5。因此 x² + 7x + 10 = (x + 2)(x + 5)。

For example, x² − 5x + 6 factorises as (x − 2)(x − 3), because −2 and −3 add to −5 and multiply to 6. When a ≠ 1, use methods such as splitting the middle term or trial and error.

例如，x² − 5x + 6 因式分解为 (x − 2)(x − 3)，因为 −2 和 −3 相加为 −5、相乘为 6。当 a ≠ 1 时，可使用拆分中项或试错法。

3. Solving Quadratic Equations by Factorising | 用因式分解解二次方程

To solve x² + 5x + 6 = 0, first factorise the left-hand side to (x + 2)(x + 3) = 0.

要解 x² + 5x + 6 = 0，先将左边因式分解为 (x + 2)(x + 3) = 0。

Since a product is zero only if at least one factor is zero, set x + 2 = 0 or x + 3 = 0, giving x = −2 or x = −3. Always set the equation to zero first; do not divide by x unless you are certain that x ≠ 0.

因为乘积为零当且仅当至少一个因式为零，所以令 x + 2 = 0 或 x + 3 = 0，得 x = −2 或 x = −3。务必先将方程整理为零；除非确定 x ≠ 0，否则不要直接除以 x。

Example: x² − 7x = 0 factorises as x(x − 7) = 0, giving x = 0 or x = 7. This type of equation appears frequently because students sometimes miss the root x = 0.

例：x² − 7x = 0 因式分解为 x(x − 7) = 0，得 x = 0 或 x = 7。这类方程经常出现，因为学生有时会漏掉 x = 0 这个根。

4. The Quadratic Formula | 二次公式

When a quadratic cannot be factorised easily, use the quadratic formula. It solves any equation of the form ax² + bx + c = 0.

当二次式不易因式分解时，可使用二次公式。该公式可解任何形如 ax² + bx + c = 0 的方程。

x = (−b ± √(b² − 4ac)) ÷ 2a

Example: Solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, c = −2. Substituting gives x = (−3 ± √(3² − 4 × 2 × (−2))) ÷ (2 × 2) = (−3 ± √(9 + 16)) ÷ 4 = (−3 ± 5) ÷ 4.

例：解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。代入得 x = (−3 ± √(3² − 4 × 2 × (−2))) ÷ (2 × 2) = (−3 ± √(9 + 16)) ÷ 4 = (−3 ± 5) ÷ 4。

Therefore x = 2/4 = 1/2 or x = −8/4 = −2. In IGCSE exams, the quadratic formula is especially useful when coefficients are not integers or the expression does not factorise neatly.

因此 x = 2/4 = 1/2 或 x = −8/4 = −2。在 IGCSE 考试中，当系数不是整数或表达式不易分解时，二次公式尤其有用。

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