IGCSE Maths: Quadratic Equations and Applications | IGCSE数学：二次方程及其应用

📚 IGCSE Maths: Quadratic Equations and Applications | IGCSE数学：二次方程及其应用

A quadratic equation is a second‑degree polynomial equation generally written as ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. The word “quadratic” comes from the Latin “quadratus” (square) because the highest power of the variable is 2. Solving quadratic equations is a core skill in IGCSE Mathematics, appearing in algebra, geometry, and real‑world problems such as projectile motion and area optimisation.

二次方程是次幂为二的多项式方程，通常写作 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。“二次”一词源于拉丁语“quadratus”（平方），因为变量的最高次数为 2。解二次方程是 IGCSE 数学中的核心技能，广泛应用于代数、几何以及抛体运动、面积优化等现实问题中。

1. Standard Form and Roots | 1. 标准形式与根

The standard form of a quadratic equation is ax² + bx + c = 0. Here, a is the coefficient of x², b is the coefficient of x, and c is the constant term. A quadratic equation has at most two real roots; these roots may be distinct or equal. The roots are the values of x that satisfy the equation, and they can be found by factorisation, the quadratic formula, completing the square, or using a graph.

二次方程的标准形式是 ax² + bx + c = 0。其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。二次方程最多有两个实数根；这些根可能不同也可能相等。根是满足方程的 x 值，可以通过因式分解、二次公式、配方法或图像来求解。

Why must a ≠ 0? If a = 0, the equation becomes linear (bx + c = 0) and changes its nature. The value of a also determines the shape of the parabola when the quadratic is graphed: a > 0 opens upward (∪), while a < 0 opens downward (∩).

为什么必须 a ≠ 0？如果 a = 0，方程就变成一次方程 (bx + c = 0)，性质随之改变。a 的值还决定了二次函数图像抛物线的形状：a > 0 时开口向上（∪），a < 0 时开口向下（∩）。

2. Solving by Factorisation | 2. 因式分解法解方程

Factorisation is the quickest method when the quadratic expression can be written as a product of two linear factors. The general procedure is:

因式分解法是在二次表达式能写成两个一次因式乘积时最快捷的方法。一般步骤如下：

Write the equation in the standard form ax² + bx + c = 0.

将方程写成标准形式 ax² + bx + c = 0。
Factorise the quadratic expression into two brackets, e.g. (px + q)(rx + s).

将二次表达式分解为两个括号的形式，例如 (px + q)(rx + s)。
Set each bracket equal to zero.

令每个括号等于零。
Solve the resulting linear equations to find the roots.

解由此得到的一次方程，即可求出根。

For example, consider x² + 5x + 6 = 0. We look for two numbers that multiply to 6 and add to 5; these are 2 and 3. Hence (x + 2)(x + 3) = 0, giving x = −2 or x = −3.

例如，考虑 x² + 5x + 6 = 0。我们寻找两个乘积为 6、和为 5 的数，即 2 和 3。因此 (x + 2)(x + 3) = 0，得到 x = −2 或 x = −3。

Special cases include the difference of squares: x² − 9 = (x − 3)(x + 3), and perfect squares: x² + 6x + 9 = (x + 3)². Recognising these patterns speeds up the process considerably.

特殊情况包括平方差：x² − 9 = (x − 3)(x + 3)，以及完全平方式：x² + 6x + 9 = (x + 3)²。识别这些模式能大大加快解题速度。

3. The Quadratic Formula | 3. 二次公式

The quadratic formula solves any quadratic equation. For ax² + bx + c = 0, the solutions are given by:

x = (−b ± √(b² − 4ac)) / (2a)

The symbol ± means that we calculate two values: one with a plus sign and one with a minus sign. This formula is derived from completing the square, and it works even when the roots are irrational or complex.

符号 ± 表示我们要计算两个值：一个用加号，一个用减号。该公式由配方法推导而来，适用于无理根或复数根的情况。

To apply the formula, identify a, b, and c carefully. For example, in 2x² − 4x − 6 = 0, a = 2, b = −4, c = −6. Substituting gives:

x = (4 ± √(16 + 48)) / 4 = (4 ± √64) / 4 = (4 ± 8) / 4

So x = (4 + 8)/4 = 3, or x = (4 − 8)/4 = −1. Always simplify the square root and fraction completely.

因此 x = (4 + 8)/4 = 3，或 x = (4 − 8)/4 = −1。务必彻底化简根号和分数。

4. Completing the Square | 4. 配方法

Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. This is extremely useful for finding the vertex of a parabola and solving equations when factorisation is difficult.

配方法将 ax² + bx + c 重写为 a(x + p)² + q 的形式。这在寻找抛物线顶点以及因式分解困难时求解方程非常有用。

The steps for a monic quadratic (a = 1) are:

对于首一二次方程 (a = 1) 的步骤如下：

Take half of the coefficient of x, i.e. b/2.

取 x 系数的一半，即 b/2。
Square this value and both add and subtract it inside the expression.

将该值的平方同时加减在表达式中。
Rewrite the first three terms as a perfect square.

将前三项改写为完全平方的形式。

For x² + 6x + 5, half of 6 is 3, and 3² = 9. So x² + 6x + 5 = x² + 6x + 9 − 9 + 5 = (x + 3)² − 4. The vertex of the corresponding parabola is (−3, −4).

对于 x² + 6x + 5，6 的一半是 3，3² = 9。所以 x² + 6x + 5 = x² + 6x + 9 − 9 + 5 = (x + 3)² − 4。对应抛物线的顶点为 (−3, −4)。

For non‑monic quadratics (where a ≠ 1), first factor out a from the x² and x terms, then complete the square inside the bracket.

对于非首一二次方程（a ≠ 1），先提出 a 到 x² 和 x 项之前，再在括号内配方。

5. The Discriminant | 5. 判别式

The discriminant, denoted Δ (delta), is the part of the quadratic formula under the square root: Δ = b² − 4ac. It determines the number and type of roots without solving the full equation.

判别式，记作 Δ（德尔塔），是二次公式中根号下的部分：Δ = b² − 4ac。它无需解完整方程即可判断根的个数和类型。

Value of Δ	Nature of Roots \| 根的性质
Δ > 0	Two distinct real roots \| 两个不相等的实数根
Δ = 0	One repeated real root \| 一个二重实数根
Δ < 0	No real roots (two complex roots) \| 无实数根（两个复数根）

For instance, in x² − 3x + 2 = 0, Δ = (−3)² − 4(1)(2) = 9 − 8 = 1 > 0, so there are two distinct real roots. In x² − 2x + 1 = 0, Δ = 4 − 4 = 0, so there is exactly one repeated root.

例如，在 x² − 3x + 2 = 0 中，Δ = (−3)² − 4(1)(2) = 9 − 8 = 1 > 0，因此有两个不相等的实数根。在 x² − 2x + 1 = 0 中，Δ = 4 − 4 = 0，因此正好有一个二重根。

6. Solving by Graphing | 6. 图像法解方程

The graph of a quadratic function y = ax² + bx + c is a parabola. The x‑intercepts of this graph are exactly the real roots of the equation ax² + bx + c = 0. If the graph does not cross the x‑axis, the equation has no real roots.

二次函数 y = ax² + bx + c 的图像是抛物线。该图像与 x 轴的交点正好是方程 ax² + bx + c = 0 的实数根。如果图像不与 x 轴相交，则方程无实数根。

The vertex is the highest or lowest point of the parabola. Its x‑coordinate can be found by x = −b/(2a); substituting this into the function gives the y‑coordinate. The axis of symmetry is the vertical line through the vertex.

顶点是抛物线的最高点或最低点。它的 x 坐标可以由 x = −b/(2a) 求得；将其代入函数可得 y 坐标。对称轴是穿过顶点的竖直线。

Graphing is especially helpful for checking answers or solving quadratic inequalities visually. A quick sketch with the correct intercepts and vertex gives a clear picture of the solution set.

图像法在检查答案或直观解二次不等式时特别有帮助。画出带有正确交点和顶点的草图，就能清晰看到解集。

7. Quadratic Inequalities | 7. 二次不等式

Solving a quadratic inequality such as ax² + bx + c > 0 (or ≥,

解二次不等式，如 ax² + bx + c > 0（或 ≥、

For example, x² − x − 6 > 0 factors as (x − 3)(x + 2) > 0. The critical points are −2 and 3. Testing x = −3 gives (−6)(−1) > 0 (true); x = 0 gives (−3)(2) < 0 (false); x = 4 gives (1)(2) > 0 (true). Hence the solution is x < −2 or x > 3.

例如，x² − x − 6 > 0 可分解为 (x − 3)(x + 2) > 0。临界点为 −2 和 3。检验 x = −3 得 (−6)(−1) > 0（成立）；x = 0 得 (−3)(2) < 0（不成立）；x = 4 得 (1)(2) > 0（成立）。因此解为 x < −2 或 x > 3。

A sign table is a quick and structured way to record these test results, especially in exam settings.

符号表是一种快速且结构化的方式来记录这些检验结果，尤其在考试环境中非常实用。

8. Applications of Quadratic Equations | 8. 二次方程的应用

Quadratic equations model many practical situations. In geometry, area problems often lead to a quadratic. Suppose a rectangle has a width of x metres and a length 3 metres longer; if the area is 20 m², then x(x + 3) = 20, which expands to x² + 3x − 20 = 0.

二次方程可以模拟许多实际情境。在几何学中，面积问题常会导出二次方程。假设一个长方形的宽为 x 米，长比宽多 3 米；若面积为 20 m²，则 x(x + 3) = 20，展开得到 x² + 3x − 20 = 0。

Solving this by the quadratic formula gives x = (−3 ± √(9 + 80)) / 2 = (−3 ± √89) / 2. Since a width cannot be negative, the positive root x = (√89 − 3)/2 ≈ 3.22 metres is the answer.

使用二次公式解这个方程，得到 x = (−3 ± √(9 + 80)) / 2 = (−3 ± √89) / 2。由于宽度不能为负，正根 x = (√89 − 3)/2 ≈ 3.22 米即为答案。

In physics, projectile motion is described by h(t) = −4.9t² + v₀t + h₀, where h is height, t is time, v₀ is initial velocity, and h₀ is initial height. Finding when the object hits the ground means

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IGCSE Maths: Quadratic Equations and Applications | IGCSE数学：二次方程及其应用