Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are a cornerstone of IGCSE Mathematics, appearing in algebra, geometry, and real-world applications. This revision guide covers the key methods, properties, and common pitfalls, ensuring you are exam-ready with clear step-by-step explanations.

二次方程是IGCSE数学的基石，出现在代数、几何以及现实应用中。本复习指南涵盖关键方法、性质和常见误区，通过清晰的逐步讲解，确保你为考试做好充分准备。

1. What Are Quadratic Equations | 什么是二次方程

A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is 2. The general form is ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. If a = 0, the equation becomes linear, not quadratic.

二次方程是次数为2的多项式方程，即变量的最高幂次为2。一般形式为 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。若 a = 0，方程变为线性方程，而非二次方程。

For example, 2x² + 3x − 5 = 0 is quadratic, while 3x + 2 = 0 is not. Quadratic equations can have zero, one, or two real solutions, depending on the value of the discriminant.

例如，2x² + 3x − 5 = 0 是二次方程，而 3x + 2 = 0 不是。二次方程可以有零个、一个或两个实数解，具体取决于判别式的值。

2. Standard Form and Key Terms | 标准形式与关键术语

The standard form is ax² + bx + c = 0. Here, ‘a’ is the coefficient of x², ‘b’ is the coefficient of x, and ‘c’ is the constant term. For example, in 3x² − 2x + 1 = 0, a = 3, b = −2, c = 1.

标准形式为 ax² + bx + c = 0。其中，’a’ 是 x² 的系数，’b’ 是 x 的系数，’c’ 是常数项。例如，在 3x² − 2x + 1 = 0 中，a = 3，b = −2，c = 1。

Before solving, always rearrange the equation into standard form. This involves expanding brackets, collecting like terms, and moving everything to one side of the equals sign. For instance, (x + 1)(x − 2) = 4 becomes x² − x − 2 = 4, then x² − x − 6 = 0.

在求解之前，务必先将方程整理为标准形式。这包括展开括号、合并同类项，并将所有项移到等号一侧。例如，(x + 1)(x − 2) = 4 变为 x² − x − 2 = 4，再变为 x² − x − 6 = 0。

3. Solving by Factorisation | 因式分解法

Factorisation is the quickest method when the quadratic can be expressed as a product of two linear factors. For ax² + bx + c = 0, find two numbers that multiply to ac and add to b. Then split the middle term and factor by grouping.

当二次方程能表示为两个线性因式的乘积时，因式分解是最快的方法。对于 ax² + bx + c = 0，找到两个数，其乘积为 ac，和为 b。然后拆分中间项并分组因式分解。

Example: Solve x² + 5x + 6 = 0. Here, a = 1, b = 5, c = 6. Two numbers multiplying to 6 and adding to 5 are 2 and 3. So, (x + 2)(x + 3) = 0.

示例：解 x² + 5x + 6 = 0。这里 a = 1，b = 5，c = 6。乘积为6、和为5的两个数是2和3。因此，(x + 2)(x + 3) = 0。
Then set each factor to zero: x + 2 = 0 → x = −2; x + 3 = 0 → x = −3. The solutions are x = −2 and x = −3.

然后将每个因式设为零：x + 2 = 0 → x = −2；x + 3 = 0 → x = −3。解为 x = −2 和 x = −3。

Always check if the quadratic can be factored simply. If a = 1, look for integer factors of c whose sum is b. If a ≠ 1, use the ac method or trial and error.

始终检查二次式是否能简单因式分解。若 a = 1，寻找 c 的整数因子，其和为 b。若 a ≠ 1，使用 ac 法或试错法。

4. The Quadratic Formula | 二次公式

The quadratic formula solves any quadratic equation, even when factorisation is impossible. Given ax² + bx + c = 0, the solutions are given by:

x = (−b ± √(b² − 4ac)) / (2a)

This formula is derived from completing the square and is essential for your exam formula sheet. Substitute values of a, b, and c carefully, paying attention to signs.

该公式由配方法推导而来，是考试公式表中的必备内容。代入 a、b、c 的值时要小心，注意符号。

Example: Solve 2x² − 4x − 3 = 0. Here a = 2, b = −4, c = −3. Then b² − 4ac = (−4)² − 4×2×(−3) = 16 + 24 = 40.

示例：解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。则 b² − 4ac = (−4)² − 4×2×(−3) = 16 + 24 = 40。
Then x = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2. These are the exact solutions.

然后 x = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2。这些是精确解。

Always simplify the square root if possible, and rationalise the denominator if necessary.

尽可能化简平方根，并在必要时有理化分母。

5. Completing the Square | 配方法

Completing the square rewrites the quadratic as a perfect square plus a constant. The general form is a(x + h)² + k = 0. For x² + bx + c, add and subtract (b/2)² to complete the square.

配方法将二次式改写为完全平方加常数。一般形式为 a(x + h)² + k = 0。对于 x² + bx + c，加法和减法 (b/2)² 以完成配平方。

Steps: Given x² + bx + c = 0, rewrite as (x + b/2)² − (b/2)² + c = 0. Then simplify and solve for x.

步骤：给定 x² + bx + c = 0，改写为 (x + b/2)² − (b/2)² + c = 0。然后化简并求解 x。

Example: Solve x² + 6x + 1 = 0. Here b = 6, so (b/2)² = 9. Rewrite as (x + 3)² − 9 + 1 = 0, i.e., (x + 3)² = 8.

示例：解 x² + 6x + 1 = 0。这里 b = 6，所以 (b/2)² = 9。改写为 (x + 3)² − 9 + 1 = 0，即 (x + 3)² = 8。
Then x + 3 = ±√8, so x = −3 ± 2√2.

然后 x + 3 = ±√8，所以 x = −3 ± 2√2。

This method is also useful for finding the turning point of a quadratic graph and for solving equations that cannot be factored.

此方法也常用于求二次函数图像的顶点，以及解无法因式分解的方程。

6. The Discriminant | 判别式

The discriminant, Δ = b² − 4ac, determines the nature of the roots without solving the equation. It is part of the quadratic formula. The value of Δ tells us how many real roots exist.

判别式 Δ = b² − 4ac 决定根的性质，无需具体求解。它是二次公式的一部分。Δ 的值告诉我们存在多少个实数根。

Δ Value	Nature of Roots
Δ > 0	Two distinct real roots
Δ = 0	One repeated real root (equal roots)
Δ < 0	No real roots (two complex roots)

For example, x² + 2x + 5 = 0 has Δ = 4 − 20 = −16, so no real solutions. This is important for determining whether a graph crosses the x-axis.

例如，x² + 2x + 5 = 0 的 Δ = 4 − 20 = −16，因此无实数解。这对于判断图像是否与 x 轴相交非常重要。

7. Solving Word Problems | 应用题解法

Many IGCSE questions involve translating a real-world situation into a quadratic equation. Key steps: define the variable, set up the equation based on given relationships, solve, and check the answer against the context.

许多IGCSE题目需要将实际问题转化为二次方程。关键步骤：定义变量，根据给定关系建立方程，求解，并对照实际情境检查答案。

Example: A rectangle’s length is 3 cm longer than its width, and its area is 40 cm². Find the width. Let width = x, length = x + 3. Then x(x + 3) = 40, giving x² + 3x − 40 = 0.

示例：长方形的长比宽长3厘米，面积为40平方厘米。求宽。设宽 = x，长 = x + 3。则 x(x + 3) = 40，得到 x² + 3x − 40 = 0。
Factorise: (x + 8)(x − 5) = 0, so x = −8 or x = 5. Since width cannot be negative, the width is 5 cm.

因式分解：(x + 8)(x − 5) = 0，所以 x = −8 或 x = 5。由于宽不能为负，因此宽为5厘米。

Always reject negative or extraneous solutions in geometric contexts, and include appropriate units in your final answer.

在几何情境中始终排除负数或无关解，并在最终答案中包含适当单位。

8. Graphs of Quadratic Functions | 二次函数图像

The graph of y = ax² + bx + c is a parabola. If a > 0, it opens upwards (smile shape); if a < 0, it opens downwards (frown shape). The roots are where the graph crosses the x-axis.

y = ax² + bx + c 的图像是抛物线。若 a > 0，开口向上；若 a < 0，开口向下。根是图像与 x 轴的交点。

The vertex (turning point) has x-coordinate −b/(2a). Substitute this x value into the equation to find the y-coordinate. The y-intercept is c, where the graph crosses the y-axis.

顶点（转折点）的 x 坐标为 −b/(2a)。将此 x 值代入方程求得 y 坐标。y 截距为 c，即图像与 y 轴的交点。

Example: For y = x² − 4x + 3, a = 1, b = −4, so x-vertex = 4/2 = 2. y = 4 − 8 + 3 = −1. Vertex is (2, −1). Roots: (x − 1)(x − 3) = 0, so x = 1 and x = 3.

示例：对于 y = x² − 4x + 3，a = 1，b = −4，所以顶点 x = 4/2 = 2。y = 4 − 8 + 3 = −1。顶点为 (2, −1)。根：(x − 1)(x − 3) = 0，所以 x = 1 和 x = 3。

Sketching graphs requires identifying roots, vertex, and y-intercept. Use the axis of symmetry to aid drawing.

绘制草图需要确定根、顶点和 y 截距。使用对称轴辅助绘图。

9. Sum and Product of Roots | 根的和与积

For a quadratic equation ax² + bx + c = 0, if α and β are the roots, then the sum of roots is α + β = −b/a, and the product is αβ = c/a. This is useful for forming equations with given roots.

对于二次方程 ax² + bx + c = 0，若 α 和 β 是根，则根的和 α + β = −b/a，根的积 αβ = c/a。这可用于构建具有给定根的方程。

For example, if roots are 2 and −5, then sum = −3 and product = −10. The equation is x² − (sum)x + (product) = 0, i.e., x² + 3x − 10 = 0.

例如，若根为 2 和 −5，则和 = −3，积 = −10。方程为 x² − (和)x + (积) = 0，即 x² + 3x − 10 = 0。

This relationship also helps check solutions. If the sum or product does not match, re-evaluate your factorisation or formula substitution.

这种关系也有助于检查解。如果和或积不匹配，请重新评估因式分解或公式代入。

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

Students often forget to set the equation to zero before solving. Always rearrange to ax² + bx + c = 0 first. Another common error is misapplying the sign in the quadratic formula, especially when b is negative.

学生常忘记先将方程设为0。务必先整理为 ax² + bx + c = 0。另一个常见错误是在二次公式中符号运用错误，尤其是当 b 为负数时。

Tip 1: Show all steps clearly. Even if your final answer is wrong, method marks are awarded in exams.

技巧1：清晰展示所有步骤。即使最终答案错误，考试中也会给方法分。
Tip 2: Check your solutions by substituting back into the original equation. This takes seconds and catches errors.

技巧2：将解代回原方程验证。这只需几秒钟，能发现错误。
Tip 3: For factorisation, if the equation has a > 1, consider using the quadratic formula to avoid missing factors.

技巧3：对于因式分解，若 a > 1，考虑使用二次公式以避免遗漏因式。

Practice with past papers, focusing on word problems and the discriminant. Speed and accuracy come from repeated exposure to different question types.

练习历年真题，重点关注应用题和判别式。速度和准确性来自对各类题型的反复接触。

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Solving Quadratic Equations | 解二次方程