Mastering Quadratic Equations | 掌握二次方程

📚 Mastering Quadratic Equations | 掌握二次方程

Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, graphs, and real-life problems, so a confident understanding of how to form, solve, and interpret them is essential for high marks.

二次方程是 IGCSE 数学中最重要的主题之一。它们出现在代数、图像和实际生活问题中，因此自信地掌握如何建立、求解和解释二次方程对于取得高分至关重要。

1. What is a Quadratic Equation? | 什么是二次方程

A quadratic equation is a polynomial equation of degree 2. This means the highest power of the variable, usually x, is 2. The general shape of its graph is a parabola.

二次方程是一个次数为 2 的多项式方程。这意味着变量（通常是 x）的最高次数是 2。它的图像形状是一条抛物线。

In IGCSE, you will usually see quadratic equations written with real-number coefficients, and you will be expected to solve them by several methods, including factorising, completing the square, and using the quadratic formula.

在 IGCSE 中，你通常会看到带有实数系数的二次方程，并且你需要掌握多种求解方法，包括因式分解法、配方法和二次公式法。

A quadratic equation must always contain an x² term. If the highest power is 1, the equation is linear; if the highest power is 3 or more, it is not quadratic.

二次方程必须始终含有 x² 项。如果最高次数是 1，则方程为一次方程；如果最高次数是 3 或更高，则它不是二次方程。

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

The standard form of a quadratic equation is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.

二次方程的标准形式是 ax² + bx + c = 0，其中 a、b 和 c 是常数，且 a ≠ 0。

a is the coefficient of x²; it must not be zero, otherwise the equation is linear. | a 是 x² 的系数；它不能为零，否则方程就是一次方程。
b is the coefficient of x. | b 是 x 的系数。
c is the constant term. | c 是常数项。

For example, in 3x² − 5x + 2 = 0, we have a = 3, b = −5 and c = 2. In −x² + 4x − 7 = 0, we have a = −1, b = 4 and c = −7.

例如，在 3x² − 5x + 2 = 0 中，a = 3，b = −5，c = 2。在 −x² + 4x − 7 = 0 中，a = −1，b = 4，c = −7。

Always rearrange an equation into standard form before identifying a, b and c. This prevents sign errors in later calculations.

在识别 a、b 和 c 之前，务必先将方程整理为标准形式。这可以防止后续计算中出现符号错误。

3. Solving by Factorising | 因式分解法

Factorising is often the fastest method when the quadratic can be written as a product of two linear factors. The key idea is that if (px + q)(rx + s) = 0, then px + q = 0 or rx + s = 0.

当二次方程可以写成两个一次因式的乘积时，因式分解通常是最快的方法。关键思想是：如果 (px + q)(rx + s) = 0，那么 px + q = 0 或 rx + s = 0。

Worked example: Solve x² − 7x + 12 = 0. We need two numbers that multiply to 12 and add to −7. These are −3 and −4, so the equation becomes (x − 3)(x − 4) = 0. Therefore x = 3 or x = 4.

示例：解方程 x² − 7x + 12 = 0。我们需要两个数，它们相乘得 12，相加得 −7。这两个数是 −3 和 −4，因此方程变为 (x − 3)(x − 4) = 0。所以 x = 3 或 x = 4。

Worked example with a leading coefficient: Solve 2x² + 7x + 3 = 0. Find two numbers that multiply to 2 × 3 = 6 and add to 7. These are 6 and 1. Split the middle term: 2x² + 6x + x + 3 = 0, then factor by grouping: 2x(x + 3) + 1(x + 3) = 0, giving (2x + 1)(x + 3) = 0. The solutions are x = −1/2 and x = −3.

首项系数不为 1 的示例：解方程 2x² + 7x + 3 = 0。找出两个数，它们相乘得 2 × 3 = 6，相加得 7。这两个数是 6 和 1。拆分中间项：2x² + 6x + x + 3 = 0，然后分组分解：2x(x + 3) + 1(x + 3) = 0，得到 (2x + 1)(x + 3) = 0。解为 x = −1/2 和 x = −3。

Always check your factors by expanding them back out. If the original equation has a negative sign or a leading coefficient other than 1, factor carefully.

务必通过展开来检查你的因式。如果原方程含有负号或首项系数不是 1，因式分解时要格外小心。

4. Completing the Square | 配方法

Completing the square is a powerful method because it works for any quadratic equation and also helps to find the vertex of a parabola. The process rewrites ax² + bx + c in the form a(x + p)² + q.

配方法是一种非常强大的方法，因为它适用于任何二次方程，并且有助于找到抛物线的顶点。该过程将 ax² + bx + c 改写为 a(x + p)² + q 的形式。

For x² + 6x + 5 = 0, first move the constant: x² + 6x = −5. Half of 6 is 3, and 3² = 9, so add 9 to both sides: x² + 6x + 9 = 4. This gives (x + 3)² = 4. Taking square roots, x + 3 = ±2, so x = −1 or x = −5.

对于 x² + 6x + 5 = 0，首先移项：x² + 6x = −5。6 的一半是 3，而且 3² = 9，因此两边同时加上 9：x² + 6x + 9 = 4。于是得到 (x + 3)² = 4。开平方得 x + 3 = ±2，所以 x = −1 或 x = −5。

When the coefficient of x² is not 1, divide the whole equation by that coefficient before completing the square, or factor it out carefully. For example, 2x² + 8x + 6 = 0 becomes x² + 4x + 3 = 0 after dividing by 2.

当 x² 的系数不是 1 时，在配方之前先将整个方程除以该系数，或者小心地将其提取出来。例如，2x² + 8x + 6 = 0 除以 2 后变为 x² + 4x + 3 = 0。

The completed square form also shows the minimum or maximum point of the graph, which is very useful for sketching and optimisation questions.

完全平方形式还能显示图像的最小值或最大值点，这对于画草图和优化问题非常有用。

5. The Quadratic Formula | 二次公式

The quadratic formula gives the solutions of ax² + bx + c = 0 directly. It is especially useful when factorising is difficult or when the coefficients are decimals or fractions.

二次公式可以直接给出 ax² + bx + c = 0 的解。当因式分解困难或系数为小数、分数时，它尤其有用。

x = (−b ± √(b² − 4ac)) ÷ 2a

To use the formula, identify a, b and c from the standard form, substitute them into the formula, and simplify carefully. Remember to calculate the discriminant b² − 4ac first because it tells you how many real solutions exist.

要使用该公式，请从标准形式中识别 a、b 和 c，将其代入公式，并仔细化简。请记住先计算判别式 b² − 4ac，因为它能告诉你存在多少个实数解。

Worked example: Solve 3x² − 2x − 8 = 0. Here a = 3, b = −2 and c = −8. The discriminant is (−2)² − 4 × 3 × (−8) = 4 + 96 = 100. Then x = (2 ± √100) ÷ 6 = (2 ± 10) ÷ 6. So x = 12 ÷ 6 = 2 or x = −8 ÷ 6 = −4/3.

示例：解方程 3x² − 2x − 8 = 0。这里 a = 3，b = −2，c = −8。判别式为 (−2)² − 4 × 3 × (−8) = 4 + 96 = 100。然后 x = (2 ± √100) ÷ 6 = (2 ± 10) ÷ 6。所以 x = 12 ÷ 6 = 2 或 x = −8 ÷ 6 = −4/3。

When a question asks for answers correct to 2 decimal places, the quadratic formula is usually the safest method, especially when the roots are irrational.

当题目要求答案精确到 2 位小数时，二次公式通常是最安全的方法，尤其是当根为无理数时。

6. Discriminant and Nature of Roots | 判别式与根的性质

The discriminant D = b² − 4ac determines the nature of the roots without actually solving the equation. This is a common IGCSE exam question.

判别式 D = b² − 4ac 可以在不实际求解方程的情况下判断根的性质。这是 IGCSE 考试中的常见题型。

If D > 0, the equation has two distinct real roots. | 如果 D > 0，方程有两个不同的实数根。
If D = 0, the equation has one repeated real root. | 如果 D = 0，方程有一个重根（实数）。
If D < 0, the equation has no real roots, but two complex roots. | 如果 D < 0，方程没有实数根，但有两个复数根。

For example, in 2x² − 4x + 3 = 0, D = (−4)² − 4 × 2 × 3 = 16 − 24 = −8, so there are no real roots.

例如，在 2x² − 4x + 3 = 0 中，D = (−4)² − 4 × 2 × 3 = 16 − 24 = −8，因此没有实数根。

If a question asks for the value of k that gives equal roots, set D = 0 and solve the resulting equation. For example, in x² + kx + 9 = 0, equal roots occur when k² − 36 = 0, so k = ±6.

如果题目要求求出使方程有两个相等根的 k 值，则令 D = 0 并求解所得方程。例如，在 x² + kx + 9 = 0 中，相等根出现在 k² − 36 = 0 时，因此 k = ±6。

7. Graphical Interpretation | 图像解释

The graph of y = ax² + bx + c is a parabola. If a > 0, the parabola opens upwards; if a < 0, it opens downwards. The solutions of ax² + bx + c = 0 are the x-coordinates where the graph crosses the x-axis.

y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上；如果 a < 0，抛物线开口向下。方程 ax² + bx + c = 0 的解就是图像与 x 轴交点的横坐标。

If the discriminant is positive, the parabola crosses the x-axis twice. If it is zero, the vertex touches the x-axis. If it is negative, the graph does not cross the x-axis at all.

如果判别式为正，抛物线与 x 轴相交两次。如果判别式为零，顶点与 x 轴相切。如果判别式为负，图像完全不与 x 轴相交。

The vertex can be found by completing the square or by using x = −b ÷ 2a. The y-intercept is always c.

顶点可以通过配方法或使用 x = −b ÷ 2a 求得。y 轴截距始终是 c。

Sketching the graph of a quadratic requires the y-intercept, the x-intercepts if they exist, and the vertex. Label all key points clearly.

画二次函数图像需要 y 轴截距、x 轴截距（如果存在）以及顶点。请清楚地标出所有关键点。

8. Word Problems Involving Quadratics | 涉及二次方程的应用题

Many IGCSE problems ask you to form a quadratic equation from a real-life situation, such as area, product of consecutive numbers, or projectile motion.

许多 IGCSE 题目要求你根据实际情境建立二次方程，例如面积、连续数的乘积或抛体运动。

Example: The product of two consecutive positive integers is 72. Let the smaller integer be n. Then n(n + 1) = 72, so n² + n − 72 = 0. Factorising gives (n + 9)(n − 8) = 0, so n = 8 or n = −9. Since n is positive, the integers are 8 and 9.

示例：两个连续正整数的乘积为 72。设较小的整数为 n。则 n(n + 1) = 72，因此 n² + n − 72 = 0。因式分解得 (n + 9)(n − 8) = 0，所以 n = 8 或 n = −9。由于 n 为正数，这两个整数是 8 和 9。

Example: A rectangle has length 2 cm more than its width, and its area is 48 cm². Let the width be x. Then x(x + 2) = 48, giving x² + 2x − 48 = 0. Factorising gives (x + 8)(x − 6) = 0, so x = 6 cm is the width. The length is 8 cm.

示例：一个矩形的长比宽多 2 cm，面积为 48 cm²。设宽为 x。则 x(x + 2) = 48，得到 x² + 2x − 48 = 0。因式分解得 (x + 8)(x − 6) = 0，所以宽为 x = 6 cm。长为 8 cm。

Always define your variable clearly, set up the equation from the conditions given, solve, and check that your answer makes sense in the original context. Negative solutions must be rejected if they represent lengths, times, or counts.

务必清楚地定义变量，根据给定条件建立方程，求解并检查答案在原始情境中

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Mastering Quadratic Equations | 掌握二次方程