Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

A quadratic equation is a polynomial equation of degree two. In this article, we will explore the standard form, three main solution methods, the discriminant, and real-world applications, with clear worked examples for your IGCSE revision.

二次方程是次数为二的多项式方程。本文将探讨其标准形式、三种主要解法、判别式及实际应用，并附清晰例题，助你备考 IGCSE 数学。

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1. The Standard Form | 标准形式

where a, b and c are constants, and a ≠ 0. The coefficient a is the quadratic coefficient, b is the linear coefficient, and c is the constant term.

每个二次方程都可以写成标准形式 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。a 称为二次项系数，b 称为一次项系数，c 称为常数项。

b = -5, c = 3. If a = 0, the equation becomes linear, so the condition a ≠ 0 is essential.

例如，2x² - 5x + 3 = 0 中 a = 2，b = -5，c = 3。若 a = 0，方程就变成一次方程，因此 a ≠ 0 是必要条件。

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2. Solving by Factorisation | 因式分解法

wo linear factors. We then use the Zero Product Property: if p·q = 0, then p = 0 or q = 0.

当二次表达式可以写成两个一次因式的乘积时，可用因式分解法。然后利用零乘积性质：若 p·q = 0，则 p = 0 或 q = 0。

uct is 12 and sum is -7. These are -3 and -4, so (x - 3)(x - 4) = 0. Hence x = 3 or x = 4.

解方程 x² - 7x + 12 = 0。寻找两个数，乘积为 12，和为 -7。这两个数是 -3 和 -4，所以 (x - 3)(x - 4) = 0。因此 x = 3 或 x = 4。

factor by grouping: x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3) = 0, giving x = -½ or x = -3.

当首项系数不为 1 时，可使用“十字相乘法”或“分组分解法”。对于 2x² + 7x + 3 = 0，先将 a 与 c 相乘：2 × 3 = 6。寻找乘积为 6、和为 7 的两个数：1 和 6。拆分中间项：2x² + x + 6x + 3。然后分组提取公因式：x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3) = 0，得 x = -½ 或 x = -3。

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3. The Quadratic Formula | 二次公式

factorisation is difficult or impossible. For ax² + bx + c = 0, the solutions are given by

二次公式可以求解任何二次方程，即使因式分解很困难或不可能。对于 ax² + bx + c = 0，解为

x = (−b ± √b² − 4ac) ÷ (2a)

= 2, b = -4, c = -3. First compute b² - 4ac = 16 - 4(2)(-3) = 16 + 24 = 40. Then x = (4 ± √40) ÷ 4 = (4 ± 2√10) ÷ 4 = (2 ± √10) ÷ 2.

用公式解 2x² - 4x - 3 = 0。这里 a = 2，b = -4，c = -3。先计算 b² - 4ac = 16 - 4(2)(-3) = 16 + 24 = 40。于是 x = (4 ± √40) ÷ 4 = (4 ± 2√10) ÷ 4 = (2 ± √10) ÷ 2。

Always write the formula in your exam paper before substituting values. This shows clear method and helps you avoid arithmetic slips.

考试时务必先写出公式再代入数值。这样能展示解题过程，避免运算失误。

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4. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + p)² + q. This reveals the vertex and helps solve equations without the formula.

配方法将二次式改写为 a(x + p)² + q 的形式。它可以揭示顶点坐标，并能不借助公式求解方程。

es (x + 3)² - 9 - 7 = 0, i.e. (x + 3)² = 16. Therefore x + 3 = ±4, giving x = 1 or x = -7.

对于 x² + 6x - 7 = 0，取 6 的一半再平方：(6/2)² = 9。于是 x² + 6x = (x + 3)² - 9。原方程变为 (x + 3)² - 9 - 7 = 0，即 (x + 3)² = 16。因此 x + 3 = ±4，得 x = 1 或 x = -7。

2)² - 4, so 2(x + 2)² - 8 + 5 = 0, giving 2(x + 2)² = 3. Hence (x + 2)² = 3/2, so x = -2 ± √3/2.

当 x² 系数不为 1 时，先提取出来。对于 2x² + 8x + 5 = 0，写成 2(x² + 4x) + 5。而 x² + 4x = (x + 2)² - 4，所以 2(x + 2)² - 8 + 5 = 0，即 2(x + 2)² = 3。因此 (x + 2)² = 3/2，x = -2 ± √3/2。

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5. The Discriminant | 判别式

The expression b² - 4ac is called the discriminant, usually denoted Δ. It tells us how many real roots a quadratic equation has without solving it.

表达式 b² - 4ac 称为判别式，通常记作 Δ。它无需解方程就能告诉我们二次方程有几个实数根。

Δ = b² - 4ac	Number of real roots	实数根个数
Δ > 0	Two distinct real roots	两个不等实根
Δ = 0	One repeated real root	一个重根
Δ < 0	No real roots	无实数根

ot: x = 2. The equation x² + 2x + 5 = 0 has Δ = 4 - 20 = -16 < 0, so it has no real roots.

例如，x² - 4x + 4 = 0 的 Δ = 16 - 16 = 0，有一个重根 x = 2。方程 x² + 2x + 5 = 0 的 Δ = 4 - 20 = -16 < 0，因此没有实数根。

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6. Sum and Product of Roots | 根的和与积

ct is αβ = c/a. These relationships are useful for checking answers and forming equations.

若 ax² + bx + c = 0 的两根为 α 和 β，则两根之和为 α + β = -b/a，两根之积为 αβ = c/a。这些关系可用于检验答案或构造方程。

t is 2/3. If you solve by formula and get approximate roots 0.423 and 1.577, their sum is indeed 2.

对于方程 3x² - 6x + 2 = 0，两根之和为 -(-6)/3 = 2，两根之积为 2/3。若用公式解得近似根 0.423 和 1.577，其和确实为 2。

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7. Solving by Graphical Methods | 图像法

x² + bx + c crosses the x-axis. This is equivalent to finding the x-intercepts of the graph.

二次方程 ax² + bx + c = 0 的实数根就是抛物线 y = ax² + bx + c 与 x 轴交点的横坐标。这等同于求图像与 x 轴的交点。

ions x = -1 and x = 3. A graph also shows the vertex and the axis of symmetry x = -b/(2a).

例如，若 y = x² - 2x - 3 的图像与 x 轴交于 x = -1 和 x = 3，则方程 x² - 2x - 3 = 0 的解为 x = -1 和 x = 3。图像还能显示顶点以及对称轴 x = -b/(2a)。

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8. Applications in Problem Solving | 实际应用题

Quadratic equations appear in geometry, physics and economics. The key is to translate the problem into an equation, solve it, and then check which solutions make sense in context.

二次方程出现在几何、物理和经济等领域。关键在于将问题转化为方程，求解后还要根据实际背景检验哪个解合理。

x = 5 or x = -8. Since width cannot be negative, the width is 5 cm and the length is 8 cm.

一个长方形的长比宽多 3 cm，面积为 40 cm²。设宽为 x cm，则 x(x + 3) = 40，即 x² + 3x - 40 = 0。因式分解得 (x + 8)(x - 5) = 0，所以 x = 5 或 x = -8。由于宽不能为负，因此宽为 5 cm，长为 8 cm。

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9. Common Mistakes | 常见错误

he equation to the standard form ax² + bx + c = 0 before factorising or using the formula.

常见错误之一是忘记先将方程化为标准形式 ax² + bx + c = 0 再进行因式分解或使用公式。

• - x = 0, so x(x - 1) = 0, giving x = 0 or x = 1. Dividing by x would lose the root x = 0.

  不要随意两边除以 x，因为 x 可能等于 0。例如 x² = x 应化为 x² - x = 0，即 x(x - 1) = 0，得 x = 0 或 x = 1。若两边除以 x 就会丢失根 x = 0。

• ing the quadratic formula, be careful with negative signs. Substitute b = -4 as -4, not 4.

  使用二次公式时，注意负号。例如 b = -4 时应代入 -4，而不是 4。

• Always simplify your answers. For instance, (4 ± √40) ÷ 4 can be simplified by noting √40 = 2√10.

  答案要化简。例如 (4 ± √40) ÷ 4 可利用 √40 = 2√10 化简。

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10. Choosing the Best Method | 选择最佳方法

In an exam, you should choose the most efficient method for each equation. The table below gives a quick guide.

考试中，你应当针对不同方程选择最高效的方法。下表提供快速指南。

Type of equation	Recommended method	推荐方法
Easily factorisable	Factorisation	因式分解
Coefficients are awkward	Quadratic formula	二次公式
Need vertex form	Completing the square	配方法

If the equation has the form x² = k, take the square root directly: x = ±√k. This is the fastest method when there is no x term.

若方程形如 x² = k，可直接开平方：x = ±√k。当没有 x 项时，这是最快的方法。

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11. Worked Exam Question | 考试例题精讲

Solve 3x² - 5x - 2 = 0 by factorisation.

用因式分解解 3x² - 5x - 2 = 0。

2) + 1(x - 2) = (x - 2)(3x + 1) = 0. Hence x - 2 = 0 or 3x + 1 = 0, so x = 2 or x = -1/3.

需要寻找两个数，乘积为 3 × (-2) = -6，和为 -5。这两个数是 -6 和 1。拆分中间项：3x² - 6x + x - 2。分组提取公因式：3x(x - 2) + 1(x - 2) = (x - 2)(3x + 1) = 0。因此 x - 2 = 0 或 3x + 1 = 0，得 x = 2 或 x = -1/3。

x = 2, 3(4) - 10 - 2 = 0. For x = -1/3, 3(1/9) + 5/3 - 2 = 1/3 + 5/3 - 2 = 0. Both check.

检验：当 x = 2 时，3(4) - 10 - 2 = 0。当 x = -1/3 时，3(1/9) + 5/3 - 2 = 1/3 + 5/3 - 2 = 0。均正确。

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12. Practice Questions | 练习自测

Attempt these questions on your own before checking the answers.

先独立完成以下练习，再核对答案。

• Solve x² - 5x - 14 = 0.

  解方程 x² - 5x - 14 = 0。

• Solve 4x² + 4x + 1 = 0 using the discriminant.

  利用判别式解 4x² + 4x + 1 = 0。

• Complete the square for x² + 10x + 20 and solve the equation x² + 10x + 20 = 0.

  用配方法完成 x² + 10x + 20 的配方，并解方程 x² + 10x + 20 = 0。

his equation has one repeated root: x = -1/2. 3. x² + 10x + 20 = (x + 5)² - 5, so x = -5 ± √5.

答案：1. x = 7 或 x = -2。2. 该方程有一个重根：x = -1/2。3. x² + 10x + 20 = (x + 5)² - 5，所以 x = -5 ± √5。

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