Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

A quadratic equation is one of the most important topics in IGCSE Mathematics. It appears in nearly every exam paper, either as a direct question or as a tool for solving geometry, motion, and financial problems.

二次方程是 IGCSE 数学中最重要的考点之一。它几乎出现在每一份试卷中，要么作为直接考题，要么作为解决几何、运动和金融问题的工具。

In this article, you will learn how to define, expand, factorise, and solve quadratic equations using four powerful methods: factorisation, the quadratic formula, completing the square, and graphical interpretation. You will also master the discriminant and avoid common mistakes.

在本文中，你将学习如何定义、展开、因式分解以及用四种强大的方法解二次方程：因式分解法、二次公式法、配方法和图像法。你还将掌握判别式并避免常见错误。

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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 x is x². The general form is

二次方程是次数为 2 的多项式方程，这意味着变量 x 的最高次数为 x²。它的一般形式为

ax² + bx + c = 0, where a ≠ 0

Here, a is the coefficient of x², b is the coefficient of x, and c is the constant term. The condition a ≠ 0 is essential, because if a = 0 the equation becomes linear, not quadratic.

其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。条件 a ≠ 0 至关重要，因为如果 a = 0，方程就变成了一次方程，而不是二次方程。

The values of x that make the equation true are called the roots or solutions of the equation. A quadratic equation can have two distinct real roots, one repeated real root, or no real roots at all.

使方程成立的 x 值称为方程的根或解。一个二次方程可以有两个不同的实数根、一个重根，或者根本没有实数根。

For example, in the equation 3x² − 2x + 5 = 0, we have a = 3, b = −2 and c = 5. Recognising a, b and c correctly is the first step to applying any solution method.

例如，在方程 3x² − 2x + 5 = 0 中，a = 3，b = −2，c = 5。正确识别 a、b、c 是应用任何解法步骤的第一步。

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2. Expanding Brackets | 展开括号

Before you can factorise or solve quadratics, you must be fluent in expanding brackets. The distributive law is the foundation of everything that follows.

在你能够因式分解或解二次方程之前，必须熟练掌握展开括号。分配律是后续一切内容的基础。

The most useful pattern for IGCSE is the product of two linear expressions:

IGCSE 中最有用的规律是两个一次式的乘积：

(x + p)(x + q) = x² + (p + q)x + pq

To expand (x + p)(x + q), multiply each term of the first bracket by each term of the second bracket. A helpful memory aid is FOIL: First, Outer, Inner, Last.

要展开 (x + p)(x + q)，需将第一个括号中的每一项乘以第二个括号中的每一项。一个有用的记忆法是 FOIL：首项、外项、内项、末项。

For example, (x + 3)(x + 5) = x² + 3x + 5x + 15 = x² + 8x + 15. Notice that the coefficient of x is 3 + 5 = 8 and the constant term is 3 × 5 = 15.

例如，(x + 3)(x + 5) = x² + 3x + 5x + 15 = x² + 8x + 15。注意 x 的系数是 3 + 5 = 8，常数项是 3 × 5 = 15。

When the leading coefficient is not 1, such as (2x + 1)(x − 4), expand carefully: 2x × x = 2x², 2x × (−4) = −8x, 1 × x = x, 1 × (−4) = −4. Combining gives 2x² − 7x − 4.

当最高次项系数不为 1 时，例如 (2x + 1)(x − 4)，需仔细展开：2x × x = 2x²，2x × (−4) = −8x，1 × x = x，1 × (−4) = −4。合并后得到 2x² − 7x − 4。

Always simplify by collecting like terms. Expanding fluently will help you check factorisation results quickly and confidently.

一定要合并同类项来化简。熟练展开有助于你快速且自信地检查因式分解的结果。

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3. Factorising Quadratics | 因式分解二次式

Factorising is the reverse process of expanding. To factorise an expression such as x² + bx + c, you must find two numbers that multiply to give c and add to give b.

因式分解是展开的逆过程。要因式分解像 x² + bx + c 这样的式子，你必须找到两个数，它们的乘积等于 c，和等于 b。

For x² + 7x + 12, look for two numbers whose product is 12 and whose sum is 7. The numbers 3 and 4 work because 3 × 4 = 12 and 3 + 4 = 7.

对于 x² + 7x + 12，寻找两个数，使其乘积为 12，和为 7。数字 3 和 4 满足条件，因为 3 × 4 = 12，且 3 + 4 = 7。

x² + 7x + 12 = (x + 3)(x + 4)

If the constant term is positive, both numbers have the same sign. If it is negative, the numbers have opposite signs. Always check the sign of the middle term.

如果常数项为正，则两个数同号；如果为负，则两个数异号。始终检查中间项的符号。

For example, x² − 5x − 14 = (x − 7)(x + 2), because −7 × 2 = −14 and −7 + 2 = −5. The larger absolute value takes the sign of the middle term.

例如，x² − 5x − 14 = (x − 7)(x + 2)，因为 −7 × 2 = −14，且 −7 + 2 = −5。绝对值较大的数取中间项的符号。

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4. Special Cases of Factorisation | 因式分解的特殊情形

Two special patterns appear frequently in IGCSE exams and can save you valuable time if you recognise them instantly.

两种特殊模式在 IGCSE 考试中频繁出现，如果能够立即识别，将为你节省宝贵的时间。

The first is the difference of two squares. Any expression in the form a² − b² factorises as (a + b)(a − b). There is no middle term because the inner and outer terms cancel.

第一种是平方差公式。任何形如 a² − b² 的式子都可以分解为 (a + b)(a − b)。由于内项与外项相互抵消，所以没有中间项。

x² − 25 = (x + 5)(x − 5)

For example, x² − 25 = x² − 5² = (x + 5)(x − 5). Similarly, 4x² − 9 = (2x)² − 3² = (2x + 3)(2x − 3).

例如，x² − 25 = x² − 5² = (x + 5)(x − 5)。同样，4x² − 9 = (2x)² − 3² = (2x + 3)(2x − 3)。

The second special case is a perfect square trinomial. An expression of the form x² ± 2ax + a² equals (x ± a)². Check whether the middle term is exactly twice the product of the square roots of the first and last terms.

第二种特殊情况是完全平方式。形如 x² ± 2ax + a² 的式子等于 (x ± a)²。检查中间项是否恰好是首项和末项平方根乘积的两倍。

For instance, x² + 10x + 25 = (x + 5)² because 2 × x × 5 = 10x. Recognising perfect squares helps you solve equations quickly by taking square roots.

例如，x² + 10x + 25 = (x + 5)²，因为 2 × x × 5 = 10x。识别完全平方式可以通过开平方快速解方程。

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5. Solving by Factorising | 用因式分解法求解

The most efficient method for many IGCSE quadratics is solving by factorisation. It relies on a simple but powerful rule called the zero-product property.

对于许多 IGCSE 二次方程，最高效的方法是因式分解法。它依赖于一个简单而强大的法则，称为零积性质。

The zero-product property states that if the product of two expressions is zero, then at least one of them must be zero. In symbols, if A × B = 0, then A = 0 or B = 0.

零积性质指出：如果两个表达式的乘积为零，那么其中至少有一个为零。用符号表示，若 A × B = 0，则 A = 0 或 B = 0。

To solve x² − 5x + 6 = 0, first factorise the left-hand side: (x − 2)(x − 3) = 0. Then apply the zero-product property: x − 2 = 0 or x − 3 = 0.

要解 x² − 5x + 6 = 0，首先对左边进行因式分解：(x − 2)(x − 3) = 0。然后应用零积性质：x − 2 = 0 或 x − 3 = 0。

x = 2 or x = 3

Always rearrange the equation so that one side equals zero before factorising. A common error is to factorise an expression that is not set equal to zero.

在因式分解前，务必重新整理方程，使一边等于零。一个常见错误是对没有化成零形式的表达式进行因式分解。

For a quadratic with a leading coefficient greater than 1, such as 2x² + 7x + 3, find two numbers whose product is a × c = 2 × 3 = 6 and whose sum is b = 7. The numbers are 1 and 6.

对于最高次项系数大于 1 的二次式，如 2x² + 7x + 3，需要找到两个数，使它们的乘积为 a × c = 2 × 3 = 6，和为 b = 7。这两个数是 1 和 6。

Split the middle term and factor by grouping

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