Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations appear in almost every IGCSE Mathematics paper, often more than once. In the Extended syllabus you are expected to solve them by factorisation, by completing the square, and with the quadratic formula. This revision guide explains each method, shows worked examples, and points out the exact mistakes that cost marks in real examinations.

一元二次方程几乎出现在每一份 IGCSE 数学试卷中，有时甚至出现多次。在 Extended 大纲中，考生需要掌握因式分解法、配方法和求根公式三种解法。本复习指南逐一讲解每种方法，给出完整例题，并指出真实考试中导致失分的典型错误。

1. What Is a Quadratic Equation | 什么是一元二次方程

A quadratic equation is a polynomial equation of degree 2, which means the highest power of the unknown \(x\) is 2. Its graph is always a parabola, and its standard shape is upward or downward depending on the leading coefficient.

一元二次方程是最高次数为 2 的多项式方程，也就是说未知数 x 的最高次幂为 2。它的图象永远是一条抛物线，其开口方向由首项系数决定。

x² − 5x + 6 = 0 — 标准的一次项与常数项齐全的方程。
2x² + 3x − 2 = 0 — 首项系数不为 1 的例子。
x² = 16 — 只需要移项即可求解的特殊方程。

Notice that every quadratic equation can be rearranged so that one side equals zero. This rearrangement is the first step of almost every solution method.

注意：每一个一元二次方程都可以整理成一边等于零的形式。这种移项整理是几乎所有解法的第一步。

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

The standard form of a quadratic equation is written as follows. The value of \(a\) must never be zero, because if \(a = 0\) the equation becomes linear and is no longer quadratic.

一元二次方程的标准形式如下所示。其中 a 一定不能为零，因为若 a = 0，方程就变成一次方程，不再是二次方程。

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 values of \(b\) and \(c\) can be zero, but the x² term must always be present.

其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。b 和 c 都可以为零，但 x² 这一项必须存在。

Equation	a	b	c
2x² + 3x − 2 = 0	2	3	−2
x² − 5x + 6 = 0	1	−5	6
3x² − 7 = 0	3	0	−7

Always read the signs carefully. A negative \(b\) or \(c\) is the most common source of sign errors in this topic.

务必仔细读出各项的符号。b 或 c 为负数时，是本题型中最常见的符号错误的来源。

3. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic can be written as a product of two linear brackets. It relies on the zero product property: if the product of two expressions is zero, then at least one of them must be zero.

当二次式能够写成两个一次括号相乘时，因式分解法是最快的解法。其依据是“零乘积性质”：若两个式子的乘积为零，则其中至少有一个式子必须为零。

If AB = 0, then A = 0 or B = 0

Follow these four steps to solve by factorisation. They must be completed in this order, and step 1 cannot be skipped.

按下面四个步骤用因式分解法求解。必须按顺序完成，第一步绝不能省略。

Step 1 — Rearrange so the equation equals 0. 第一步：移项使方程一边等于零。
Step 2 — Factor the quadratic into two brackets. 第二步：把二次式分解成两个括号。
Step 3 — Set each bracket equal to 0. 第三步：令每个括号分别等于零。
Step 4 — Solve the two linear equations. 第四步：解出两个一次方程。

Worked example: solve x² − 5x + 6 = 0. We need two numbers that multiply to give 6 and add to give −5. These numbers are −2 and −3, so the factorisation is:

例题：解方程 x² − 5x + 6 = 0。我们需要找两个数，相乘得 6，相加得 −5。这两个数是 −2 和 −3，因此因式分解为：

(x − 2)(x − 3) = 0

Setting each bracket to zero gives x = 2 or x = 3. Always state the solutions in a clear form, because marks are awarded for the final answer line.

令每个括号等于零，得到 x = 2 或 x = 3。务必以清晰的形式写出解集，因为最终答案行是有分的。

4. Special Factorisation Cases | 特殊因式分解情形

Some quadratic equations do not have all three terms, or they fit a special pattern. Recognising these patterns saves time and reduces errors in the examination.

有些一元二次方程并不是三项俱全，或者恰好符合某种特殊模式。识别这些模式可以节省时间并减少考试中的错误。

Case 1: Difference of two squares — 情形一：平方差公式

x² − a² = (x − a)(x + a)

For example, x² − 49 = 0 becomes (x − 7)(x + 7) = 0, so x = 7 or x = −7. Notice there is no middle term at all.

例如，x² − 49 = 0 可化为 (x − 7)(x + 7) = 0，所以 x = 7 或 x = −7。注意这里根本没有一次项。

Case 2: Perfect square trinomial — 情形二：完全平方式

x² ± 2ax + a² = (x ± a)²

For example, x² + 6x + 9 = 0 is (x + 3)² = 0, giving only one solution, x = −3.

例如，x² + 6x + 9 = 0 就是 (x + 3)² = 0，只给出一个解 x = −3。

Case 3: Common factor first — 情形三：先提取公因式

For 2x² + 8x = 0, first take out 2x: 2x(x + 4) = 0. Then x = 0 or x = −4. Do not divide both sides by x, because you would lose the solution x = 0.

对于 2x² + 8x = 0，先提取公因式 2x：2x(x + 4) = 0。于是 x = 0 或 x = −4。千万不要两边同时除以 x，否则会失去 x = 0 这个解。

5. Solving by Completing the Square | 配方法

Completing the square converts a quadratic into the form (x + p)² + q. This method always works, and it also reveals the turning point of the parabola, so it is doubly useful in graph questions.

配方法把二次式化成 (x + p)² + q 的形式。这种方法永远可行，而且还能直接显示抛物线的顶点，因此在图象类题目中格外有用。

For a quadratic of the form x² + bx, add and subtract the square of half of b:

对于形如 x² + bx 的二次式，需要加上并减去“b 的一半”的平方：

x² + bx = (x + b⁄2)² − (b⁄2)²

Example: x² + 6x + 2 = 0 — 例题：x² + 6x + 2 = 0
Rewrite: (x + 3)² − 9 + 2 = 0 — 改写： (x + 3)² − 9 + 2 = 0
Simplify: (x + 3)² − 7 = 0 — 化简： (x + 3)² − 7 = 0
Solve: (x + 3)² = 7, so x = −3 + √7 or x = −3 − √7 — 求解： (x + 3)² = 7，所以 x = −3 + √7 或 x = −3 − √7

The square root step requires the ± sign. If you forget it, you will only find one of the two solutions.

开平方这一步必须带上“±”号。如果漏掉它，就只能求出一个解而不是两个解。

6. The Quadratic Formula | 求根公式

When factorisation is difficult or impossible, the quadratic formula gives the solutions directly. In the IGCSE examination this formula is printed in the formula booklet, but you must know exactly how to substitute into it.

当因式分解困难甚至无法进行时，求根公式可以直接给出解。在 IGCSE 考试中，这条公式印在公式表里，但你必须清楚如何准确代入。

x = (−b ± √(b² − 4ac)) ÷ 2a

Worked example: solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, and c = −2. Substitute these values carefully, paying attention to the negative c.

例题：解方程 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。代入时要格外小心，注意 c 是负数。

x = (−3 ± √(3² − 4 × 2 × (−2))) ÷ (2 × 2)

The discriminant part is 9 + 16 = 25, so the formula gives x = (−3 + 5) ÷ 4 = 0.5 and x = (−3 − 5) ÷ 4 = −2. The solutions are x = 0.5 or x = −2.

判别式部分为 9 + 16 = 25，因此公式给出 x = (−3 + 5) ÷ 4 = 0.5，以及 x = (−3 − 5) ÷ 4 = −2。方程的解为 x = 0.5 或 x = −2。

7. The Discriminant | 判别式

The expression b² − 4ac inside the square root is called the discriminant. Its value tells you the number and type of solutions without solving the equation fully.

根号内的表达式 b² − 4ac 称为判别式。它的值可以在不完全解方程的情况下，直接判断解的个数与类型。

Discriminant	Type of roots	Graph interpretation
b² − 4ac > 0	Two distinct real roots	Parabola crosses the x-axis twice
b² − 4ac = 0	One repeated real root	Parabola just touches the x-axis
b² − 4ac < 0	No real roots	Parabola does not meet the x-axis

Chinese table summary: 判别式大于零时有两个不同实根；等于零时有一个重根；小于零时没有实根。对应的图象分别是穿过、相切、不相交 x 轴。This is a regular Extended-style question, often stated as “show that the equation has no real roots”.

判别式大于零时有两个不同实根；等于零时有一个重根；小于零时没有实根。对应的图象分别是穿过、相切、不相交 x 轴。这是 Extended 常见的题型，例如“证明该方程没有实数根”。

8. Solving Word Problems | 应用题

Examination problems often disguise quadratics inside geometry or number problems. The key is to translate the words into an equation, solve it, and then interpret the answers in the context of the question.

考试题常把二次方程隐藏在几何或数字问题中。关键是先把文字翻译成方程，再求解，最后结合题意解释所得的解。

Worked example: a rectangle has length 4 cm longer than its width, and its area is 60 cm². Let the width be x cm. Then the length is x + 4 cm, so x(x + 4) = 60.

例题：一个长方形的长比宽长 4 cm，面积为 60 cm²。设宽为 x cm，则长为 x + 4 cm，于是 x(x + 4) = 60。

x² + 4x − 60 = 0, so (x + 10)(x − 6) = 0

This gives x = −10 or x = 6. A length cannot be negative, so we reject −10. The width is 6 cm and the length is 10 cm. Always reject invalid roots in word problems and always include units in your final answer.

由此得到 x = −10 或 x = 6。长度不能为负数，因此舍去 −10。宽为 6 cm，长为 10 cm。在应用题中一定要舍去不合理的根，并记得在最终答案中加上单位。

9. Graphs and Roots | 图象与根

The solutions of the equation ax² + bx + c = 0 are exactly the x-coordinates where the parabola y = ax² + bx + c crosses the x-axis. This connection is frequently tested in graph-sketching questions.

方程 ax² + bx + c = 0 的解，正是抛物线 y = ax² + bx + c 与 x 轴交点的横坐标。这一联系在画图类题目中经常被考查。

Once the equation is written in completed square form, the coordinates of the turning point can be read directly:

一旦方程写成完全平方形式，顶点的坐标就可以直接读出：

y = (x − h)² + k has vertex (h, k)

For example, y = x² − 4x + 3 can be written as (x − 2)² − 1. The vertex is (2, −1), and solving (x − 2)² − 1 = 0 gives roots x = 1 and x = 3, which match

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Solving Quadratic Equations | 解一元二次方程