Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations appear in nearly every IGCSE Mathematics paper, whether tested directly as a solving exercise or embedded in problem-solving questions. Mastering this topic is essential for achieving top grades.

二次方程几乎出现在每一份 IGCSE 数学试卷中，既会以直接解方程的题型出现，也会融入应用题中考查。掌握这一主题对于冲击高分至关重要。

1. The Standard Form of a Quadratic Equation | 二次方程的标准形式

A quadratic equation is an equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0. The highest power of the variable is 2.

二次方程是指可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。未知数的最高次数为 2。

The coefficient a is the coefficient of x², which must not be zero.
The coefficient b is the coefficient of x.
The term c is the constant term.

系数 a 是 x² 的系数，不能为零。
系数 b 是 x 的系数。
项 c 是常数项。

For example, 2x² – 5x + 3 = 0 is a quadratic equation in standard form. In this case, a = 2, b = -5 and c = 3.

例如，2x² – 5x + 3 = 0 就是一个标准形式的二次方程。此处 a = 2，b = -5，c = 3。

Before solving, always rewrite the equation so that one side equals zero. If the equation is given as 2x² + 1 = 5x, subtract 5x from both sides to obtain 2x² – 5x + 1 = 0.

在求解之前，务必先将方程整理为一边等于零的形式。若方程给出为 2x² + 1 = 5x，则两边同时减去 5x，得到 2x² – 5x + 1 = 0。

2. Solving by Factorisation: When a = 1 | 因式分解法：当 a = 1 时

Factorisation is often the fastest method when the quadratic expression factorises easily. When a = 1, look for two numbers that multiply to give c and add to give b.

当二次表达式容易分解时，因式分解通常是最快的方法。当 a = 1 时，需要找到两个数，使它们的乘积等于 c，和等于 b。

Solve x² + 7x + 12 = 0.

解方程 x² + 7x + 12 = 0。

We need two numbers whose product is 12 and whose sum is 7. The numbers 3 and 4 satisfy both conditions. Therefore:

我们需要找到两个数，其乘积为 12，和为 7。数字 3 和 4 同时满足这两个条件。因此：

x² + 7x + 12 = (x + 3)(x + 4) = 0

Now use the zero product property: if a product of factors equals zero, then at least one factor must be zero. So x + 3 = 0 or x + 4 = 0, giving x = -3 or x = -4.

现在运用零积性质：若几个因子的乘积等于零，则至少有一个因子为零。所以 x + 3 = 0 或 x + 4 = 0，因此 x = -3 或 x = -4。

Always check your answers by substituting them back into the original equation. For x = -3: (-3)² + 7(-3) + 12 = 9 – 21 + 12 = 0, correct.

务必通过代回原方程来检验答案。对 x = -3：(-3)² + 7(-3) + 12 = 9 – 21 + 12 = 0，正确。

3. Factorisation When a ≠ 1 | 当 a ≠ 1 时的因式分解

When a is not equal to 1, the factorisation requires more care. One systematic method is to multiply a by c, then find two numbers that multiply to give ac and add to give b. This method is sometimes called the ‘ac method’ or ‘splitting the middle term’.

当 a 不等于 1 时，分解需要更加细心。一种系统的方法是将 a 乘以 c，然后找到两个数，使它们的乘积等于 ac，和等于 b。这种方法有时被称为”ac 法”或”拆中项法”。

Solve 2x² + 7x + 3 = 0.

解方程 2x² + 7x + 3 = 0。

Here a = 2, b = 7 and c = 3. Multiply a and c: 2 × 3 = 6. We need two numbers whose product is 6 and whose sum is 7. The numbers 1 and 6 work. Now split the middle term:

此处 a = 2，b = 7，c = 3。将 a 和 c 相乘：2 × 3 = 6。我们需要两个数，其乘积为 6，和为 7。数字 1 和 6 符合要求。现在拆分中间项：

2x² + x + 6x + 3 = 0

Group the terms in pairs: (2x² + x) + (6x + 3). Factor each pair separately: x(2x + 1) + 3(2x + 1). Now factor out the common bracket (2x + 1):

将项成对分组：(2x² + x) + (6x + 3)。分别对每对进行因式分解：x(2x + 1) + 3(2x + 1)。然后提取公因式 (2x + 1)：

(2x + 1)(x + 3) = 0

Therefore x = -1/2 or x = -3.

因此 x = -1/2 或 x = -3。

Alternatively, you may use the ‘cross-multiplication’ or ‘trial and error’ approach by testing factor pairs of a and c. Practice both methods and use whichever you find more reliable.

你也可以尝试”十字相乘”法，或通过枚举 a 和 c 的因子对来试错。两种方法都要多加练习，选择你更可靠的那一种。

4. Solving by Taking Square Roots | 开平方法

When a quadratic equation has no x term (b = 0), it can be solved directly by isolating x² and taking square roots. This is often the fastest approach.

当二次方程不含 x 项（即 b = 0）时，可以直接通过将 x² 隔离并开平方来求解。这往往是最快的解法。

Solve 3x² – 27 = 0.

解方程 3x² – 27 = 0。

Add 27 to both sides: 3x² = 27. Divide by 3: x² = 9. Now take the square root of both sides. Remember that there are two solutions: one positive and one negative.

两边同时加 27：3x² = 27。除以 3：x² = 9。然后对两边开平方。记住有两个解：一个正数，一个负数。

x = √9 or x = -√9, so x = 3 or x = -3

The solution can also be written as x = ±3. Omitting the negative root is one of the most common errors in IGCSE exams.

解也可以写作 x = ±3。漏掉负根是 IGCSE 考试中最常见的错误之一。

5. Completing the Square | 配方法

Completing the square is a powerful algebraic technique that transforms a quadratic expression into the form p(x + q)² + r. This method is particularly useful when the equation does not factorise easily.

配方法是一种强大的代数技巧，它能将二次表达式转化为 p(x + q)² + r 的形式。当方程不易因式分解时，这种方法尤其有用。

Solve x² + 6x + 2 = 0 by completing the square.

用配方法解方程 x² + 6x + 2 = 0。

Start with x² + 6x. Take half of 6, which is 3, and square it to get 9. Add and subtract 9 inside the equation:

从 x² + 6x 开始。取 6 的一半，即 3，平方得到 9。在方程中加入并减去 9：

(x + 3)² – 9 + 2 = 0

Simplify: (x + 3)² – 7 = 0. Then add 7 to both sides and take the square root:

化简：(x + 3)² – 7 = 0。然后两边加 7 并开平方：

x + 3 = ±√7

x = -3 ± √7

So the two solutions are x = -3 + √7 and x = -3 – √7. These are exact values, which examiners often prefer over rounded decimals.

因此两个解为 x = -3 + √7 和 x = -3 – √7。这些是精确值，考官通常更倾向于精确值而非四舍五入的小数。

6. The Quadratic Formula | 二次求根公式

The quadratic formula provides a universal method for solving any quadratic equation. For ax² + bx + c = 0, the solutions are given by:

二次求根公式提供了解任意二次方程的通用方法。对于 ax² + bx + c = 0，解由以下公式给出：

x = (-b ± √(b² – 4ac)) / 2a

This formula is provided in the IGCSE formula sheet, so you do not need to memorise it, but you must know how to substitute values correctly and simplify the result.

这个公式在 IGCSE 公式表中会提供，因此无需记忆，但你必须知道如何正确代入数值并化简结果。

Solve 2x² – 4x – 3 = 0 using the quadratic formula.

使用二次求根公式解方程 2x² – 4x – 3 = 0。

Identify a = 2, b = -4 and c = -3. Substitute into the formula:

确定 a = 2，b = -4，c = -3。代入公式：

x = (4 ± √((-4)² – 4 × 2 × (-3))) / (2 × 2)

Simplify step by step. First, b² = 16. Next, -4ac = -4 × 2 × (-3) = 24. So the discriminant b² – 4ac = 16 + 24 = 40.

逐步化简。首先，b² = 16。其次，-4ac = -4 × 2 × (-3) = 24。因此判别式 b² – 4ac = 16 + 24 = 40。

x = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2

So x = (2 + √10)/2 or x = (2 – √10)/2. When the question asks for answers to a certain number of decimal places, substitute √10 ≈ 3.162 to obtain x ≈ 2.581 or x ≈ -0.581.

因此 x = (2 + √10)/2 或 x = (2 – √10)/2。当题目要求答案保留若干位小数时，代入 √10 ≈ 3.162，得到 x ≈ 2.581 或 x ≈ -0.581。

7. The Discriminant and the Nature of Roots | 判别式与根的性质

The expression b² – 4ac is called the discriminant. It determines the number and type of roots without solving the equation completely.

表达式 b² – 4ac 称为判别式。它能在不完全解方程的情况下判断根的个数和类型。

If b² – 4ac > 0, the equation has two distinct real roots.
If b² – 4ac = 0, the equation has exactly one real root (repeated root).
If b² – 4ac < 0, the equation has no real roots (two complex roots).

若 b² – 4ac > 0，方程有两个不相等的实数根。
若 b² – 4ac = 0，方程恰有一个实数根（重根）。
若 b² – 4ac < 0，方程没有实数根（两个复数根）。

For example, in the equation x² + 4x + 5 = 0, the discriminant is 16 – 20 = -4, which is negative. Hence the equation has no real roots, and the graph of y = x² + 4x + 5 never crosses the x-axis.

例如，在方程 x² + 4x + 5 = 0 中，判别式为 16 – 20 = -4，为负数。因此方程没有实数根，且 y = x² + 4x + 5 的图像永远不会与 x 轴相交。

8. Rationalising Surds in Quadratic Solutions | 二次方程解中的分母有理化

When the quadratic formula produces a surd in the numerator, you may need to simplify the surd or rationalise a denominator. This is a common skill tested alongside solving quadratics.

当二次求根公式得出带根号的分子时，你可能需要化简根式或对分母进行有理化。这是与解二次方程相伴的常见考点。

Solve x² + 2x – 1 = 0 using the formula, and simplify your answers.

使用公式法解方程 x² + 2x – 1 = 0，并化简答案。

Here a = 1, b = 2 and c = -1. Substitute:

此处 a = 1，b = 2，c = -1。代入：

x = (-2 ± √(4 + 4)) / 2 = (-2 ± √8) / 2

Since √8 = 2√2, the expression simplifies to:

因为 √8 = 2√2，表达式化简为：

x = (-2 ± 2√2) / 2 = -1 ± √2

So the two exact solutions are x = -1 + √2 and x = -1 – √2. Always cancel common factors, just as we divided every term by 2 above.

因此两个精确解为 x = -1 + √2 和 x = -1 – √2。务必约去公因子，就像上面每一项都除以 2 一样。

9. Quadratic Equations with Fractions | 含分数的二次方程

Sometimes a quadratic equation contains fractions. The most reliable strategy is to multiply through by the common denominator to clear the fractions first, then solve using the usual methods.

有时二次方程中含有分数。最可靠的策略是先乘以公分母以消去分数，再使用常规方法求解。

Solve (x²/2) + (x/3) = 1.

解方程 (x²/2) + (x/3) = 1。

The denominators are 2 and 3, so the common denominator is 6. Multiply every term by 6:

分母为 2 和 3，因此公分母为 6。将每一项乘以 6：

3x² + 2x = 6

Rearrange to standard form: 3x² + 2x – 6 = 0. Since this does not factorise simply, use the quadratic formula with a = 3, b = 2 and c = -6:

移项化为标准形式：3x² + 2x – 6 = 0。由于该式不易因式分解，使用 a = 3，b = 2，c = -6 的二次求根公式：

x = (-2 ± √(4 + 72)) / 6 = (-2 ± √76) / 6 = (-2 ± 2√19) / 6 = (-1 ± √19) / 3

Be especially careful when clearing fractions: every term on both sides of the equation must be multiplied by the common denominator, including constant terms.

消去分数时要格外小心：方程两边每一项都必须乘以公分母，包括常数项。

10. Forming Quadratic Equations from Word Problems | 由应用题建立二次方程

IGCSE often tests your ability to translate a word problem into a quadratic equation and then solve it. Read the question carefully, define a variable, and use the given conditions to construct the equation.

IGCSE 经常考查将文字题转化为二次方程并求解的能力。仔细读题，定义未知数，利用给定条件建立方程。

A rectangle has a length that is 4 cm longer than its width. Its area is 60 cm². Find the dimensions of the rectangle.

一个矩形的长比宽长 4 cm。其面积为 60 cm²。求矩形的尺寸。

Let the width be x cm. Then the length is (x + 4) cm. Since area = length × width:

设宽为 x cm，则长为 (x + 4) cm。由于面积 = 长 × 宽：

x(x + 4) = 60

Expand and rearrange: x² + 4x – 60 = 0. Factorise: (x + 10)(x – 6) = 0. Hence x = -10 or x = 6.

展开并整理：x² + 4x – 60 = 0。因式分解：(x + 10)(x – 6) = 0。因此 x = -10 或 x = 6。

Since a width cannot be negative, reject x = -10. The width is 6 cm and the length is 10 cm. Always check that your final answer makes sense in the context of the question.

因为宽度不可能为负，舍去 x = -10。宽为 6 cm，长为 10 cm。始终检查最终答案是否符合题目的实际语境。

11. Common Mistakes to Avoid | 常见错误提醒

Many IGCSE students lose marks on quadratic equations for avoidable reasons. The following list highlights the most frequent traps.

许多 IGCSE 学生在二次方程题上因可避免的原因丢分。以下列表突出最常见的陷阱。

Forgetting to rearrange the equation into standard form before factorising or using the formula.
Omitting the negative root when taking square roots.
Sign errors when substituting negative values into the quadratic formula.
Incorrectly cancelling fractions, such as dividing only one term by the common factor.
Rejecting a negative solution in word problems without checking whether it is physically impossible.
Forgetting to verify solutions by substituting them back into the original equation.

在因式分解或使用公式之前，忘记将方程整理为标准形式。
开平方时漏掉负根。
将负值代入二次求根公式时出现符号错误。
错误地约分分数，例如只将其中一项除以公因子。
在应用题中未检查负解是否符合实际就轻易舍去。
忘记将解代回原方程进行验证。

12. Exam Strategy and Summary | 考试策略与总结

When you encounter a quadratic equation in the exam, first check whether the equation can be factorised. If it factorises easily, that is usually the quickest route. If not, use the quadratic formula. If the equation has no x term, take square roots directly. If the question asks for exact answers, leave surds in simplified form; if it asks for decimal approximations, round to the specified degree of accuracy.

在考试中遇到二次方程时，先检查方程是否可以因式分解。如果容易分解，这通常是最快的路径。如果不能分解，则使用二次求根公式。如果方程不含 x 项，直接开平方。如果题目要求精确值，则将根式化简保留；如果要求小数近似值，则按指定精度四舍五入。

A solid command of quadratic equations supports many other topics, including graphing parabolas, inequalities and optimisation problems. Practise every method until it becomes automatic, and always double-check your final answers in the context of the question.

熟练掌握二次方程为许多其他主题奠定基础，包括抛物线作图、不等式和优化问题。请将每种方法练到自动化的程度，并始终结合题目语境仔细检查最终答案。

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Solving Quadratic Equations | 解一元二次方程