Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations appear throughout the IGCSE Mathematics syllabus and are essential for solving problems involving areas, projectile motion, and many real-world scenarios. In this revision guide, we will explore the three main methods for solving quadratic equations: factorising, using the quadratic formula, and completing the square. We will also cover solving equations that need rearrangement and common exam pitfalls.

一元二次方程贯穿IGCSE数学大纲，是解决面积问题、抛体运动以及许多现实情境问题的关键。在本复习指南中，我们将探索解一元二次方程的三种主要方法：因式分解法、公式法和配方法。我们还会涉及需要先整理变形的一元二次方程，以及考试中常见的易错点。

1. The General Form of a Quadratic Equation | 一元二次方程的一般形式

A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0. The highest power of x is 2, which is why it is called a quadratic equation. For example, 2x² − 3x + 5 = 0 is a quadratic equation with a = 2, b = −3 and c = 5.

一元二次方程是形如 ax² + bx + c = 0 的方程，其中a、b、c是常数，且a ≠ 0。x的最高次数是2，因此称为二次方程。例如，2x² − 3x + 5 = 0 是一元二次方程，其中a = 2，b = −3，c = 5。

A quadratic equation can have two real roots, one repeated root, or no real roots. When the left-hand side can be written as a product of two linear factors, factorisation is usually the quickest method.

一元二次方程可以有两个实数根、一个重根，或者没有实数根。当左边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。

2. Solving by Factorising | 因式分解法

The principle of factorising is based on the zero product property: if the product of two expressions is zero, then at least one of the expressions must be zero. For example, if (x − 2)(x + 3) = 0, then x − 2 = 0 or x + 3 = 0, so x = 2 or x = −3.

因式分解法的依据是零积性质：如果两个表达式的乘积为零，那么至少其中一个表达式必须为零。例如，如果 (x − 2)(x + 3) = 0，则 x − 2 = 0 或 x + 3 = 0，所以 x = 2 或 x = −3。

To solve a quadratic equation by factorising, follow these steps:

用因式分解法解一元二次方程，请按以下步骤操作：

First, rearrange the equation so that it is in the form ax² + bx + c = 0.

首先，将方程整理为 ax² + bx + c = 0 的形式。
Then, factorise the quadratic expression into two brackets.

然后，将二次表达式因式分解为两个括号。
Finally, set each bracket equal to zero and solve the resulting linear equations.

最后，令每个括号等于零，并解所得的线性方程。

Example 1: Solve x² − 5x + 6 = 0.

例1：解 x² − 5x + 6 = 0。

We look for two numbers that multiply to give 6 and add to give −5. These are −2 and −3, because (−2) × (−3) = 6 and (−2) + (−3) = −5. Therefore:

我们寻找两个数，它们相乘得6，相加得−5。这两个数是−2和−3，因为(−2) × (−3) = 6，且(−2) + (−3) = −5。因此：

x² − 5x + 6 = (x − 2)(x − 3) = 0

So x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.

所以 x − 2 = 0 或 x − 3 = 0，得 x = 2 或 x = 3。

Example 2: Solve 2x² + 7x + 3 = 0.

例2：解 2x² + 7x + 3 = 0。

Here a ≠ 1, so we look for factors of the form (px + q)(rx + s). Since 2x² must come from 2x × x, we try (2x + 1)(x + 3). Expanding gives 2x² + 6x + x + 3 = 2x² + 7x + 3, which is correct.

这里a ≠ 1，因此我们寻找形如 (px + q)(rx + s) 的因式。由于2x²必须来自2x × x，我们尝试 (2x + 1)(x + 3)。展开得 2x² + 6x + x + 3 = 2x² + 7x + 3，正确。

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

Hence 2x + 1 = 0 or x + 3 = 0, so x = −½ or x = −3.

因此 2x + 1 = 0 或 x + 3 = 0，所以 x = −½ 或 x = −3。

If the coefficient of x² is 1, remember to check whether the constant term c is positive or negative. This will determine the signs inside the brackets.

如果x²的系数为1，请检查常数项c的正负。这将决定括号内的符号。

3. Difference of Two Squares | 平方差公式

A special factorisation is the difference of two squares: a² − b² = (a + b)(a − b). This is particularly useful when the quadratic has no x term.

一种特殊的因式分解是平方差公式：a² − b² = (a + b)(a − b)。当二次方程没有x项时，这一公式特别有用。

Example: Solve x² − 9 = 0.

例：解 x² − 9 = 0。

We write x² − 9 = x² − 3² = (x + 3)(x − 3). Therefore (x + 3)(x − 3) = 0, giving x = −3 or x = 3.

我们将 x² − 9 写为 x² − 3² = (x + 3)(x − 3)。因此 (x + 3)(x − 3) = 0，得 x = −3 或 x = 3。

x² − 9 = (x + 3)(x − 3)

Note that x² + 9 cannot be factorised over real numbers, because it is a sum of squares, not a difference. Similarly, 4x² − 25 = (2x)² − 5² = (2x + 5)(2x − 5).

注意 x² + 9 在实数范围内不能因式分解，因为它是平方和而非平方差。同样地，4x² − 25 = (2x)² − 5² = (2x + 5)(2x − 5)。

4. The Quadratic Formula | 求根公式

When a quadratic expression cannot be factorised easily, we use the quadratic formula. For ax² + bx + c = 0, the solutions are given by:

当二次表达式不易因式分解时，我们使用求根公式。对于 ax² + bx + c = 0，其解为：

x = (−b ± √(b² − 4ac)) / (2a)

This formula works for every quadratic equation, provided b² − 4ac ≥ 0. It is essential to substitute a, b and c carefully, paying attention to negative signs.

该公式适用于所有一元二次方程，前提是 b² − 4ac ≥ 0。代入a、b、c时必须仔细，尤其要注意负号。

Example: Solve x² + 4x − 7 = 0, giving your answers correct to 3 significant figures.

例：解 x² + 4x − 7 = 0，答案保留3位有效数字。

Here a = 1, b = 4, c = −7. Substituting into the formula:

这里a = 1，b = 4，c = −7。代入公式：

x = (−4 ± √(4² − 4 × 1 × (−7))) / (2 × 1)

x = (−4 ± √(16 + 28)) / 2 = (−4 ± √44) / 2

Since √44 = 6.633…, we get x = (−4 + 6.633…) / 2 ≈ 1.32 or x = (−4 − 6.633…) / 2 ≈ −5.32.

由于 √44 = 6.633…，得 x = (−4 + 6.633…) / 2 ≈ 1.32 或 x = (−4 − 6.633…) / 2 ≈ −5.32。

x ≈ 1.32 or x ≈ −5.32

When using a calculator, make sure to use brackets properly when entering the expression. Also, when asked for exact answers, leave the square root in its surd form, such as (−4 ± √44) / 2.

使用计算器时，要确保正确使用括号。如果要求精确值，请将平方根保留为根式形式，如 (−4 ± √44) / 2。

5. The Discriminant b² − 4ac | 判别式 b² − 4ac

The value of b² − 4ac, called the discriminant, tells us how many real roots a quadratic equation has.

b² − 4ac 的值称为判别式，它告诉我们一元二次方程有多少个实数根。

Discriminant 判别式	Number of real roots 实数根个数
b² − 4ac > 0, and is a perfect square	Two rational roots 两个有理根
b² − 4ac > 0, but not a perfect square	Two irrational roots 两个无理根
b² − 4ac = 0	One repeated root 一个重根
b² − 4ac < 0	No real roots 没有实数根

Example: Determine the number of real roots of the equation 3x² − 2x + 5 = 0.

例：判断方程 3x² − 2x + 5 = 0 的实数根个数。

Here a = 3, b = −2, c = 5. The discriminant is (−2)² − 4 × 3 × 5 = 4 − 60 = −56. Since −56 < 0, the equation has no real roots.

这里a = 3，b = −2，c = 5。判别式为 (−2)² − 4 × 3 × 5 = 4 − 60 = −56。由于 −56 < 0，方程没有实数根。

b² − 4ac = −56 < 0

This means the graph of y = 3x² − 2x + 5 does not intersect the x-axis. In exam questions, if you are asked to show that an equation has two distinct real roots, show that the discriminant is positive.

这意味着 y = 3x² − 2x + 5 的图象不与x轴相交。在考试中，如果要求证明方程有两个不同的实数根，只需证明判别式为正。

6. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic expression in the form a(x + p)² + q. This method is useful for solving equations and also for finding the vertex of a parabola.

配方法将二次表达式改写为 a(x + p)² + q 的形式。这一方法可用于解方程，也可用于求抛物线的顶点。

For x² + bx, we add and subtract (b/2)² to create a perfect square. The rule is:

对于 x² + bx，我们加上并减去 (b/2)² 以构造完全平方。规则是：

x² + bx = (x + b/2)² − (b/2)²

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

例：用配方法解 x² + 6x + 2 = 0。

First, move the constant term to the other side:

首先，将常数项移到等号另一边：

x² + 6x = −2

Now add (6/2)² = 9 to both sides:

然后两边同时加上 (6/2)² = 9：

x² + 6x + 9 = −2 + 9

(x + 3)² = 7

Taking the square root of both sides:

两边开平方：

x + 3 = ±√7

So x = −3 + √7 or x = −3 − √7.

因此 x = −3 + √7 或 x = −3 − √7。

If the coefficient of x² is not 1, first divide the whole equation by that coefficient. For example, to complete the square for 2x² − 4x + 1 = 0, divide by 2 to get x² − 2x + ½ = 0.

如果x²的系数不是1，先将整个方程除以该系数。例如，要对 2x² − 4x + 1 = 0 配方，先除以2得 x² − 2x + ½ = 0。

7. Solving Equations that Require Rearrangement | 需先整理变形的方程

Sometimes the quadratic equation is not given in standard form. We must expand brackets, remove fractions, and collect all terms on one side before solving.

有时一元二次方程并非标准形式。我们必须先展开括号、去分母，并将所有项移到同一侧，然后再求解。

Example 1: Solve x(x + 5) = 14.

例1：解 x(x + 5) = 14。

Expand the bracket: x² + 5x = 14. Subtract 14 from both sides: x² + 5x − 14 = 0.

展开括号：x² + 5x = 14。两边减去14：x² + 5x − 14 = 0。

x² + 5x − 14 = 0

Factorise: (x + 7)(x − 2) = 0. Hence x = −7 or x = 2.

因式分解：(x + 7)(x − 2) = 0。因此 x = −7 或 x = 2。

Example 2: Solve (x + 1)² = 3x + 5.

例2：解 (x + 1)² = 3x + 5。

Expand: x² + 2x + 1 = 3x + 5. Bring all terms to the left: x² + 2x + 1 − 3x − 5 = 0, so x² − x − 4 = 0.

展开：x² + 2x + 1 = 3x + 5。将所有项移到左边：x² + 2x + 1 − 3x − 5 = 0，即 x² − x − 4 = 0。

This does not factorise, so we use the quadratic formula with a = 1, b = −1, c = −4:

该式不能因式分解，因此使用求根公式，a = 1，b = −1，c = −4：

x = (1 ± √(1 + 16)) / 2 = (1 ± √17) / 2

Example 3: Solve 3/(x + 1) = x − 1.

例3：解 3/(x + 1) = x − 1。

Multiply both sides by (x + 1), remembering that x ≠ −1:

两边乘以 (x + 1)，注意 x ≠ −1：

3 = (x − 1)(x + 1) = x² − 1

Thus x² − 1 = 3, so x² = 4, giving x = 2 or x = −2. Both values are valid because neither is −1.

因此 x² − 1 = 3，即 x² = 4，得 x = 2 或 x = −2。两个值都成立，因为它们都不等于−1。

8. Word Problems Involving Quadratics | 一元二次方程应用题

Quadratic equations often arise from geometric problems. In such questions, read the problem carefully, define a variable, form a quadratic equation, solve it, and then check whether both solutions make sense in context.

一元二次方程常出现在几何问题中。解此类题时，要仔细读题、设定变量、列出二次方程、求解，然后检查两个解在情境中是否都合理。

Example: A rectangle has length 3 cm longer than its width. Its area is 40 cm². Find the dimensions of the rectangle.

例：一个长方形的长比宽长3 cm，面积为40 cm²。求长方形的尺寸。

Let the width be x cm. Then the length is (x + 3) cm. The area is x(x + 3) = 40.

设宽为 x cm，则长为 (x + 3) cm。面积为 x(x + 3) = 40。

x² + 3x = 40

x² + 3x − 40 = 0

Factorise: (x + 8)(x − 5) = 0. Hence x = −8 or x = 5. Since a length cannot be negative, x = 5. Therefore the width is 5 cm and the length is 8 cm.

因式分解：(x + 8)(x − 5) = 0。因此 x = −8 或 x = 5。由于长度不能为负数，x = 5。所以宽为5 cm，长为8 cm。

Always reject negative or impossible solutions in practical problems. It is also good practice to verify your answer by substituting it back into the original equation.

在实际问题中，务必舍去负数或不可能的根。同时，最好将答案代回原方程进行验证。

9. Graphs of Quadratic Functions | 二次函数图象

The solutions of a quadratic equation ax² + bx + c = 0 correspond to the x-intercepts of the graph y = ax² + bx + c. The graph of a quadratic function is a parabola, which opens upward if a > 0 and downward if a < 0.

一元二次方程 ax² + bx + c = 0 的解对应函数 y = ax² + bx + c 的图象与x轴的交点。二次函数的图象是抛物线，当a > 0时开口向上，当a < 0时开口向下。

The vertex of the parabola can be found using x = −b/(2a). Substituting this value into the equation gives the y-coordinate of the turning point. This is also the axis of symmetry of the parabola.

抛物线的顶点可通过 x = −b/(2a) 求得。将该值代入方程即可得到顶点的y坐标。这也是抛物线的对称轴。

For example, for y = x² − 4x + 3, the axis of symmetry is x = 4/2 = 2. Substituting x = 2 gives y = 4 − 8 + 3 = −1, so the vertex is (2, −1). The x-intercepts are found by solving x² − 4x + 3 = 0, which factorises as (x − 1)(x − 3) = 0, giving x = 1 and x = 3.

例如，对于 y = x² − 4x + 3，对轴称为 x = 4/2 = 2。代入x = 2得 y = 4 − 8 + 3 = −1，因此顶点为 (2, −1)。x轴交点通过解 x² − 4x + 3 = 0 求得，因式分解得 (x − 1)(x − 3) = 0，即x = 1和x = 3。

10. Common Exam Mistakes | 常见考试错误

Many students lose marks on quadratic equations because of small but avoidable errors. Here are the most frequent mistakes and how to avoid them.

许多学生在二次方程题目中失分，是因为一些细小但可以避免的错误。以下是最常见的错误及避免方法。

Forgetting to set the equation equal to zero before factorising. Always rearrange to ax² + bx + c = 0 first.

在因式分解前忘记将方程整理为等于零的形式。务必先整理为 ax² + bx + c = 0。
Making sign errors when substituting negative values into the quadratic formula. Write every substitution carefully.

将负值代入求根公式时出现符号错误。请仔细写出每一步代入过程。
Losing solutions when taking square roots. If x² = k, then x = +√k or x = −√k.

开平方时丢失解。如果 x² = k，则 x = +√k 或 x = −√k。
Dividing both sides of an equation by x, which removes the root x = 0. Instead, factorise by taking x out as a common factor.

将方程两边同时除以x，从而丢失了x = 0这个根。正确做法是提取公因式x。
Forgetting to reject negative solutions in real-world problems.

在实际问题中忘记舍去负数解。

Example of the fourth mistake: Solve x² = 5x. If we divide by x, we get x = 5, missing x = 0. The correct method is:

关于第四个错误的例子：解 x² = 5x。如果两边除以x，得到x = 5，却丢失了x = 0。正确方法是：

x² − 5x = 0

x(x − 5) = 0

Hence x = 0 or x = 5.

因此 x = 0 或 x = 5。

11. Summary | 总结

To solve a quadratic equation, first rearrange it into the form ax² + bx + c = 0. Then choose an appropriate method:

解一元二次方程时，首先要整理为 ax² + bx + c = 0 的形式，然后选择合适的方法：

Method 方法	When to use 适用情况
Factorising	When the expression factorises easily, especially when b and c are small integers
Quadratic formula	When factorising is difficult or impossible, or when exact surd answers are required
Completing the square	When the question explicitly asks for it, or when finding the turning point

Remember to use the discriminant to check the number of roots, and always interpret your answers in the context of the problem.

记住用判别式判断根的个数，并始终结合题目情境解释你的答案。

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Solving Quadratic Equations | 解一元二次方程