Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations are one of the most frequently tested topics in IGCSE Mathematics. Whether you are solving by factorisation, completing the square, or applying the quadratic formula, a solid understanding of this topic is essential for both Paper 2 and Paper 4. In this article, we will break down every method, highlight common pitfalls, and provide exam-style practice to help you secure full marks.

一元二次方程是IGCSE数学中最常考的知识点之一。无论你是用因式分解、配方法，还是套用求根公式，扎实掌握这一主题对Paper 2和Paper 4都至关重要。本文将逐一拆解每种解法，指出常见陷阱，并提供考试风格的练习题，助你稳稳拿分。

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

A quadratic equation in one variable is any equation that can be written in the general form:

只含一个未知数的一元二次方程，可以写成以下一般形式：

ax² + bx + c = 0

where a, b and c are real numbers, and a ≠ 0. The coefficient a is called the leading coefficient, b is the linear coefficient, and c is the constant term.

其中 a、b、c 为实数，且 a ≠ 0。系数 a 称为二次项系数，b 称为一次项系数，c 称为常数项。

For example, 2x² – 5x + 3 = 0 is a quadratic equation with a = 2, b = -5 and c = 3. However, 3x + 7 = 0 is linear, not quadratic, because it has no x² term.

例如，2x² – 5x + 3 = 0 是一个二次方程，其中 a = 2，b = -5，c = 3。而 3x + 7 = 0 是一次方程而非二次方程，因为它不含 x² 项。

Before solving, always rearrange the equation so that one side equals zero and the terms are written in descending order of degree. This is known as writing the equation in standard form.

解题前，务必先将方程整理为一边等于零的形式，并按次数从高到低排列各项，这称为标准形式。

2. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when the quadratic expression can be factorised easily. The principle is based on the zero product property: if the product of two factors equals zero, then at least one of the factors must be zero.

当二次表达式能够被轻松分解时，因式分解法通常是最快的解法。其原理基于零乘积性质：若两个因式的乘积为零，则至少有一个因式为零。

Step-by-step example: Solve x² + 7x + 12 = 0.

分步示例：解方程 x² + 7x + 12 = 0。

Find two numbers that multiply to give c = 12 and add to give b = 7. These numbers are 3 and 4.

找出两个数，使它们的乘积等于 c = 12，和等于 b = 7。这两个数是 3 和 4。
Factorise: (x + 3)(x + 4) = 0.

因式分解：(x + 3)(x + 4) = 0。
Apply the zero product property: x + 3 = 0 or x + 4 = 0.

应用零乘积性质：x + 3 = 0 或 x + 4 = 0。
Solve each linear equation: x = -3 or x = -4.

解每个一次方程：x = -3 或 x = -4。

For quadratics where the leading coefficient is not 1, such as 2x² + 5x – 3 = 0, you may need the method of splitting the middle term. Multiply a and c (2 × -3 = -6), then find two numbers that multiply to -6 and add to 5. These are 6 and -1. Rewrite the equation as 2x² + 6x – x – 3 = 0, then factor by grouping:

当二次项系数不为 1 时，例如 2x² + 5x – 3 = 0，可以使用十字相乘法。先将 a 与 c 相乘（2 × -3 = -6），再找到两个数，使它们的积为 -6，和为 5。这两个数是 6 和 -1。将方程改写为 2x² + 6x – x – 3 = 0，然后分组分解：

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

Hence x = ½ or x = -3.

因此 x = ½ 或 x = -3。

3. Solving by Completing the Square | 配方法

Completing the square transforms a quadratic equation into the form (x + p)² = q, from which the equation can be solved by taking square roots. This method works for all quadratic equations and is also useful for finding turning points of graphs.

配方法将二次方程转化为 (x + p)² = q 的形式，然后通过开平方求解。这种方法适用于所有二次方程，也常用于求抛物线的顶点坐标。

Example: Solve x² + 6x + 1 = 0.

示例：解方程 x² + 6x + 1 = 0。

Rewrite the equation as x² + 6x = -1.

将方程改写为 x² + 6x = -1。
Take half of the coefficient of x: (6/2) = 3, then square it: 3² = 9.

取 x 系数的一半：(6/2) = 3，再平方：3² = 9。
Add 9 to both sides: x² + 6x + 9 = -1 + 9 = 8.

两边同时加 9：x² + 6x + 9 = -1 + 9 = 8。
Express the left side as a perfect square: (x + 3)² = 8.

将左边写成完全平方形式：(x + 3)² = 8。
Take the square root: x + 3 = ±√8 = ±2√2.

开平方：x + 3 = ±√8 = ±2√2。
Solve: x = -3 + 2√2 or x = -3 – 2√2.

求解：x = -3 + 2√2 或 x = -3 – 2√2。

When the coefficient of x² is not 1, first divide every term by that coefficient before completing the square. For instance, to solve 2x² – 8x + 5 = 0, divide everything by 2 to obtain x² – 4x + 2.5 = 0, then proceed as above.

当 x² 的系数不为 1 时，先将所有项除以该系数，再进行配方。例如，解 2x² – 8x + 5 = 0 时，先将各项除以 2 得到 x² – 4x + 2.5 = 0，然后按上述步骤操作。

4. The Quadratic Formula | 求根公式

The quadratic formula is a universal tool that solves any quadratic equation, even when factorisation is difficult or impossible. The formula is derived from completing the square and is stated as:

求根公式是解任何一元二次方程的通用工具，即使因式分解困难或不可能时也能使用。该公式由配方法推导而来，表述如下：

x = (-b ± √(b² – 4ac)) / (2a)

To use the formula, identify the values of a, b and c from the standard form ax² + bx + c = 0, substitute them into the formula, and simplify.

使用公式时，先从标准形式 ax² + bx + c = 0 中确定 a、b、c 的值，代入公式并化简即可。

Example: Solve 3x² – 5x – 2 = 0 using the quadratic formula.

示例：用求根公式解 3x² – 5x – 2 = 0。

Here a = 3, b = -5, c = -2. Substitute into the formula:

这里 a = 3，b = -5，c = -2。代入公式：

x = (5 ± √((-5)² – 4 × 3 × (-2))) / (2 × 3) = (5 ± √(25 + 24)) / 6 = (5 ± √49) / 6 = (5 ± 7) / 6

Therefore x = (5 + 7) / 6 = 2, or x = (5 – 7) / 6 = -1/3.

因此 x = (5 + 7) / 6 = 2，或 x = (5 – 7) / 6 = -1/3。

Always check that the equation is in standard form before substituting. A common mistake is using the wrong signs for b or c, so write the equation out carefully.

代入前务必确认方程是标准形式。常见错误是弄错 b 或 c 的符号，因此要仔细写出方程。

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

The expression b² – 4ac under the square root sign is called the discriminant, often denoted by Δ. It determines the nature of the roots without actually solving the equation.

根号下的表达式 b² – 4ac 称为判别式，通常记为 Δ。它可以在不解方程的情况下判断根的性质。

Discriminant Δ = b² – 4ac	Nature of Roots	根的性质
Δ > 0	Two distinct real roots	两个不相等的实数根
Δ = 0	One repeated real root	一个重根（两个相等实数根）
Δ < 0	No real roots	无实数根

Example: Determine the nature of the roots of 2x² – 4x + 3 = 0.

示例：判断 2x² – 4x + 3 = 0 的根的性质。

Compute Δ = (-4)² – 4 × 2 × 3 = 16 – 24 = -8. Since Δ < 0, the equation has no real roots.

计算 Δ = (-4)² – 4 × 2 × 3 = 16 – 24 = -8。因为 Δ < 0，所以方程无实数根。

Examiners often ask you to show that a quadratic equation has two distinct real roots by calculating the discriminant and stating its sign. Make sure you simplify the discriminant fully before drawing a conclusion.

考试中常要求你通过计算判别式并说明其符号，来证明一个二次方程有两个不相等的实数根。务必先完整化简判别式，再下结论。

6. Applications in Word Problems | 应用题中的二次方程

Quadratic equations frequently arise in geometry, physics and economics. A typical IGCSE problem involves the area of a rectangle, the motion of a projectile, or the product of two consecutive integers.

二次方程经常出现在几何、物理和经济学问题中。IGCSE 的典型题目涉及矩形面积、抛体运动或两个连续整数的乘积。

Example: The length of a rectangle is 3 cm longer than its width, and its area is 40 cm². Find the dimensions of the rectangle.

示例：一个矩形的长比宽长 3 cm，面积为 40 cm²。求该矩形的尺寸。

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

设宽为 x cm，则长为 (x + 3) cm。面积方程为：

x(x + 3) = 40 → x² + 3x – 40 = 0

Factorise: (x + 8)(x – 5) = 0, so x = -8 or x = 5. Since a length cannot be negative, x = 5. Therefore the rectangle is 5 cm by 8 cm.

因式分解：(x + 8)(x – 5) = 0，所以 x = -8 或 x = 5。由于长度不能为负，x = 5。因此该矩形的尺寸为 5 cm × 8 cm。

When solving word problems, always define your variable clearly, write down the equation, solve it, and then reject any solution that does not make sense in the context of the problem. Finally, state your answer with the correct units.

解应用题时，要明确定义变量、写出方程、求解，然后舍去不符合题意的解，最后带上正确单位写出答案。

7. Graphs of Quadratic Functions | 二次函数的图像

The graph of y = ax² + bx + c is a parabola. Understanding the connection between the equation and its graph helps you solve equations graphically and interpret the discriminant visually.

y = ax² + bx + c 的图像是一条抛物线。理解方程与图像之间的联系，可以帮助你通过图形解方程，并从视觉上理解判别式的含义。

When a > 0, the parabola opens upwards (U-shape); when a < 0, it opens downwards (n-shape).

当 a > 0 时，抛物线开口向上（U 形）；当 a < 0 时，开口向下（n 形）。
The roots of the equation ax² + bx + c = 0 are the x-coordinates where the parabola crosses the x-axis.

方程 ax² + bx + c = 0 的根，就是抛物线与 x 轴交点的横坐标。
If Δ > 0, the parabola crosses the x-axis at two points; if Δ = 0, it touches the x-axis at one point (the vertex); if Δ < 0, it does not intersect the x-axis at all.

若 Δ > 0，抛物线与 x 轴有两个交点；若 Δ = 0，它与 x 轴相切于一点（顶点）；若 Δ < 0，则与 x 轴无交点。
The axis of symmetry is x = -b / (2a), and the vertex has x-coordinate -b / (2a).

对称轴为 x = -b / (2a)，顶点的横坐标为 -b / (2a)。

For example, y = x² – 4x + 3 has roots x = 1 and x = 3, so its graph crosses the x-axis at (1, 0) and (3, 0). The axis of symmetry is x = 2, and the vertex is at (2, -1).

例如，y = x² – 4x + 3 的根为 x = 1 和 x = 3，因此其图像与 x 轴交于 (1, 0) 和 (3, 0)。对称轴为 x = 2，顶点为 (2, -1)。

8. Common Mistakes and Exam Tips | 常见错误与考试技巧

Students frequently lose marks on quadratic equations due to careless algebraic errors or incomplete answers. Here are the most common pitfalls and how to avoid them.

学生在二次方程题上失分，通常是因为粗心的代数错误或答案不完整。以下是最常见的陷阱及规避方法。

Forgetting to rearrange the equation into standard form before factorising or using the formula. Always set one side to zero first.

在因式分解或套用公式前，忘记将方程整理为标准形式。务必先将一边化为零。
Dropping a solution when taking square roots. Remember that x² = k has two solutions: x = √k and x = -√k.

开平方时遗漏一个解。记住 x² = k 有两个解：x = √k 和 x = -√k。
Sign errors when substituting negative values into the quadratic formula. Write brackets around negative numbers, e.g. (-5)² instead of -5².

将负数代入求根公式时出现符号错误。给负数加上括号，例如写 (-5)² 而不是 -5²。
Leaving answers in unsimplified form. Always simplify surds such as √12 = 2√3, and cancel fractions where possible.

答案未化简。务必化简根式，如 √12 = 2√3，并尽可能约分。
Not checking solutions in the original equation, especially for word problems. Substitute your answers back to verify.

未将解代回原方程检验，尤其是在应用题中。将答案代回验证一下。

In exams, show all your working clearly. Even if your final answer is wrong, you can earn method marks for correct steps. Underline your final answers and write them to an appropriate degree of accuracy, usually 3 significant figures unless stated otherwise.

考试中要清晰地写出所有步骤。即使最终答案有误，正确的步骤也能获得方法分。在最终答案下划线，并按题目要求保留精度，通常为 3 位有效数字（除非另有说明）。

9. Practice Questions | 练习题目

Test your understanding with these exam-style questions. Try each one before looking at the hints.

用以下考试风格题目检验你的理解。先独立尝试，再看提示。

Question 1: Solve x² – 9x + 20 = 0 by factorisation.

问题 1：用因式分解法解 x² – 9x + 20 = 0。

Hint: Find two numbers whose product is 20 and sum is 9.

提示：找出乘积为 20、和为 9 的两个数。

Question 2: Solve 2x² + 3x – 5 = 0 using the quadratic formula. Give your answers correct to 2 decimal places.

问题 2：用求根公式解 2x² + 3x – 5 = 0，答案精确到小数点后两位。

Hint: a = 2, b = 3, c = -5. Compute the discriminant first: Δ = 9 + 40 = 49.

提示：a = 2，b = 3，c = -5。先计算判别式：Δ = 9 + 40 = 49。

Question 3: Find the value(s) of k such that the equation x² + kx + 9 = 0 has exactly one real root.

问题 3：求 k 的值，使得方程 x² + kx + 9 = 0 恰有一个实数根。

Hint: Set the discriminant equal to zero: k² – 36 = 0, so k = ±6.

提示：令判别式为零：k² – 36 = 0，因此 k = ±6。

Question 4: A right-angled triangle has a hypotenuse of 13 cm and one leg of length x cm. The other leg is (x – 7) cm. Form a quadratic equation and solve it to find x.

问题 4：一个直角三角形的斜边为 13 cm，一条直角边为 x cm，另一条直角边为 (x – 7) cm。列出一元二次方程并求解 x。

Hint: Use Pythagoras: x² + (x – 7)² = 13². Simplify to 2x² – 14x – 120 = 0, then x² – 7x – 60 = 0, giving x = 12 or x = -5. Reject the negative value, so x = 12.

提示：使用勾股定理：x² + (x – 7)² = 13²。化简得 2x² – 14x – 120 = 0，即 x² – 7x – 60 = 0，解得 x = 12 或 x = -5。舍去负值，所以 x = 12。

If you found any question difficult, return to the corresponding section above and review the worked example. Regular practice is the key to mastering quadratic equations.

如果你觉得某道题困难，请回到上文对应小节复习示例。定期练习是掌握一元二次方程的关键。

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Solving Quadratic Equations | 解一元二次方程