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 almost every paper, from factorisation to graphing and problem-solving. Mastering this topic will significantly boost your score.

二次方程是IGCSE数学中最重要的话题之一。它几乎出现在每一份试卷中，从因式分解到作图再到应用题。掌握好这个知识点将显著提高你的分数。

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

A quadratic equation is any equation that can be written in the standard form:

二次方程是指可以写成标准形式的方程：

ax² + bx + c = 0, where a ≠ 0

The letters a, b and c are constants (numbers), and x is the unknown variable. The term ax² makes it quadratic; if a = 0, the equation becomes linear.

字母 a、b 和 c 是常数（数字），x 是未知变量。ax² 项使其成为二次方程；如果 a = 0，方程就变成了一次方程（线性方程）。

Examples: 2x² + 3x − 5 = 0, x² − 4 = 0, and 3x² = 6x all can be rearranged into standard form.
例子：2x² + 3x − 5 = 0、x² − 4 = 0 和 3x² = 6x 都可以整理成标准形式。

Before solving, always rearrange your equation so that one side equals zero. This is the first step in every method.

在求解之前，始终将方程整理成一边等于零的形式。这是所有解法的第一步。

2. Solving by Factorisation | 用因式分解法求解

Factorisation is the fastest method when the quadratic has simple integer factors. We look for two numbers whose product equals ac and whose sum equals b (for ax² + bx + c).

当二次方程有简单的整数因式时，因式分解是最快的方法。我们寻找两个数，使它们的乘积等于 ac，和等于 b（对于 ax² + bx + c）。

Worked Example 1 | 示例 1: Solve x² + 5x + 6 = 0.

Find two numbers multiplying to 6 and adding to 5: these are 2 and 3. Hence (x + 2)(x + 3) = 0. By the zero-product rule, either x + 2 = 0 or x + 3 = 0. Therefore x = −2 or x = −3.

找两个数相乘为6、相加为5：这两个数是2和3。因此 (x + 2)(x + 3) = 0。根据零乘积法则，x + 2 = 0 或 x + 3 = 0。所以 x = −2 或 x = −3。

Worked Example 2 | 示例 2: Solve 2x² + 7x + 3 = 0.

Here ac = 2 × 3 = 6. We need two numbers with product 6 and sum 7: 1 and 6. Split the middle term: 2x² + x + 6x + 3 = 0. Group: x(2x + 1) + 3(2x + 1) = 0, giving (2x + 1)(x + 3) = 0. Hence x = −½ or x = −3.

这里 ac = 2 × 3 = 6。我们需要乘积为6、和为7的两个数：1和6。拆中间项：2x² + x + 6x + 3 = 0。分组：x(2x + 1) + 3(2x + 1) = 0，得到 (2x + 1)(x + 3) = 0。因此 x = −½ 或 x = −3。

Always check your factorisation by expanding the brackets back.
务必通过重新展开括号来检查因式分解是否正确。

3. The Quadratic Formula | 二次公式法

When factorisation is difficult or impossible, use the quadratic formula. It works for every quadratic equation.

当因式分解困难或无法进行时，使用二次公式。它适用于所有二次方程。

x = (−b ± √(b² − 4ac)) / (2a)

This formula gives the solutions directly. The symbol ± means you calculate twice: once with plus, once with minus.

这个公式直接给出解。符号 ± 表示要计算两次：一次用加号，一次用减号。

Worked Example | 示例: Solve 2x² − 4x − 3 = 0.

Here a = 2, b = −4, c = −3. Substitute: x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4. Since √40 = 2√10, we get x = (4 ± 2√10) / 4 = 1 ± (√10)/2. These are the two exact solutions.

这里 a = 2，b = −4，c = −3。代入：x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4。因为 √40 = 2√10，所以 x = (4 ± 2√10) / 4 = 1 ± (√10)/2。这就是两个精确解。

On the calculator paper, you may be asked to give answers correct to 3 significant figures.
在允许使用计算器的试卷上，可能会要求你保留3位有效数字。

4. Completing the Square | 配方法

Completing the square rewrites the quadratic in a perfect-square form. This method is useful for finding turning points and for solving equations without the formula.

配方法将二次方程改写为完全平方形式。这一方法对于求顶点坐标以及在不使用公式的情况下解方程非常有用。

Procedure | 步骤: For x² + bx + c = 0, write:

(x + b/2)² = (b/2)² − c

Worked Example | 示例: Solve x² + 6x + 2 = 0.

Write (x + 3)² − 9 + 2 = 0 → (x + 3)² = 7. Take square roots: x + 3 = ±√7, so x = −3 ± √7.

写成 (x + 3)² − 9 + 2 = 0 → (x + 3)² = 7。两边开平方：x + 3 = ±√7，所以 x = −3 ± √7。

If b is odd, you will get fractions. Keep them as exact values.
如果 b 是奇数，结果会出现分数。请保留精确值。

5. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root: Δ = b² − 4ac. It tells us the nature of the roots without solving.

判别式是二次公式中根号下的部分：Δ = b² − 4ac。它无需解方程就能告诉我们根的性质。

Value of Δ \| Δ的值	Nature of Roots \| 根的性质
Δ > 0	Two distinct real roots \| 两个不同的实数根
Δ = 0	One repeated root \| 一个重根（两个相等的实数根）
Δ < 0	No real roots \| 无实数根

Worked Example | 示例: For 3x² + 2x + 4 = 0, Δ = 2² − 4 × 3 × 4 = 4 − 48 = −44 < 0. Therefore the equation has no real roots.

示例：对于 3x² + 2x + 4 = 0，Δ = 2² − 4 × 3 × 4 = 4 − 48 = −44 < 0。因此该方程没有实数根。

If Δ is a perfect square, the roots are rational and factorisation is possible.
如果 Δ 是完全平方数，则根为有理数，可以使用因式分解。

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

The graph of y = ax² + bx + c is a parabola. If a > 0, it opens upward (a smile shape). If a < 0, it opens downward (a frown shape).

y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，开口向上；如果 a < 0，开口向下。

The y-intercept is always (0, c).
y 轴截距始终是 (0, c)。
The x-intercepts are the roots of the equation ax² + bx + c = 0.
x 轴截距是方程 ax² + bx + c = 0 的根。
The axis of symmetry is the vertical line x = −b/(2a).
对称轴是竖直线 x = −b/(2a)。

Sketching the graph uses these features: the intercepts, the vertex, and the direction of opening. You do not need to plot every point.

画草图时利用这些特征：截距、顶点和开口方向。不需要逐点描图。

7. Finding the Turning Point | 求顶点（极值点）

The turning point (vertex) of a quadratic is the maximum or minimum point on the graph. Use completing the square to find it exactly.

二次函数的顶点（极值点）是图像上的最高点或最低点。使用配方法可以精确求出。

Worked Example | 示例: Find the turning point of y = x² − 4x + 5.

Complete the square: y = (x − 2)² − 4 + 5 = (x − 2)² + 1. Since the bracket squared is never negative, the minimum value of y is 1, occurring when x = 2. Thus the vertex is (2, 1).

配方：y = (x − 2)² − 4 + 5 = (x − 2)² + 1。因为平方项永不为负，所以 y 的最小值为1，此时 x = 2。因此顶点坐标为 (2, 1)。

For y = (x − h)² + k, the vertex is (h, k).
对于 y = (x − h)² + k，顶点为 (h, k)。
If the coefficient of x² is negative, the vertex is the maximum point.
如果 x² 的系数为负，则顶点是最大值点。

8. Word Problems | 应用题

Many exam questions set up a quadratic equation from a real-life context. You will need to translate the words into symbols, solve, and then interpret the answer.

许多考题会从实际情境中建立二次方程。你需要将文字转化为符号，解方程，然后解释答案。

Worked Example | 示例: A rectangle has length (x + 3) cm and width (x − 1) cm. Its area is 15 cm². Find x.

Area equation: (x + 3)(x − 1) = 15. Expand and rearrange: x² + 2x − 3 = 15 → x² + 2x − 18 = 0. Use the formula: x = (−2 ± √(4 + 72)) / 2 = (−2 ± √76) / 2 = (−2 ± 2√19) / 2 = −1 ± √19. Since length cannot be negative, x = −1 + √19 ≈ 3.36 cm.

面积方程：(x + 3)(x − 1) = 15。展开并整理：x² + 2x − 3 = 15 → x² + 2x − 18 = 0。使用公式：x = (−2 ± √(4 + 72)) / 2 = (−2 ± √76) / 2 = (−2 ± 2√19) / 2 = −1 ± √19。由于长度不能为负，x = −1 + √19 ≈ 3.36 cm。

Always reject any solution that does not make sense in the problem context.
务必舍弃在问题情境中不合理的解。

9. Exam Tips and Common Mistakes | 考试技巧与常见错误

Avoid these frequent errors to secure full marks on quadratic questions.

避免以下常见错误，确保在二次方程题目上拿到满分。

Do not forget to rearrange the equation to equal zero before factorising or using the formula.
不要忘记在因式分解或使用公式前将方程整理为等于零的形式。
When using the formula, be careful with negative signs: b² is always positive, but 4ac may be negative.
使用公式时注意符号：b² 始终为正，但 4ac 可能为负。
If the question asks for exact answers, leave √ in surd form; do not round.
如果题目要求精确答案，保留根号形式，不要四舍五入。
Check your solutions by substituting them back into the original equation.
将解代回原方程进行验证。
On graph questions, label the intercepts and the turning point clearly.
在作图题中，清晰标注截距和顶点。

10. Practice Questions | 练习题目

Try these questions on your own, then check the answers below.

先独立完成以下题目，再对照答案。

Q1. Solve x² − 9 = 0.
Q2. Solve x² + 3x − 10 = 0 by factorisation.
Q3. Solve 2x² + 5x − 1 = 0 using the formula, giving answers to 2 decimal places.
Q4. Find the vertex of y = x² + 8x + 7.
Q5. Determine the nature of roots of x² − 4x + 4 = 0.

Answers | 答案:

Q1: x = 3 or x = −3 (difference of two squares). Q2: (x + 5)(x − 2) = 0 → x = −5 or x = 2. Q3: x = (−5 ± √33) / 4 ≈ 0.19 or −2.69. Q4: y = (x + 4)² − 9, so vertex is (−4, −9). Q5: Δ = 16 − 16 = 0, so one repeated root.

Q1：x = 3 或 x = −3（平方差公式）。Q2：(x + 5)(x − 2) = 0 → x = −5 或 x = 2。Q3：x = (−5 ± √33) / 4 ≈ 0.19 或 −2.69。Q4：y = (x + 4)² − 9，顶点为 (−4, −9)。Q5：Δ = 16 − 16 = 0，所以有一个重根。

11. Summary and Revision Checklist | 总结与复习清单

You should now be able to handle any quadratic equation problem in the IGCSE exam. Use this checklist to confirm your readiness.

现在你应该能够应对IGCSE考试中任何二次方程问题。使用这份清单确认你是否已准备就绪。

I can identify the standard form ax² + bx + c = 0.
我能识别标准形式 ax² + bx + c = 0。
I can solve by factorisation quickly and accurately.
我能快速准确地用因式分解法求解。
I can apply the quadratic formula with correct signs and simplification.
我能正确运用二次公式，注意符号和化简。
I can complete the square and interpret the result.
我能完成配方并解释结果。
I can use the discriminant to describe the roots.
我能用判别式描述根的性质。
I can sketch a quadratic graph with key features labelled.
我能画出二次函数草图并标注关键特征。

Review this guide before your exam, practise past paper questions on quadratics, and you will gain confidence. Remember: every quadratic can be solved, but choosing the fastest method saves valuable exam time.

考试前复习本指南，练习历年真题中的二次函数题目，你就会充满信心。记住：每个二次方程都能解出，但选择最快的方法可以节省宝贵的考试时间。

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Value of Δ \| Δ的值	Nature of Roots \| 根的性质
Δ > 0	Two distinct real roots \| 两个不同的实数根
Δ = 0	One repeated root \| 一个重根（两个相等的实数根）
Δ < 0	No real roots \| 无实数根

Solving Quadratic Equations | 解二次方程