Solving Quadratic Equations | 一元二次方程的求解

📚 Solving Quadratic Equations | 一元二次方程的求解

Quadratic equations appear throughout the IGCSE Mathematics syllabus. Mastering their solution methods is essential for success in both non-calculator and calculator papers. This revision guide covers the definition, standard forms, solution techniques, and common pitfalls.

一元二次方程贯穿整个 IGCSE 数学大纲。掌握其求解方法对于非计算器试卷和计算器试卷都至关重要。本复习指南将涵盖定义、标准形式、求解技巧以及常见易错点。

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. Its highest power of the unknown variable is two. The general form is written as \( ax^2 + bx + c = 0 \) — but since we do not use LaTeX here, we write it as ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.

一元二次方程是次数为 2 的多项式方程，其未知数的最高次数为二。一般形式写作 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。

If a = 0, the equation becomes linear, not quadratic. So the condition a ≠ 0 is fundamental.

如果 a = 0，方程变为一次方程，而非二次方程。因此 a ≠ 0 这一条件至关重要。

2. The General Form and Standard Form | 一般形式与标准形式

Any quadratic equation can be rearranged into the standard form ax² + bx + c = 0. For example, \( 2x^2 = 3x – 1 \) becomes \( 2x^2 – 3x + 1 = 0 \).

任何一元二次方程都可以整理成标准形式 ax² + bx + c = 0。例如，2x² = 3x – 1 可化为 2x² – 3x + 1 = 0。

When solving, always write the equation in this form first. This avoids sign errors and makes the coefficients easy to read.

求解时，应先将方程写成这种形式。这样可以避免符号错误，并让系数一目了然。

ax² + bx + c = 0

3. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic can be factorised into two linear brackets. For example, solve x² – 5x + 6 = 0.

当二次式可以分解成两个一次因式时，因式分解法是最快捷的方法。例如，解 x² – 5x + 6 = 0。

We look for two numbers that multiply to give +6 and add to give -5. Those numbers are -2 and -3. Hence:

我们寻找两个数，它们相乘等于 +6，相加等于 -5。这两个数是 -2 和 -3。因此：

(x – 2)(x – 3) = 0

If the product of two factors is zero, at least one factor must be zero. So x – 2 = 0 or x – 3 = 0, giving x = 2 or x = 3.

如果两个因式的乘积为零，则至少有一个因式为零。所以 x – 2 = 0 或 x – 3 = 0，得到 x = 2 或 x = 3。

For quadratics where the coefficient a > 1, the process is slightly longer. Solve 2x² + 5x – 3 = 0 by splitting the middle term.

当二次项系数 a > 1 时，过程稍长。用拆中项法解 2x² + 5x – 3 = 0。

Multiply a and c: 2 × (-3) = -6.
Find factors of -6 that add to +5: -1 and +6.
Rewrite the middle term: 2x² – x + 6x – 3.
Factor by grouping: x(2x – 1) + 3(2x – 1).
Final factorisation: (2x – 1)(x + 3) = 0.

将 a 和 c 相乘：2 × (-3) = -6。
找到 -6 的两个因数，使其和等于 +5：-1 和 +6。
重写中间项：2x² – x + 6x – 3。
分组分解：x(2x – 1) + 3(2x – 1)。
最终分解：(2x – 1)(x + 3) = 0。

Therefore x = ½ or x = -3.

因此 x = ½ 或 x = -3。

4. The Quadratic Formula | 求根公式

The quadratic formula works for any quadratic equation, even when factorisation is difficult or impossible. The formula is:

求根公式适用于任何一元二次方程，即使在分解困难或无法分解的情况下也有效。公式如下：

x = (-b ± √(b² – 4ac)) / (2a)

Here the symbol ± means that there are two values: one with a plus sign and one with a minus sign.

这里的 ± 符号表示有两个值：一个取加号，一个取减号。

Example: Solve 3x² – 4x – 2 = 0 using the formula. Here a = 3, b = -4, c = -2.

例如：用求根公式解 3x² – 4x – 2 = 0。此时 a = 3，b = -4，c = -2。

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

x = (4 ± √(16 + 24)) / 6 = (4 ± √40) / 6

Since √40 = 2√10, the solutions are x = (4 + 2√10)/6 and x = (4 – 2√10)/6. These can be simplified to x = (2 ± √10)/3.

由于 √40 = 2√10，解为 x = (4 + 2√10)/6 和 x = (4 – 2√10)/6。可化简为 x = (2 ± √10)/3。

Always check whether you are asked to give exact answers or answers correct to a certain number of decimal places.

注意题目要求精确答案还是保留到小数位的近似答案。

5. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + p)² + q. This is useful for finding turning points and solving equations without using the formula.

配方法将二次式改写为 a(x + p)² + q 的形式。这有助于求顶点坐标，也可用于解方程而不使用求根公式。

Example: Write x² + 6x + 5 in the form (x + p)² + q.

例如：将 x² + 6x + 5 改写为 (x + p)² + q 的形式。

Take half of 6, which is 3, and square it to get 9. So:

取 6 的一半，即 3，平方得 9。因此：

x² + 6x + 5 = (x + 3)² – 9 + 5 = (x + 3)² – 4

To solve (x + 3)² – 4 = 0, add 4 to both sides and take the square root: x + 3 = ±2, so x = -1 or x = -5.

要解 (x + 3)² – 4 = 0，两边同时加 4，再开平方：x + 3 = ±2，因此 x = -1 或 x = -5。

If the coefficient a is not 1, factorise it out first before completing the square.

如果二次项系数 a 不是 1，需要先提取公因数，再进行配方。

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

The expression b² – 4ac is called the discriminant. It tells us how many real roots a quadratic equation has.

代数式 b² – 4ac 称为判别式。它告诉我们一元二次方程有多少个实数根。

Value of b² – 4ac	Nature of roots
Positive and a perfect square	Two distinct rational roots
Positive but not a perfect square	Two distinct irrational roots
Zero	Two equal real roots (one repeated root)
Negative	No real roots

判别式 b² – 4ac 的值为正且是完全平方数时，方程有两个不相等的有理数根；为正但不是完全平方数时，方程有两个不相等的无理数根；为零时，方程有两个相等的实数根（一个重根）；为负时，方程没有实数根。

Example: For kx² + 4x + 1 = 0, find the set of values of k for which the equation has two distinct real roots.

例如：对于 kx² + 4x + 1 = 0，求 k 的取值范围，使方程有两个不相等的实数根。

4² – 4 × k × 1 > 0

16 – 4k > 0

So k < 4, but since the equation is quadratic we also need k ≠ 0. Therefore the set of values is k < 4 and k ≠ 0.

所以 k < 4，但因为是二次方程，k 还必须满足 k ≠ 0。因此取值范围为 k < 4 且 k ≠ 0。

7. Sum and Product of Roots | 根与系数的关系

For a quadratic equation ax² + bx + c = 0 with roots α and β (alpha and beta), we can write:

对于二次方程 ax² + bx + c = 0，设其两根为 α 和 β，则有：

Sum of roots: α + β = -b/a

Product of roots: α × β = c/a

Example: A quadratic equation has roots 3 and -5. Find the equation.

例如：已知某二次方程的两根为 3 和 -5，求该方程。

Sum = 3 + (-5) = -2, so -b/a = -2. Product = 3 × (-5) = -15, so c/a = -15. Choosing a = 1 gives b = 2 and c = -15. The equation is x² + 2x – 15 = 0.

两根和 = 3 + (-5) = -2，所以 -b/a = -2。两根积 = 3 × (-5) = -15，所以 c/a = -15。令 a = 1，得 b = 2，c = -15。方程为 x² + 2x – 15 = 0。

This relationship is particularly useful in questions that give the roots without providing the equation itself.

这种关系特别适用于题目直接给出两根而要求构造方程的题型。

8. Graphs of Quadratic Functions | 二次函数图像

The graph of y = ax² + bx + c is a parabola. If a > 0, the parabola opens upwards; if a < 0, it opens downwards.

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

The x-intercepts of the graph are the real roots of the equation ax² + bx + c = 0. The y-intercept is the value of c.

抛物线与 x 轴的交点就是方程 ax² + bx + c = 0 的实数根。与 y 轴的交点为 (0, c)。

The vertex (turning point) has x-coordinate given by x = -b/(2a). You can then substitute this value back into the equation to find the y-coordinate.

顶点（转折点）的横坐标为 x = -b/(2a)，再将该横坐标代入原式即可求出纵坐标。

If the equation is written in completed square form y = a(x + p)² + q, the vertex is simply (-p, q).

若方程已写成配方法形式 y = a(x + p)² + q，则顶点坐标直接为 (-p, q)。

9. Word Problems Leading to Quadratics | 二次方程应用题

Many exam problems require you to set up a quadratic equation from a real-world situation. Common contexts include areas, product of consecutive numbers, and projectile motion.

许多考试题需要你根据现实情境建立二次方程。常见背景包括面积、连续整数乘积和抛体运动。

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

例如：一个长方形的长比宽多 3 cm，面积为 40 cm²。求宽。

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

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

x² + 3x – 40 = 0

Factorise: (x + 8)(x – 5) = 0. Since width cannot be negative, x = 5. So the width is 5 cm.

分解因式：(x + 8)(x – 5) = 0。因为宽不能为负数，所以 x = 5，即宽为 5 cm。

Always reject negative solutions when the variable represents a physical length or a count.

当变量表示实际长度或数量时，一定要舍去负数解。

10. Common Mistakes | 常见错误

Here are frequent errors that cost marks in exams. Avoid them by practising carefully.

以下是在考试中常见的失分错误。通过仔细练习来避免它们。

Forgetting to rearrange the equation into standard form before factorising or using the formula.
Dropping the negative sign when substituting b into the quadratic formula.
Writing x = ±√k without first isolating the squared term.
Dividing both sides of an equation by x when x may equal zero, thus losing a root.
Confusing the sum and product of roots formulas.

在分解或使用求根公式前，忘记将方程整理为标准形式。
在代入 b 时漏掉负号。
没有先分离平方项就写出 x = ±√k。
在 x 可能等于零的情况下两边同时除以 x，导致丢根。
混淆两根之和与两根之积的公式。

Check your answers by substituting them back into the original equation. This only takes a few seconds and catches most errors.

将解代回原方程进行验算。这只需几秒钟，却能发现大部分错误。

11. Exam Tips for IGCSE | IGCSE 考试技巧

In the non-calculator paper, factorisation and completing the square are often expected. In the calculator paper, the quadratic formula can be used freely, but show all working clearly.

在非计算器试卷中，通常期望使用因式分解法和配方法。在计算器试卷中，可以自由使用求根公式，但必须写出清晰的解题过程。

If a question asks for answers “correct to 2 decimal places”, use the quadratic formula and round at the very end. If it asks for “exact values”, use surds or factorisation.

如果题目要求”保留两位小数”，应使用求根公式并在最后一步四舍五入。如果要求”精确值”，则使用根式或因式分解。

Read the question carefully to decide how many solutions are needed. Sometimes only positive values are meaningful.

仔细审题以确定需要多少个解。有时只有正数解才具有实际意义。

Remember that the graph of a quadratic function can help you estimate the number of real roots before you solve. Use this as a sanity check.

记住，二次函数图像可以帮助你在求解前估计实数根的个数。可以用它来检验答案是否合理。

12. Practice Questions | 练习题目

Attempt these questions on your own, then check your answers by substituting back.

请独立尝试以下题目，然后通过代回验算检查答案。

Solve x² – 7x + 10 = 0.
Solve 2x² + x – 6 = 0.
Solve x² + 8x + 2 = 0 by completing the square.
Use the quadratic formula to solve 5x² – 3x – 1 = 0, giving answers correct to 2 decimal places.
Find the value(s) of k such that x² + kx + 9 = 0 has two equal roots.

解方程 x² – 7x + 10 = 0。
解方程 2x² + x – 6 = 0。
用配方法解 x² + 8x + 2 = 0。
用求根公式解 5x² – 3x – 1 = 0，答案保留两位小数。
求 k 的值，使 x² + kx + 9 = 0 有两个相等的实数根。

Answers: 1) x = 2 or 5. 2) x = 3/2 or -2. 3) x = -4 ± √14. 4) x ≈ 0.87 or -0.23. 5) k = 6 or -6.

答案：1) x = 2 或 5。2) x = 3/2 或 -2。3) x = -4 ± √14。4) x ≈ 0.87 或 -0.23。5) k = 6 或 -6。

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Solving Quadratic Equations | 一元二次方程的求解