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Solving Quadratic Equations by Factorisation | 因式分解法求解二次方程

2026-08-30 00:27 · 本文部分内容由AI辅助创作

📚 Solving Quadratic Equations by Factorisation | 因式分解法求解二次方程

Quadratic equations are one of the most important topics in IGCSE mathematics. They appear in many exam papers and in later topics such as graphs and functions. In this article, we will focus on solving quadratic equations by factorisation, the most direct method for many exam questions.

二次方程是IGCSE数学中最重要的主题之一。它出现在许多试卷中,以及后续的图象与函数等知识点中。本文将集中讨论利用因式分解法求解二次方程,这是应对许多考试题目最直接的方法。

1. The Standard Form | 标准形式

nts, and x is the unknown variable. For example, 3x² - 5x + 2 = 0 is a quadratic equation.

任何二次方程都可以写成 ax² + bx + c = 0 的形式,其中 a ≠ 0。这里 a、b 和 c 为常数,x 为未知数。例如,3x² - 5x + 2 = 0 就是一个二次方程。

In IGCSE exams, you must be able to rearrange equations into this standard form before solving. If the equation contains fractions or brackets, simplify first.

在IGCSE考试中,你必须能够先将方程变形为标准形式再求解。如果方程含有分数或括号,应先化简。


2. Factorising Quadratic Expressions | 因式分解二次表达式

ing an expression as a product of two brackets. For example, x² + 5x + 6 = (x + 2)(x + 3).

因式分解是指将一个表达式写成两个括号相乘的形式。例如,x² + 5x + 6 = (x + 2)(x + 3)。

that multiply to give c and add to give b. In the example above, 2 × 3 = 6 and 2 + 3 = 5.

要因式分解 x² + bx + c,找到两个数,乘积等于 c,和等于 b。在上例中,2 × 3 = 6,且 2 + 3 = 5。


3. The Zero Product Property | 零乘积性质

If the product of two factors is zero, then at least one of the factors must be zero. This is the key idea used to solve factorised quadratic equations.

如果两个因式的乘积为零,那么至少有一个因式为零。这是解因式分解后二次方程的关键原理。

So if (x - 2)(x + 3) = 0, then x - 2 = 0 or x + 3 = 0. Hence x = 2 or x = -3.

因此,若 (x - 2)(x + 3) = 0,则 x - 2 = 0 或 x + 3 = 0,所以 x = 2 或 x = -3。


4. Worked Example 1: Solve x² - 5x + 6 = 0 | 例题1:解 x² - 5x + 6 = 0

Step 1: Factorise the left-hand side. Find two numbers whose product is 6 and whose sum is -5. The numbers are -2 and -3.

第一步:因式分解左边。找到两个数,乘积为6,和为-5。这两个数是-2和-3。

Therefore x² - 5x + 6 = (x - 2)(x - 3) = 0.

因此 x² - 5x + 6 = (x - 2)(x - 3) = 0。

Step 2: Apply the zero product property: x - 2 = 0 or x - 3 = 0.

第二步:应用零乘积性质:x - 2 = 0 或 x - 3 = 0。

Step 3: Solve each equation: x = 2 or x = 3. So the solutions are x = 2 and x = 3.

第三步:分别解方程:x = 2 或 x = 3。因此解为 x = 2 和 x = 3。


5. Worked Example 2: Solve 2x² + 7x + 3 = 0 | 例题2:解 2x² + 7x + 3 = 0

When the coefficient of x² is not 1, factorisation requires more care. Here we look for two brackets of the form (px + q)(rx + s).

当 x² 的系数不为1时,因式分解需要更仔细。这里我们要找形如 (px + q)(rx + s) 的两个括号。

e factor pairs for 2 are 2 and 1. For 3, the pairs are 3 and 1 (both positive, since the middle term is positive).

我们需要 p × r = 2,且 q × s = 3。2 的因数对有 2 和 1;3 的因数对有 3 和 1(由于中间项为正,取正数)。

Trying (2x + 1)(x + 3) gives 2x² + 6x + x + 3 = 2x² + 7x + 3. This is correct.

尝试 (2x + 1)(x + 3) 得到 2x² + 6x + x + 3 = 2x² + 7x + 3。这是正确的。

Thus (2x + 1)(x + 3) = 0, so 2x + 1 = 0 or x + 3 = 0. Hence x = -½ or x = -3.

因此 (2x + 1)(x + 3) = 0,所以 2x + 1 = 0 或 x + 3 = 0。因此 x = -½ 或 x = -3。


6. The Difference of Two Squares | 平方差公式

l type of quadratic expression is the difference of two squares: a² - b² = (a - b)(a + b).

一种特殊的二次表达式是平方差:a² - b² = (a - b)(a + b)。

For example, solve x² - 9 = 0. Here a = x and b = 3, so x² - 9 = (x - 3)(x + 3) = 0.

例如,解 x² - 9 = 0。这里 a = x,b = 3,所以 x² - 9 = (x - 3)(x + 3) = 0。

Thus x = 3 or x = -3. This method is faster than expanding brackets.

因此 x = 3 或 x = -3。此方法比展开括号更快。


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7. Rearranging Into Standard Form | 化为标准形式

Sometimes the equation is given with terms on both sides. You must collect all terms on one side before factorising.

有时方程的两边都有项。你必须先将所有项移到一边,再因式分解。

x and lose a solution. Correct method: x² - 4x = 0, so x(x - 4) = 0. Hence x = 0 or x = 4.

例:解 x² = 4x。许多学生两边同除以 x 而丢失一个解。正确方法:x² - 4x = 0,于是 x(x - 4) = 0,所以 x = 0 或 x = 4。


8. Common Mistakes | 常见错误


9. Exam Tips | 考试技巧

Always rearrange to the form ax² + bx + c = 0 before factorising.

在因式分解之前,务必整理成 ax² + bx + c = 0 的形式。

Check your factors by expanding the brackets in your head or on paper.

通过心算或在纸上展开括号来检查你的因式分解是否正确。

If you cannot factorise, use the quadratic formula or completing the square. For IGCSE, factorisation is often expected when it is possible.

如果你无法因式分解,可以使用求根公式或配方法。对于IGCSE,当可能时通常期望使用因式分解法。


10. Practice Questions | 练习

Try these questions and check your answers below.

尝试以下问题,并在下方核对答案。

  1. x² - 7x + 10 = 0

    解方程 x² - 7x + 10 = 0

  2. x² - 36 = 0

    解方程 x² - 36 = 0

  3. 2x² - 5x - 3 = 0

    解方程 2x² - 5x - 3 = 0

  4. 3x² = 6x

    解方程 3x² = 6x


11. Solutions to Practice | 练习答案

1. x² - 7x + 10 = (x - 2)(x - 5) = 0, so x = 2 or x = 5.

1. x² - 7x + 10 = (x - 2)(x - 5) = 0,因此 x = 2 或 x = 5。

2. x² - 36 = (x - 6)(x + 6) = 0, so x = 6 or x = -6.

2. x² - 36 = (x - 6)(x + 6) = 0,因此 x = 6 或 x = -6。

3. 2x² - 5x - 3 = (2x + 1)(x - 3) = 0, so x = -½ or x = 3.

3. 2x² - 5x - 3 = (2x + 1)(x - 3) = 0,因此 x = -½ 或 x = 3。

4. 3x² = 6x → 3x² - 6x = 0 → 3x(x - 2) = 0, so x = 0 or x = 2.

4. 3x² = 6x → 3x² - 6x = 0 → 3x(x - 2) = 0,因此 x = 0 或 x = 2。

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