Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most frequently tested topics in IGCSE Mathematics. They appear in Paper 2 and Paper 4, often in multiple questions, and serve as the foundation for more advanced algebraic work. Mastering quadratic equations is not optional — it is essential for achieving a high grade.

二次方程是 IGCSE 数学中最常考的话题之一。在 Paper 2 和 Paper 4 中都会出现，而且经常不止一题，同时它也是更高级代数知识的基础。掌握二次方程不是可选项——而是取得高分的关键。

1. Expanding Double Brackets | 展开双括号

Before we can factorise quadratics, we must understand how to expand them. Expanding double brackets means multiplying every term in the first bracket by every term in the second bracket. A helpful mnemonic is FOIL: First, Outer, Inner, Last.

在因式分解二次式之前，我们必须先理解如何展开它们。展开双括号就是将第一个括号中的每一项乘以第二个括号中的每一项。一个有用的助记法是 FOIL：首项（First）、外项（Outer）、内项（Inner）、末项（Last）。

For example, (x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15.

例如，(x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15。

Notice a key pattern: the constant term (15) is the product of 3 and 5, while the coefficient of x (8) is the sum of 3 and 5. This relationship between product and sum is exactly what we reverse when factorising.

注意一个关键规律：常数项（15）是 3 和 5 的乘积，而 x 的系数（8）是 3 和 5 的和。这种乘积与和的关系正是我们在因式分解时要反向运用的。

(x + a)(x + b) = x² + (a + b)x + ab

Always check your expansion by multiplying your final answer back out again. A quick check can prevent careless errors.

永远通过重新展开你的最终答案来检查结果。快速检查可以避免粗心错误。

2. Factorising x² + bx + c | 因式分解二次式 x² + bx + c

To factorise x² + bx + c, we look for two numbers that multiply to give c and add to give b. Once the numbers are found, we can write the quadratic as the product of two binomials.

要将 x² + bx + c 因式分解，我们需要找到两个数，它们的乘积为 c，和为 b。找到这两个数后，就可以把二次式写成两个二项式的乘积。

Example 1: Factorise x² + 7x + 12. We need two numbers that multiply to 12 and add to 7. These numbers are 3 and 4, since 3 × 4 = 12 and 3 + 4 = 7.

例 1：因式分解 x² + 7x + 12。我们需要两个数，乘积为 12，和为 7。这两个数是 3 和 4，因为 3 × 4 = 12 且 3 + 4 = 7。

x² + 7x + 12 = (x + 3)(x + 4)

Example 2: Factorise x² − 5x + 6. We need two numbers that multiply to 6 and add to −5. The correct pair is −2 and −3, because (−2) × (−3) = 6 and (−2) + (−3) = −5.

例 2：因式分解 x² − 5x + 6。我们需要两个数，乘积为 6，和为 −5。正确的两个数是 −2 和 −3，因为 (−2) × (−3) = 6，且 (−2) + (−3) = −5。

x² − 5x + 6 = (x − 2)(x − 3)

If the constant term is negative, the two numbers will have opposite signs. For example, x² − 2x − 8: we need two numbers that multiply to −8 and add to −2. Those are −4 and 2.

如果常数项为负，那么这两个数一正一负。例如，x² − 2x − 8：我们需要两个数乘积为 −8，和为 −2。这两个数是 −4 和 2。

x² − 2x − 8 = (x − 4)(x + 2)

3. Factorising ax² + bx + c | 因式分解二次式 ax² + bx + c

When the coefficient of x² is not 1, factorising requires an extra step. The method involves finding two numbers that multiply to a × c and add to b, then splitting the middle term.

当 x² 的系数不是 1 时，因式分解需要额外一步。方法是找到两个数，它们的乘积为 a × c，和为 b，然后拆分中间项。

Example: Factorise 2x² + 7x + 3. Here a = 2, b = 7, c = 3. We need two numbers that multiply to 2 × 3 = 6 and add to 7. These are 6 and 1.

例：因式分解 2x² + 7x + 3。这里 a = 2，b = 7，c = 3。我们需要两个数乘积为 2 × 3 = 6，和为 7。这两个数是 6 和 1。

Now split the middle term 7x into 6x + x, giving 2x² + 6x + x + 3. Next, factor in pairs:

现在把中间项 7x 拆成 6x + x，得到 2x² + 6x + x + 3。然后分组因式分解：

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

Always remember: if the coefficient of x² is negative, factor out −1 first. For example, −x² + 5x − 6 = −(x² − 5x + 6) = −(x − 2)(x − 3).

始终记住：如果 x² 的系数为负，先提取 −1。例如，−x² + 5x − 6 = −(x² − 5x + 6) = −(x − 2)(x − 3)。

4. Difference of Two Squares | 平方差公式

The difference of two squares is a special pattern that appears frequently in exams. It states that a² − b² = (a + b)(a − b). This identity holds because the middle terms cancel out during expansion.

平方差公式是考试中频繁出现的特殊模式。它表明 a² − b² = (a + b)(a − b)。这个恒等式成立是因为展开时中间项相互抵消。

Example 1: x² − 49 = x² − 7² = (x + 7)(x − 7).

例 1：x² − 49 = x² − 7² = (x + 7)(x − 7)。

Example 2: 4x² − 25 = (2x)² − 5² = (2x + 5)(2x − 5). The key is recognising perfect squares.

例 2：4x² − 25 = (2x)² − 5² = (2x + 5)(2x − 5)。关键是识别完全平方数。

This pattern also works with algebraic terms. For instance, x⁴ − 16 = (x² + 4)(x² − 4) = (x² + 4)(x + 2)(x − 2), by applying the pattern twice.

平方差公式也适用于代数项。例如，x⁴ − 16 = (x² + 4)(x² − 4) = (x² + 4)(x + 2)(x − 2)，即连续使用两次公式。

5. Solving Quadratics by Factorisation | 用因式分解法解方程

To solve a quadratic equation by factorising, we rely on the zero product property: if a × b = 0, then either a = 0 or b = 0 (or both). This property allows us to break a product equation into two simpler linear equations.

用因式分解法解二次方程的关键是零乘积性质：如果 a × b = 0，那么 a = 0 或 b = 0（或两者同时成立）。这个性质让我们把一个乘积方程拆成两个更简单的一次方程。

Example: Solve x² − 5x + 6 = 0. Factorising gives (x − 2)(x − 3) = 0. Therefore x − 2 = 0 or x − 3 = 0, so x = 2 or x = 3.

例：解 x² − 5x + 6 = 0。因式分解得 (x − 2)(x − 3) = 0。因此 x − 2 = 0 或 x − 3 = 0，所以 x = 2 或 x = 3。

Critical reminder: always rearrange the equation so that one side is zero before factorising. For example, to solve x² = 3x, do not divide both sides by x (this loses the solution x = 0). Instead, rearrange to x² − 3x = 0, then factor x(x − 3) = 0, giving x = 0 or x = 3.

重要提醒：在因式分解之前，务必整理方程使一侧为零。例如，解 x² = 3x 时，不要两边同时除以 x（这样会丢失解 x = 0）。正确做法是化为 x² − 3x = 0，然后因式分解 x(x − 3) = 0，得到 x = 0 或 x = 3。

6. The Quadratic Formula | 求根公式

When factorisation is difficult or impossible, we use the quadratic formula. For any quadratic equation ax² + bx + c = 0, the solutions are given by:

当因式分解困难或无法进行时，我们使用求根公式。对于任意二次方程 ax² + bx + c = 0，解由以下公式给出：

x = (−b ± √(b² − 4ac)) / 2a

This formula is provided in the IGCSE formula booklet, but you must know how to substitute values correctly and simplify the result.

这个公式会在 IGCSE 公式册中提供，但你必须掌握如何正确代入数值并化简结果。

Example: Solve 2x² + 3x − 5 = 0. Here a = 2, b = 3, c = −5. Substituting into the formula:

例：解 2x² + 3x − 5 = 0。这里 a = 2，b = 3，c = −5。代入公式：

x = (−3 ± √(3² − 4 × 2 × (−5))) / (2 × 2) = (−3 ± √(9 + 40)) / 4 = (−3 ± √49) / 4

Since √49 = 7, we have x = (−3 + 7)/4 = 1 or x = (−3 − 7)/4 = −2.5. Note that the symbol ± means we calculate both the positive and negative cases.

因为 √49 = 7，所以 x = (−3 + 7)/4 = 1 或 x = (−3 − 7)/4 = −2.5。注意 ± 符号表示我们要分别计算正负两种情况。

7. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. This form is especially useful for finding the turning point of a parabola and for solving equations that cannot be factorised easily.

配方法将二次式改写为 (x + p)² + q 的形式。这种形式特别适用于求抛物线的顶点坐标，以及解不易因式分解的方程。

To complete the square for x² + bx, we add (b/2)². Since we cannot change the value of the expression, we must also subtract it.

对于 x² + bx 配方，我们加上 (b/2)²。由于不能改变表达式的值，还必须同时减去它。

Example: Write x² + 6x + 2 in completed square form. Here b = 6, so (b/2)² = 9. Therefore:

例：将 x² + 6x + 2 写成配方法形式。这里 b = 6，所以 (b/2)² = 9。因此：

x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7

From this form, the turning point of y = x² + 6x + 2 is (−3, −7). The x-coordinate is the value that makes the bracket zero, and the y-coordinate is the constant term.

从这个形式可以看出，y = x² + 6x + 2 的顶点坐标为 (−3, −7)。x 坐标是使括号为零的值，y 坐标是常数项。

If the leading coefficient is not 1, for example 2x² + 8x + 5, factor out 2 first: 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3.

如果首项系数不是 1，例如 2x² + 8x + 5，先提取 2：2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3。

8. The Discriminant | 判别式

The discriminant is the expression b² − 4ac inside the square root of the quadratic formula. It determines the nature of the roots without solving the full equation.

判别式是求根公式中根号内的表达式 b² − 4ac。它无需解完整方程就能判断根的性质。

If Δ > 0, the quadratic has two distinct real roots. This means the parabola crosses the x-axis at two different points.

如果 Δ > 0，二次方程有两个不相等的实数根。这意味着抛物线在 x 轴上穿过两个不同的点。

If Δ = 0, the quadratic has exactly one repeated root. The parabola touches the x-axis at exactly one point, called a tangent point.

如果 Δ = 0，二次方程有一个重根。抛物线在 x 轴上恰好接触一个点，称为切点。

If Δ < 0, the quadratic has no real roots. The parabola never touches the x-axis.

如果 Δ < 0，二次方程没有实数根。抛物线从不接触 x 轴。

Example: For 3x² − 2x + 5 = 0, Δ = (−2)² − 4 × 3 × 5 = 4 − 60 = −56 < 0, so there are no real roots.

例：对于 3x² − 2x + 5 = 0，Δ = (−2)² − 4 × 3 × 5 = 4 − 60 = −56 < 0，所以没有实数根。

9. Word Problems | 应用题

Many exam questions require setting up a quadratic equation from a real-world context. The key steps are: read carefully, define a variable, translate the conditions into an equation, solve it, and then check whether the answer makes sense in context.

许多考试题目需要从实际情境中建立二次方程。关键步骤是：仔细阅读、定义变量、将条件转化为方程、求解，然后检查答案在情境中是否有意义。

Example: The length of a rectangle is 3 cm longer than its width, and the area is 40 cm². Find the width. Let width = x, so length = x + 3. Then x(x + 3) = 40, giving x² + 3x − 40 = 0.

例：一个长方形的长比宽长 3 cm，面积为 40 cm²。求宽。设宽为 x，则长为 x + 3。于是 x(x + 3) = 40，即 x² + 3x − 40 = 0。

Factorising gives (x + 8)(x − 5) = 0, so x = −8 or x = 5. Since a width cannot be negative, the only valid answer is x = 5 cm. Always reject negative solutions if the context demands a positive value.

因式分解得 (x + 8)(x − 5) = 0，所以 x = −8 或 x = 5。由于宽度不能为负，唯一有效的答案是 x = 5 cm。当情境要求正数时，务必舍去负数解。

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

One common mistake is forgetting to rearrange the equation to zero before factorising. Another is misapplying sign rules when expanding double brackets, for example writing (x − 3)(x + 4) incorrectly as x² − 7x − 12.

一个常见错误是忘记在因式分解前将方程整理为零。另一个错误是展开双括号时符号规则运用不当，例如把 (x − 3)(x + 4) 错误地写成 x² − 7x − 12。

Take care with the quadratic formula: write out a, b, c clearly first and substitute step by step. Rushing often leads to sign errors, especially when b or c is negative.

使用求根公式时要小心：先清楚写出 a、b、c，然后逐步

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Solving Quadratic Equations | 解二次方程