Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations appear throughout the IGCSE Edexcel Mathematics syllabus. They are polynomials of degree 2, and mastering their solution methods is essential for success in algebra, graphs, and many word problems.

二次方程是 IGCSE Edexcel 数学课程中的核心内容。它们是最高次数为 2 的多项式方程，掌握其解法对代数、函数图像以及各类应用题都至关重要。

1. The General Form | 二次方程的一般形式

A quadratic equation can be written in the standard form ax² + bx + c = 0, where a, b and c are real numbers, and a ≠ 0. Here x is the unknown variable, and we wish to find its value(s) that satisfy the equation.

二次方程通常写成标准形式 ax² + bx + c = 0，其中 a、b、c 是实数，且 a ≠ 0。x 是未知数，我们的目标是找到满足等式的x值。

The highest power of x is 2, which is why it is called ‘quadratic’. A quadratic equation can have two distinct real roots, one repeated real root, or no real roots, depending on the value of the discriminant.

x 的最高次数是 2，因此称为“二次”方程。二次方程可能有两个不同的实数根、一个重根，也可能没有实数根，这取决于判别式的值。

2. Solving by Factorisation: Monic Quadratics | 用因式分解法解首一二次方程

When the coefficient of x² is 1 (a = 1), we call it a monic quadratic. To factorise x² + bx + c, we look for two integers m and n such that m + n = b and mn = c. The equation then becomes (x + m)(x + n) = 0.

当 x² 的系数为 1（即 a = 1）时，我们称其为“首一”二次式。要分解 x² + bx + c，我们需要寻找两个整数 m 和 n，使得 m + n = b，且 mn = c。于是原方程化为 (x + m)(x + n) = 0。

Once factorised, we use the zero-product property: if the product of two factors is zero, then at least one of the factors must be zero. Therefore, either x + m = 0 or x + n = 0, giving the solutions x = −m and x = −n.

分解后，我们利用零积性质：如果两个因子的乘积为零，那么至少有一个因子为零。因此，要么 x + m = 0，要么 x + n = 0，从而得到 x = −m 和 x = −n。

x² + 5x + 6 = 0 ⇒ (x + 2)(x + 3) = 0 ⇒ x = −2 或 x = −3

3. Difference of Two Squares | 平方差公式

A special factorisation pattern is the difference of two squares: a² − b² = (a + b)(a − b). This is very useful when solving equations like x² − 16 = 0. In this case, x² − 4² = (x + 4)(x − 4) = 0, so x = 4 or x = −4.

一个特殊的因式分解模式是平方差公式：a² − b² = (a + b)(a − b)。这在解如 x² − 16 = 0 的方程时非常有用。此时 x² − 4² = (x + 4)(x − 4) = 0，因此 x = 4 或 x = −4。

This pattern also helps when a is not 1. For example, 4x² − 9 = 0 can be written as (2x)² − 3² = 0, giving (2x + 3)(2x − 3) = 0, and hence x = −3/2 or x = 3/2.

当 a 不是 1 时，这个模式同样适用。例如，4x² − 9 = 0 可以写成 (2x)² − 3² = 0，得到 (2x + 3)(2x − 3) = 0，因此 x = −3/2 或 x = 3/2。

4. Factorising Non-Monic Quadratics | 分解非首一二次式

When a ≠ 1, for example 2x² + 7x + 3, factorisation requires more care. One method is to find two integers whose product is a × c and whose sum is b. Then split the middle term and group terms.

当 a ≠ 1 时，例如 2x² + 7x + 3，因式分解需要更加细心。一种方法是找到两个整数，它们的乘积等于 a × c，且它们的和等于 b。然后拆分中间项，再分组提取公因式。

For 2x² + 7x + 3, we need two numbers with product 2 × 3 = 6 and sum 7. These numbers are 6 and 1. We rewrite the equation as 2x² + 6x + x + 3 = 0, then group: 2x(x + 3) + 1(x + 3) = 0, giving (2x + 1)(x + 3) = 0.

对于 2x² + 7x + 3，我们需要找到两个数，它们的乘积为 2 × 3 = 6，和为 7。这两个数是 6 和 1。我们将方程改写为 2x² + 6x + x + 3 = 0，然后分组：2x(x + 3) + 1(x + 3) = 0，得到 (2x + 1)(x + 3) = 0。

2x² + 7x + 3 = 0 ⇒ (2x + 1)(x + 3) = 0 ⇒ x = −1/2 或 x = −3

5. Using the Quadratic Formula | 使用求根公式

Some quadratic equations cannot be factorised easily using integers. In such cases, we can use the quadratic formula:

有些二次方程不容易用整数因式分解。此时我们可以使用求根公式：

x = (−b ± √(b² − 4ac))/(2a)

Here the ‘±’ sign means that we add and subtract the square root term, yielding two possible values of x. This formula works for all quadratic equations, provided the discriminant b² − 4ac is not negative.

这里的“±”符号表示我们分别加上和减去根号项，从而得到两个可能的 x 值。这个公式适用于所有二次方程，只要判别式 b² − 4ac 不为负数。

Example: Solve 2x² − 4x − 3 = 0. Here a = 2, b = −4, c = −3. Then b² − 4ac = (−4)² − 4 × 2 × (−3) = 16 + 24 = 40. Therefore x = (4 ± √40)/4 = (4 ± 2√10)/4 = (2 ± √10)/2.

例如：解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。则 b² − 4ac = (−4)² − 4 × 2 × (−3) = 16 + 24 = 40。因此 x = (4 ± √40)/4 = (4 ± 2√10)/4 = (2 ± √10)/2。

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

The expression b² − 4ac is called the discriminant, often denoted by Δ. It tells us how many real roots a quadratic equation has without solving it fully.

表达式 b² − 4ac 称为判别式，通常用 Δ 表示。它可以在不完全解方程的情况下告诉我们方程有多少个实数根。

If Δ > 0, the equation has two distinct real roots.
If Δ = 0, the equation has exactly one repeated real root.
If Δ < 0, the equation has no real roots (the roots are complex).

若 Δ > 0，方程有两个不同的实数根。
若 Δ = 0，方程有一个二重实数根。
若 Δ < 0，方程没有实数根（根为复数）。

For example, x² − 2x + 1 = 0 has Δ = (−2)² − 4 × 1 × 1 = 0, so it has a single root x = 1. Meanwhile, x² + x + 1 = 0 has Δ = 1 − 4 = −3, meaning no real solutions.

例如，x² − 2x + 1 = 0 中 Δ = (−2)² − 4 × 1 × 1 = 0，所以它只有一个根 x = 1。而 x² + x + 1 = 0 中 Δ = 1 − 4 = −3，意味着没有实数解。

7. Completing the Square | 配方法

Another method for solving quadratics is completing the square. For x² + bx + c = 0, we rewrite the quadratic as (x + b/2)² − (b/2)² + c = 0. This is especially useful for deriving the quadratic formula and for solving without factorising.

另一种解二次方程的方法是配方法。对于 x² + bx + c = 0，我们将二次式改写为 (x + b/2)² − (b/2)² + c = 0。这在推导求根公式以及无法因式分解时特别有用。

Example: Solve x² + 6x + 2 = 0. First, x² + 6x = (x + 3)² − 9. So the equation becomes (x + 3)² − 9 + 2 = 0, i.e. (x + 3)² = 7. Taking square roots: x + 3 = ±√7, hence x = −3 ± √7.

例如：解 x² + 6x + 2 = 0。首先，x² + 6x = (x + 3)² − 9。于是方程变为 (x + 3)² − 9 + 2 = 0，即 (x + 3)² = 7。两边开平方：x + 3 = ±√7，因此 x = −3 ± √7。

For a ≠ 1, we first divide the whole equation by a, then complete the square on the resulting monic quadratic.

当 a ≠ 1 时，我们先将整个方程除以 a，然后对所得的首一二次式配方。

8. Solving Quadratic Equations from Word Problems | 用二次方程解应用题

Many real-world problems, such as projectile motion, area optimisation, and number puzzles, lead to quadratic equations. The key is to translate the language into an algebraic equation, solve it, and then check that the solutions make sense in the context.

许多实际问题，如抛物线运动、面积优化和数字谜题，都会产生二次方程。关键是将其转化为代数方程，求解，然后检查解在原始情境中是否合理。

Example: A rectangle has a length 4 cm longer than its width, and its area is 21 cm². Let the width be w. Then w(w + 4) = 21, so w² + 4w − 21 = 0. Factorising gives (w + 7)(w − 3) = 0, so w = 3 or w = −7. Since width cannot be negative, w = 3 cm, and the length is 7 cm.

例如：一个长方形的长比宽长 4 cm，面积为 21 cm²。设宽为 w，则 w(w + 4) = 21，即 w² + 4w − 21 = 0。因式分解得 (w + 7)(w − 3) = 0，于是 w = 3 或 w = −7。由于宽不能为负数，所以 w = 3 cm，长为 7 cm。

9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Students often make avoidable errors when solving quadratics. Being aware of these can save valuable marks in exams.

学生在解二次方程时常犯一些可以避免的错误。了解这些错误能在考试中为你节省宝贵的分数。

Forgetting to set the equation to 0 before factorising.
Applying the zero-product property incorrectly when the expression is not fully factorised.
Misusing the quadratic formula signs, especially when b is negative.
Ignoring negative solutions in word problems even when they are mathematically valid but physically impossible.

因式分解前忘记将方程化为等于 0 的形式。
在表达式未完全分解时就错误地使用零积性质。
使用求根公式时符号出错，特别是 b 为负数的情况。
在应用题中忽略负数解，即使它在数学上成立但在实际情境中无意义。

Always check your solutions by substituting them back into the original equation. This simple step eliminates many mechanical mistakes.

始终将解代回原方程进行检验。这个简单步骤可以消除许多机械性错误。

10. Practice Questions | 练习题目

Try solving these quadratics using an appropriate method. Write down each step clearly.

请用适当的方法解下列二次方程，并清晰写出每一步骤。

x² − 7x + 12 = 0
3x² + 5x − 2 = 0
4x² − 25 = 0
x² + 8x + 1 = 0 (give answers in surd form)

x² − 7x + 12 = 0
3x² + 5x − 2 = 0
4x² − 25 = 0
x² + 8x + 1 = 0（答案用根式表示）

Answers: 1) x = 3, 4. 2) x = 1/3, −2. 3) x = ±5/2. 4) x = −4 ± √15.

答案：1) x = 3, 4。2) x = 1/3, −2。3) x = ±5/2。4) x = −4 ± √15。

11. Graphical Interpretations | 图像意义

The roots of a quadratic equation ax² + bx + c = 0 correspond to the x-intercepts of the parabola y = ax² + bx + c. If the parabola touches the x-axis at exactly one point, the equation has a repeated root. If it never crosses the x-axis, the equation has no real roots.

二次方程 ax² + bx + c = 0 的根对应于抛物线 y = ax² + bx + c 与 x 轴的交点。如果抛物线与 x 轴仅相切于一点，则方程有重根。如果它始终不与 x 轴相交，则方程无实数根。

The y-coordinate of the vertex can also be found by completing the square. For a positive a, the parabola opens upward; for negative a, it opens downward. This visual understanding helps in checking solutions.

顶点的 y 坐标也可以通过配方法求得。当 a 为正时，抛物线开口向上；当 a 为负时，开口向下。这种直观理解有助于检验解的合理性。

12. Summary | 总结

To solve quadratic equations effectively, you should be fluent in four methods: factorisation, the quadratic formula, completing the square, and using the discriminant to anticipate the nature of roots. Each method has its strengths, and exam questions may require you to choose the most efficient one.

要有效解二次方程，你需要熟练掌握四种方法：因式分解、求根公式、配方法，以及利用判别式预判根的性质。每种方法各有优势，考试题目可能会要求你选择最有效的方法。

Remember to always rearrange the equation into the standard quadratic form before solving, and to interpret your final answers in the context of the problem. With consistent practice, quadratic equations will become a confident topic in your IGCSE preparation.

请记住，求解前务必把方程整理成标准形式，并在实际情境中解读你的最终答案。通过持续练习，二次方程将成为你 IGCSE 备考中的得心应手的内容。

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Solving Quadratic Equations | 解一元二次方程