Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most important topics in IGCSE Mathematics. You will encounter them in algebra, coordinate geometry, and problem-solving questions across both Papers 1 and 2. Mastering the different methods of solving quadratics will give you a strong foundation for higher-level mathematics.

二次方程是 IGCSE 数学中最重要的主题之一。你会在代数、坐标几何以及试卷 1 和试卷 2 的应用题中遇到它们。掌握解二次方程的不同方法，将为你的高阶数学学习打下坚实的基础。

1. What Is a Quadratic Equation? | 什么是二次方程

A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is 2. Its standard form is:

二次方程是最高次数为 2 的多项式方程，意味着变量的最高次幂是 2。它的标准形式是：

ax² + bx + c = 0, where a ≠ 0

Here, a, b, and c are constants, and a cannot be zero. If a = 0, the equation becomes linear. The values of x that satisfy the equation are called the roots or solutions of the equation. A quadratic equation can have at most two real roots.

其中 a、b、c 是常数，且 a 不能为零。如果 a = 0，方程就变成了一次方程。满足方程的 x 值称为方程的根或解。一个二次方程最多可以有两个实数根。

2. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic has simple integer roots. For the simpler case where a = 1, we find two numbers that multiply to give c and add to give b. For the general form ax² + bx + c, we look for two numbers that multiply to give ac and add to give b.

当二次方程具有简单的整数根时，因式分解是最快的方法。当 a = 1 时，我们找到两个数相乘等于 c、相加等于 b。对于一般形式 ax² + bx + c，我们需要找到两个数相乘等于 ac、相加等于 b。

For example, solve x² – 5x + 6 = 0. We need two numbers that multiply to 6 and add to -5. These numbers are -2 and -3. Therefore:

例如，解 x² – 5x + 6 = 0。我们需要两个数相乘等于 6、相加等于 -5。这两个数是 -2 和 -3。因此：

(x – 2)(x – 3) = 0

Using the zero product property, either x – 2 = 0 or x – 3 = 0. Hence, x = 2 or x = 3. Always check your answers by substituting them back into the original equation.

根据零乘积性质，要么 x – 2 = 0，要么 x – 3 = 0。因此 x = 2 或 x = 3。记得将解代回原方程进行验算。

3. Difference of Two Squares | 平方差

A special factoring case is the difference of two squares: a² – b² = (a – b)(a + b). This is extremely useful when solving equations of the form x² – k = 0, where k is a positive constant.

一个特殊的因式分解情况是平方差公式：a² – b² = (a – b)(a + b)。这在解 x² – k = 0（k 为正数）形式的方程时非常有用。

For example, solve x² – 9 = 0. Since 9 = 3², we can write:

例如，解 x² – 9 = 0。因为 9 = 3²，我们可以写成：

(x – 3)(x + 3) = 0

Thus, x = 3 or x = -3. Notice that for x² = k where k > 0, the solutions are x = √k and x = -√k. Do not forget the negative root!

因此，x = 3 或 x = -3。注意对于 x² = k（k > 0），解为 x = √k 和 x = -√k。千万不要忘记负根！

4. Completing the Square | 配方法

Completing the square is a technique that works for any quadratic equation, even when factorisation is not possible. For an expression x² + bx, we add and subtract (b/2)² to form a perfect square.

配方法是一种适用于任何二次方程的技巧，即使在无法因式分解时也能使用。对于表达式 x² + bx，我们通过加上和减去 (b/2)² 来构造一个完全平方。

Example: solve x² + 6x + 8 = 0. First, rewrite the first two terms:

例：解 x² + 6x + 8 = 0。首先重写前两项：

x² + 6x = (x + 3)² – 9

So the equation becomes (x + 3)² – 9 + 8 = 0, which simplifies to (x + 3)² – 1 = 0. Hence (x + 3)² = 1, so x + 3 = ±1, giving x = -2 or x = -4.

因此方程变为 (x + 3)² – 9 + 8 = 0，化简为 (x + 3)² – 1 = 0。所以 (x + 3)² = 1，x + 3 = ±1，得到 x = -2 或 x = -4。

For a general quadratic ax² + bx + c = 0, first divide through by a before completing the square. The result can be written in the form a(x – h)² + k, which is also useful for graph sketching.

对于一般的二次方程 ax² + bx + c = 0，先除以 a 再进行配方。结果可写成 a(x – h)² + k 的形式，这对画图也非常有用。

5. The Quadratic Formula | 二次公式

The quadratic formula is a universal method that can solve any quadratic equation. It is derived by completing the square on the general form ax² + bx + c = 0.

二次公式是一种通用方法，可以解任何二次方程。它是通过对一般形式 ax² + bx + c = 0 配方推导得出的。

x = (-b ± √(b² – 4ac)) / 2a

Simply substitute the values of a, b, and c into the formula. For example, solve 2x² – 4x – 6 = 0. Here, a = 2, b = -4, c = -6. Substituting gives:

只需将 a、b、c 的值代入公式即可。例如，解 2x² – 4x – 6 = 0。这里 a = 2，b = -4，c = -6。代入得：

x = (4 ± √(16 + 48)) / 4 = (4 ± √64) / 4 = (4 ± 8) / 4

Therefore, x = (4 + 8)/4 = 3 or x = (4 – 8)/4 = -1. The formula works for every quadratic equation, but be careful with negative values of b — substituting into (-b) correctly is a common source of errors.

因此，x = (4 + 8)/4 = 3 或 x = (4 – 8)/4 = -1。公式适用于所有二次方程，但代入负数 b 时要特别小心——正确计算 (-b) 是常见的易错点。

6. The Discriminant | 判别式

The discriminant, denoted by Δ = b² – 4ac, tells us the nature of the roots without solving the full equation. It is the part under the square root in the quadratic formula.

判别式，记作 Δ = b² – 4ac，可以帮助我们在不完全解方程的情况下判断根的性质。它是二次公式中根号下的部分。

If Δ > 0: two distinct real roots
If Δ = 0: one repeated real root (two equal roots)
If Δ < 0: no real roots

如果 Δ > 0：两个不相等的实数根
如果 Δ = 0：一个重根（两个相等的实数根）
如果 Δ < 0：无实数根

For example, the equation x² + 4x + 5 = 0 has Δ = 16 – 20 = -4 < 0, so it has no real roots. Its graph never crosses the x-axis. In exam questions, the phrases "equal roots", "real roots", and "no real roots" directly correspond to these three cases.

例如，方程 x² + 4x + 5 = 0 的判别式 Δ = 16 – 20 = -4 < 0，所以它没有实数根。其图像不与 x 轴相交。考试中 "等根"、"实数根"、"无实数根" 等表述分别对应这三种情况。

7. Graphical Solutions | 图像解法

The roots of a quadratic equation correspond to the x-intercepts of its graph y = ax² + bx + c. The graph is a parabola with a vertical axis of symmetry.

二次方程的根对应其图像 y = ax² + bx + c 的 x 截距。该图像是抛物线，具有竖直的对称轴。

The vertex (turning point) can be found by completing the square: the equation y = a(x – h)² + k has vertex (h, k). Alternatively, the x-coordinate of the vertex is given by x = -b/(2a), and the axis of symmetry is the vertical line through that point.

顶点（转向点）可以通过配方法找到：y = a(x – h)² + k 的顶点为 (h, k)。另外，顶点坐标的 x 值可以用 x = -b/(2a) 计算，对称轴就是经过该点的竖直线。

If a > 0, the parabola opens upwards and the vertex is a minimum point; if a < 0, it opens downwards and the vertex is a maximum point. This is often tested alongside inequality questions.

如果 a > 0，抛物线开口向上，顶点是最小值点；如果 a < 0，开口向下，顶点是最大值点。这一点经常与不等式问题结合考查。

8. Word Problems | 应用题

Many exam questions require you to form and solve a quadratic equation from a given context. For example: “The area of a rectangle is 20 cm². Its length is 4 cm more than its width. Find the width.”

许多考试题目要求你从给定情境中建立并求解二次方程。例如：”一个矩形的面积是 20 平方厘米。它的长比宽多 4 厘米。求宽。”

Let the width be x cm. Then the length is (x + 4) cm. The area gives:

设宽为 x 厘米。则长为 (x + 4) 厘米。由面积得：

x(x + 4) = 20 → x² + 4x – 20 = 0

Using the quadratic formula:

使用二次公式：

x = (-4 ± √(16 + 80)) / 2 = (-4 ± √96) / 2 = (-4 ± 4√6) / 2 = -2 ± 2√6

Since the width must be positive, x = -2 + 2√6 ≈ 2.90 cm. Always reject negative solutions when the context requires positive values, and state the final answer with the correct unit.

由于宽度必须为正数，所以 x = -2 + 2√6 ≈ 2.90 厘米。如果题意要求正值，必须舍去负数解，并带上正确单位写出最终答案。

9. Common Mistakes and Exam Tips | 常见错误与考试建议

Always rearrange the equation into the standard form ax² + bx + c = 0 before solving.
Do not forget the ± sign when using square roots or the quadratic formula.
Check your answers by substitution — this takes only a few seconds.
If factorisation is difficult, switch to the quadratic formula immediately to save time.
For completing the square, make sure the coefficient of x² is 1 before you begin.
Read discriminant questions carefully: “equal roots” means Δ = 0, “no real roots” means Δ < 0.

解题前务必将方程整理为标准形式 ax² + bx + c = 0。
使用平方根或二次公式时不要忘记 ± 号。
通过代入验算检查答案——只需几秒钟。
如果因式分解困难，立即改用二次公式以节省时间。
配方时，确保 x² 的系数为 1 再开始。
仔细审读判别式题目：”等根”表示 Δ = 0，”无实数根”表示 Δ < 0。

By mastering these five methods — factorisation, difference of two squares, completing the square, the quadratic formula, and the discriminant — you will be well-prepared for any quadratic equation question on your IGCSE exam.

通过掌握这五种方法——因式分解、平方差、配方法、二次公式和判别式——你将能够从容应对 IGCSE 考试中任何关于二次方程的题目。

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Solving Quadratic Equations | 解二次方程