Solving Quadratic Equations | 二次方程解法详解

📚 Solving Quadratic Equations | 二次方程解法详解

Quadratic equations appear in nearly every IGCSE Mathematics paper. They test your algebraic manipulation, your understanding of graphs, and your ability to apply logical methods. Mastering them is not optional — it is essential for a high grade.

二次方程几乎出现在每一份 IGCSE 数学试卷中。它考查代数变形能力、对图像的理解以及运用逻辑方法的能力。熟练掌握二次方程不是可选项，而是冲击高分的关键。

1. Standard Form of a Quadratic Equation | 二次方程的标准形式

A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. The coefficient a is called the leading coefficient, b is the linear coefficient, and c is the constant term.

二次方程是指可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为实数，且 a ≠ 0。系数 a 称为首项系数，b 称为一次项系数，c 称为常数项。

ax² + bx + c = 0

For example, 2x² + 5x − 3 = 0 is a quadratic equation, while x³ − 4x = 0 is not because the highest power of x is 3. In IGCSE exams, you must first rearrange any given equation into standard form before applying a solution method.

例如，2x² + 5x − 3 = 0 是二次方程，而 x³ − 4x = 0 不是，因为 x 的最高次数是 3。在 IGCSE 考试中，你需要先将给定方程整理成标准形式，再选择解法。

2. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic expression factorises cleanly. The idea is to rewrite ax² + bx + c as a product of two binomials, then use the zero product property: if pq = 0, then p = 0 or q = 0.

因式分解法在二次表达式能干净地分解时是最快的方法。其核心是把 ax² + bx + c 改写为两个二项式的乘积，然后利用零乘积性质：若 pq = 0，则 p = 0 或 q = 0。

Consider x² + 7x + 12 = 0. We look for two numbers that multiply to 12 and add to 7. Those numbers are 3 and 4, so we write:

考虑方程 x² + 7x + 12 = 0。我们寻找两个数，它们相乘得 12，相加得 7。这两个数是 3 和 4，因此可写成：

(x + 3)(x + 4) = 0

Using the zero product property, either x + 3 = 0 or x + 4 = 0, giving x = −3 or x = −4. Always check your answers by substituting them back into the original equation.

利用零乘积性质，要么 x + 3 = 0，要么 x + 4 = 0，因此 x = −3 或 x = −4。求得解后务必代回原方程验算。

3. The Method of Completing the Square | 配方法

Completing the square transforms a quadratic into the form a(x + p)² + q. This method is particularly useful when factorisation is not obvious, and it is also the foundation for deriving the quadratic formula.

配方法将二次式转化为 a(x + p)² + q 的形式。当因式分解不明显时，这个方法尤其有用，它也是推导求根公式的基础。

Take x² + 6x + 5 = 0. Half of 6 is 3, so we write (x + 3)² and adjust:

以 x² + 6x + 5 = 0 为例。6 的一半是 3，因此写成 (x + 3)² 并进行修正：

(x + 3)² − 9 + 5 = 0

(x + 3)² = 4

Taking square roots gives x + 3 = ±2, so x = −1 or x = −5. Remember: when you take a square root, you must include both the positive and negative results.

两边开平方得 x + 3 = ±2，因此 x = −1 或 x = −5。请记住：开平方时必须同时取正负两个结果。

4. The Quadratic Formula | 求根公式

The quadratic formula works for every quadratic equation, including those that cannot be factorised. You are expected to memorise it for the IGCSE examination.

求根公式适用于所有二次方程，包括无法因式分解的情况。IGCSE 考试中要求你必须记住这个公式。

x = (−b ± √(b² − 4ac)) / (2a)

For example, solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, c = −2. Substitute into the formula:

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

x = (−3 ± √(9 + 16)) / 4 = (−3 ± 5) / 4

Thus x = (−3 + 5)/4 = 1/2, or x = (−3 − 5)/4 = −2. The two solutions are x = 0.5 and x = −2.

因此 x = (−3 + 5)/4 = 1/2，或 x = (−3 − 5)/4 = −2。两个解为 x = 0.5 和 x = −2。

5. The Discriminant: How Many Solutions? | 判别式：解的数量如何判断

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

求根公式中的 b² − 4ac 称为判别式，记作 Δ。它无需完整求解，就能告诉我们二次方程实根的个数。

If Δ > 0, there are two distinct real roots.
如果 Δ > 0，方程有两个不相等的实数根。
If Δ = 0, there is exactly one repeated real root.
如果 Δ = 0，方程有两个相等的实数根（即一个重根）。
If Δ < 0, there are no real roots.
如果 Δ < 0，方程没有实数根。

For the equation 2x² + 5x + 6 = 0, we have Δ = 25 − 48 = −23 < 0. This means the curve y = 2x² + 5x + 6 never crosses the x-axis.

对于方程 2x² + 5x + 6 = 0，计算得 Δ = 25 − 48 = −23 < 0。这意味着曲线 y = 2x² + 5x + 6 与 x 轴没有交点。

6. Graphs of Quadratic Functions | 二次函数的图像

The graph of a quadratic function y = ax² + bx + c is a parabola. When a > 0, the parabola opens upward, like a U shape. When a < 0, it opens downward, like an inverted U.

二次函数 y = ax² + bx + c 的图像是一条抛物线。当 a > 0 时，抛物线开口向上，呈 U 形；当 a < 0 时，抛物线开口向下，呈倒 U 形。

Vertex form: y = a(x − h)² + k

In this vertex form, the point (h, k) is the vertex of the parabola — the highest point if a < 0, or the lowest point if a > 0. The line x = h is the axis of symmetry.

在顶点式中，(h, k) 即为抛物线的顶点——当 a < 0 时是最高点，当 a > 0 时是最低点。直线 x = h 就是对称轴。

7. Finding the Vertex and Intercepts | 求顶点与交点

To find the vertex from standard form, use x = −b/(2a) to locate the axis of symmetry, then substitute back into the equation to find the y-coordinate of the vertex.

从标准形式求顶点时，先用公式 x = −b/(2a) 求出对称轴位置，再代回原方程得到顶点的纵坐标。

The y-intercept is easy: it is simply c, the value when x = 0. The x-intercepts, if they exist, are exactly the roots of the equation ax² + bx + c = 0.

求 y 截距很简单：它就是 c，即 x = 0 时的值。x 截距如果存在，恰好就是方程 ax² + bx + c = 0 的根。

Consider y = x² − 4x + 3. The axis of symmetry is x = −(−4)/(2 × 1) = 2. Substituting x = 2 gives y = 4 − 8 + 3 = −1, so the vertex is (2, −1). The y-intercept is 3, and the roots are x = 1 and x = 3, so the x-intercepts are (1, 0) and (3, 0).

考虑 y = x² − 4x + 3。对称轴为 x = −(−4)/(2 × 1) = 2。代 x = 2 入方程得 y = 4 − 8 + 3 = −1，所以顶点为 (2, −1)。y 截距为 3，根为 x = 1 和 x = 3，所以 x 截距为 (1, 0) 和 (3, 0)。

8. Real-World Applications | 实际应用问题

Quadratic equations appear in many real-world contexts, and IGCSE problems often frame them in worded questions involving area, projectile motion or profit maximisation.

二次方程出现在许多现实场景中，IGCSE 的题目常以面积、抛体运动或利润最大化等文字应用题的形式出现。

Example: A rectangular garden has a length 3 metres longer than its width, and its area is 40 m². If the width is w metres, then w(w + 3) = 40, which expands to w² + 3w − 40 = 0. Solving gives w = 5 or w = −8. Since width cannot be negative, the width is 5 m and the length is 8 m.

例如：一块矩形花园的长比宽多 3 米，面积为 40 平方米。若宽为 w 米，则 w(w + 3) = 40，展开得 w² + 3w − 40 = 0。解得 w = 5 或 w = −8。由于宽度不能为负，因此宽为 5 米，长为 8 米。

Always check whether solutions make sense in the context of the problem. Discard any solutions that give negative lengths, distances or quantities when the context requires positive values.

务必检查解在题目语境中是否合理。当实际情境要求正数值时，要舍去产生负长度、负距离或负数量的解。

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

Many students lose marks on quadratic questions for predictable reasons. Below are the most frequent errors, along with strategies to avoid them.

很多学生在二次方程题目上丢分，原因往往是可以预见的。下面列出最常见的错误以及规避策略。

Forgetting to set the equation to zero before factorising or using the formula.
在因式分解或使用公式前忘记将方程化为等于零的形式。
Dropping the ± sign when taking square roots.
在开平方时遗漏±号。
Incorrectly substituting negative values of b and c into the quadratic formula.
在代入求根公式时，对 b 和 c 的负号处理出错。
Miscalculating the discriminant, especially when b is negative.
计算判别式出错，尤其是当 b 为负数时。
Equating x + 2 squared to x² + 4 instead of x² + 4x + 4.
将 (x + 2)² 误写成 x² + 4，而不是 x² + 4x + 4。

To avoid these errors, slow down when substituting into the formula, write down every step, and use a mental checklist: is the equation in standard form? Did I include both ± solutions? Do my answers satisfy the original equation?

要避免这些错误，请在代入公式时放慢速度，写下每一步，并建议一个心理检查清单：方程是否已经是标准形式？是否包含了正负两个解？求出的答案是否满足原方程？

10. Choosing the Best Method | 如何选择最优解法

In the IGCSE examination, time management is critical. Choosing the right method can save minutes and reduce errors.

在 IGCSE 考试中，时间管理至关重要。选择合适的方法可以节省几分钟时间并降低出错率。

Method	Best used when
Factorisation	Expression factorises easily, small coefficients
Completing the square	When asked for vertex form, or a = 1
Quadratic formula	Any equation, especially when a ≠ 1 or roots are not integers
Graphical method	When the question explicitly asks to estimate roots

方法	适用情形
因式分解法	表达式容易分解，系数较小时
配方法	题目要求顶点式，或 a = 1 时
求根公式	任何方程，尤其当 a ≠ 1 或根不为整数时
图像法	题目明确要求估算根时

As a rule of thumb: try factorisation first. If you cannot spot the factors within about ten seconds, switch to the quadratic formula. In problems asking about the vertex or axis of symmetry, completing the square is usually the most direct route.

经验法则：优先尝试因式分解。如果 10 秒左右看不出因式组合，就改用求根公式。当题目要求顶点或对称轴时，配方法通常是最直接的路径。

11. Working with the Sum and Product of Roots | 根的和与积

A useful property connects the coefficients of a quadratic to its roots. If the roots of the equation ax² + bx + c = 0 are α and β, then:

二次方程的系数与根之间存在一个非常有用的关系。若方程 ax² + bx + c = 0 的根为 α 和 β，则：

Sum of roots: α + β = −b / a

Product of roots: α × β = c / a

For instance, if a quadratic has roots 2 and −5, then the sum is −3 and the product is −10, so the equation can be written as x² + 3x − 10 = 0 after multiplying by a = 1. This technique can quickly check your answers or construct quadratics with given roots.

例如，若一个二次方程的根为 2 和 −5，则根之和为 −3，根之积为 −10，因此方程可以写成 x² + 3x − 10 = 0（取 a = 1）。这个技巧既能快速验算答案，也能用于构造指定根的二次方程。

If a is not 1, multiply the entire expression by the desired value of a. For example, if a = 2, the equation becomes 2x² + 6x − 20 = 0, which is equivalent to the same roots.

若 a 不为 1，只需将整个表达式乘以所需的 a 值。例如，若 a = 2，方程变为 2x² + 6x − 20 = 0，它仍然保持同样的根。

12. Revision Checklist | 复习清单

Before the exam, use this checklist to confirm that you have mastered every skill related to quadratic equations.

考试前，请使用这份清单确认你已经掌握与二次方程相关的每项技能。

I can rearrange any equation into the standard form ax² + bx + c = 0.
我可以将任意方程化为标准形式 ax² + bx + c = 0。
I can factorise quadratics of the form x² + bx + c.
我可以分解形如 x² + bx + c 的二次式。
I can factorise quadratics of the form ax² + bx + c where a ≠ 1.
我可以分解形如 ax² + bx + c（a ≠ 1）的二次式。
I can complete the square and write the vertex form of a quadratic.
我会使用配方法并写出二次式的顶点形式。
I can apply the quadratic formula correctly, including negative coefficients.
我能正确运用求根公式，包括负系数的情况。
I can use the discriminant Δ to determine the number of real roots.
我会利用判别式 Δ 判断实数根的个数。
I can sketch the graph of a quadratic using the vertex, axis and intercepts.
我会利用顶点、对称轴和交点来绘制二次函数的草图。
I can solve word problems involving quadratic equations.
我能解决与二次方程相关的文字应用题。

Work through past papers and time yourself. In the real exam, accuracy beats speed — but both are achievable with consistent practice.

认真过一遍历年真题并给自己计时。在真正的考试中，准确率比速度更重要——但通过持续练习，两者都可以兼得。

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Solving Quadratic Equations | 二次方程解法详解