Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

A quadratic equation is a polynomial equation of degree 2, written in the general form \( ax^2 + bx + c = 0 \) (but without LaTeX, we write ax² + bx + c = 0). In the Edexcel IGCSE syllabus, mastering the solution of quadratic equations is essential because it appears in pure mathematics, algebra, graphs, and real‑life problem solving. This article provides a complete guide to the methods you need, with worked examples and common pitfalls.

二次方程是次数为 2 的多项式方程，一般形式为 ax² + bx + c = 0。在 Edexcel IGCSE 考纲中，掌握二次方程的解法至关重要，因为它出现在纯数学、代数、函数图像以及实际应用题中。本文将系统讲解所需的全部解法，配以例题和常见错误提醒。

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

A quadratic equation contains an x² term and can be written in the standard form ax² + bx + c = 0, where a, b, c are constants and a ≠ 0. The highest power of x is 2. Examples include 2x² – 5x + 3 = 0 and x² – 9 = 0.

二次方程含有 x² 项，可以写成标准形式 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。未知数 x 的最高次数为 2。例如 2x² – 5x + 3 = 0 和 x² – 9 = 0。

If a = 0, the equation becomes linear. A quadratic equation can have two real roots, one repeated root, or no real roots, depending on the value of the discriminant.

若 a = 0，方程退化为一次方程。二次方程可以有两个实数根、一个重根，或者没有实数根，这取决于判别式的值。

2. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the expression can be written as a product of two linear factors. We use the rule: if p × q = 0, then p = 0 or q = 0. For example, solve x² + 5x + 6 = 0.

当表达式可以写成两个一次因式的乘积时，因式分解是最快捷的方法。我们利用规则：若 p × q = 0，则 p = 0 或 q = 0。例如，解 x² + 5x + 6 = 0。

We look for two numbers that multiply to 6 and add to 5, which are 2 and 3. So x² + 5x + 6 = (x + 2)(x + 3) = 0. Hence x + 2 = 0 or x + 3 = 0, giving x = –2 or x = –3.

我们找两个数，它们的乘积为 6，和为 5，即 2 和 3。所以 x² + 5x + 6 = (x + 2)(x + 3) = 0。于是 x + 2 = 0 或 x + 3 = 0，得到 x = –2 或 x = –3。

When the coefficient of x² is not 1, use the “ac” method: multiply a and c, find two factors whose sum is b, then split the middle term and factor by grouping. For example, solve 2x² + 7x + 3 = 0.

当 x² 的系数不为 1 时，使用“ac”法：将 a 和 c 相乘，找到两个因数，其和为 b，然后拆中项并分组因式分解。例如，解 2x² + 7x + 3 = 0。

a = 2, c = 3 ⇒ ac = 6. Factors of 6 that add to 7 are 1 and 6. Write 7x as 1x + 6x.

2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3) = 0.

Thus 2x + 1 = 0 or x + 3 = 0, so x = –½ or x = –3. Always check your factors by expanding.

因此 2x + 1 = 0 或 x + 3 = 0，所以 x = –½ 或 x = –3。务必通过展开来检验因式是否正确。

3. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form p(x + q)² + r. This method is useful for solving equations that do not factorise easily, and it also reveals the turning point of the graph. The key formula is:

配方法将二次式重写为 p(x + q)² + r 的形式。此方法适用于不易因式分解的方程，同时还能揭示图像的顶点。关键公式是：

x² + bx = (x + b/2)² – (b/2)²

To solve x² + 6x – 7 = 0, we complete the square on x² + 6x: add and subtract (6/2)² = 9. This gives (x + 3)² – 9 – 7 = 0, i.e. (x + 3)² – 16 = 0.

解 x² + 6x – 7 = 0，对 x² + 6x 配方：加上并减去 (6/2)² = 9。得到 (x + 3)² – 9 – 7 = 0，即 (x + 3)² – 16 = 0。

Now isolate the square: (x + 3)² = 16. Taking square roots gives x + 3 = ±4. Hence x = 1 or x = –7.

然后分离平方项：(x + 3)² = 16。开平方得 x + 3 = ±4。因此 x = 1 或 x = –7。

For a quadratic with a ≠ 1, first factor out a from the x² and x terms. For example, 2x² – 8x + 1 = 0 ⇒ 2(x² – 4x) + 1 = 0. Complete the square inside: 2[(x – 2)² – 4] + 1 = 0 ⇒ 2(x – 2)² – 8 + 1 = 0 ⇒ 2(x – 2)² – 7 = 0.

当 a ≠ 1 时，先从 x² 和 x 项提取 a。例如，2x² – 8x + 1 = 0 ⇒ 2(x² – 4x) + 1 = 0。在括号内配方：2[(x – 2)² – 4] + 1 = 0 ⇒ 2(x – 2)² – 8 + 1 = 0 ⇒ 2(x – 2)² – 7 = 0。

Then (x – 2)² = 7/2 ⇒ x = 2 ± √(7/2)

此式给出了精确的根，通常写成 x = 2 ± √(7/2)。这种形式在考试中是被接受的。

4. The Quadratic Formula | 二次公式

The quadratic formula works for every quadratic equation. For ax² + bx + c = 0, the solutions are given by:

二次公式适用于所有二次方程。对于 ax² + bx + c = 0，解由下式给出：

x = (–b ± √(b² – 4ac)) / (2a)

This formula is derived by completing the square. You should memorise it and use it whenever factorisation is difficult or impossible. For example, solve 3x² – 5x + 1 = 0.

此公式由配方法推导而来。你应该记住它，并在因式分解困难或不可能时使用。例如，解 3x² – 5x + 1 = 0。

Here a = 3, b = –5, c = 1. Substitute into the formula:

这里 a = 3，b = –5，c = 1。代入公式：

x = (5 ± √(25 – 12)) / 6 = (5 ± √13) / 6

The two roots are x = (5 + √13)/6 and x = (5 – √13)/6. Leave the answer in surd form for exact values.

两个根为 x = (5 + √13)/6 和 x = (5 – √13)/6。若要求精确值，应保留根号形式。

5. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root: Δ = b² – 4ac. Its value determines the nature of the roots without solving the equation.

判别式是二次公式根号下的部分：Δ = b² – 4ac。它的值决定了方程的根的性质，无需解方程。

If b² – 4ac > 0: two distinct real roots.

若 b² – 4ac > 0：有两个不相等的实数根。
If b² – 4ac = 0: one repeated real root (the graph touches the x‑axis).

若 b² – 4ac = 0：有一个重根（图像与 x 轴相切）。
If b² – 4ac < 0: no real roots (the graph does not cross the x‑axis).

若 b² – 4ac < 0：没有实数根（图像不与 x 轴相交）。

Example: For what values of k does x² + 4x + k = 0 have two distinct real roots? Here b² – 4ac = 16 – 4k. We require 16 – 4k > 0 ⇒ k < 4.

例：k 为何值时，x² + 4x + k = 0 有两个不相等的实数根？这里 b² – 4ac = 16 – 4k。需满足 16 – 4k > 0 ⇒ k < 4。

If the discriminant is a perfect square and the equation has rational coefficients, the quadratic factorises over the integers; otherwise the roots are surds.

若判别式是完全平方数，且方程系数为有理数，则二次式可在整数范围内因式分解；否则根为无理式（根号形式）。

6. Forming a Quadratic from its Roots | 由根构造二次方程

If the roots of a quadratic equation are α and β, then the equation can be written as (x – α)(x – β) = 0. Expanding this gives x² – (α + β)x + αβ = 0. This shows the sum and product of roots directly:

若二次方程的两个根为 α 和 β，则方程可写成 (x – α)(x – β) = 0。展开得到 x² – (α + β)x + αβ = 0。这直接显示了根的和与积：

Sum of roots = –b/a, Product of roots = c/a

根的和 = –b/a，根的积 = c/a

For example, find a quadratic equation whose roots are 3 and –5. The sum is 3 + (–5) = –2, and the product is 3 × (–5) = –15. Hence the equation is x² – (–2)x + (–15) = 0, i.e. x² + 2x – 15 = 0.

例如，求一个二次方程，其根为 3 和 –5。根的和为 3 + (–5) = –2，根的积为 3 × (–5) = –15。因此方程为 x² – (–2)x + (–15) = 0，即 x² + 2x – 15 = 0。

This method is particularly useful in coordinate geometry and when dealing with intersections of curves and lines.

此方法在坐标几何和曲线与直线交点问题中特别有用。

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

The graph of y = ax² + bx + c is a parabola. If a > 0 it opens upwards; if a < 0 it opens downwards. The solutions of ax² + bx + c = 0 are the x‑intercepts of the parabola.

函数 y = ax² + bx + c 的图像是抛物线。若 a > 0，开口向上；若 a < 0，开口向下。方程 ax² + bx + c = 0 的解就是抛物线与 x 轴交点的横坐标。

The x‑coordinate of the vertex (turning point) is x = –b/(2a). Substituting this value into the equation gives the y‑coordinate. For example, for y = x² – 4x + 1, the vertex is at x = –(–4)/(2×1) = 2, y = 4 – 8 + 1 = –3, so the turning point is (2, –3).

顶点（转折点）的 x 坐标为 x = –b/(2a)。将此值代入方程得到 y 坐标。例如，对于 y = x² – 4x + 1，顶点 x = –(–4)/(2×1) = 2，y = 4 – 8 + 1 = –3，所以转折点为 (2, –3)。

When the quadratic is written in completed‑square form y = a(x – h)² + k, the vertex is immediately (h, k). The line x = h is the axis of symmetry.

当二次式写成配方法形式 y = a(x – h)² + k 时，顶点直接就是 (h, k)。直线 x = h 是对称轴。

8. Vertex and Line of Symmetry | 顶点与对称轴

The axis of symmetry of a parabola is a vertical line that passes through the vertex. It divides the parabola into two mirror images. Its equation is x = –b/(2a).

抛物线的对称轴是经过顶点的竖直直线，它将抛物线分为左右对称的两部分。其方程为 x = –b/(2a)。

For y = 2x² – 12x + 5, the axis of symmetry is x = –(–12)/(2×2) = 3. To find the vertex, substitute x = 3: y = 2(9) – 36 + 5 = 18 – 36 + 5 = –13. So the vertex is (3, –13).

对于 y = 2x² – 12x + 5，对称轴为 x = –(–12)/(2×2) = 3。求顶点需将 x = 3 代入：y = 2(9) – 36 + 5 = 18 – 36 + 5 = –13。所以顶点是 (3, –13)。

You can also complete the square to find the vertex: y = 2(x² – 6x) + 5 = 2[(x – 3)² – 9] + 5 = 2(x – 3)² – 18 + 5 = 2(x – 3)² – 13.

你也可以用配方法求顶点：y = 2(x² – 6x) + 5 = 2[(x – 3)² – 9] + 5 = 2(x – 3)² – 18 + 5 = 2(x – 3)² – 13。读出的顶点坐标同样是 (3, –13)。

9. Solving Quadratic Inequalities | 解二次不等式

Quadratic inequalities such as x² – 5x + 6 < 0 are often tested in IGCSE. To solve, first factorise: (x – 2)(x – 3) < 0. The critical values are 2 and 3. Since the graph of y = (x – 2)(x – 3) opens upwards, the expression is negative between the roots.

IGCSE 常考二次不等式，例如 x² – 5x + 6 < 0。首先因式分解：(x – 2)(x – 3) < 0。临界值为 2 和 3。因为 y = (x – 2)(x – 3) 的图像开口向上，在两根之间表达式的值为负。

Thus the solution is 2 < x < 3.

因此解为 2 < x < 3。

For x² – 5x + 6 > 0, the solution is x < 2 or x > 3. Always sketch a quick graph or use a sign table to avoid mistakes.

对于 x² – 5x + 6 > 0，解为 x < 2 或 x > 3。解题时快速画草图或使用符号表可避免出错。

10. Quadratic Equations in Context | 二次方程的应用题

Many exam problems require you to form a quadratic equation from a real‑life situation. For example, the area of a rectangle is 30 cm², its length is 7 cm more than its width. Find the dimensions.

许多考试题需要从实际情境建立二次方程。例如，一个长方形的面积为 30 cm²，长比宽多 7 cm，求它的尺寸。

Let the width be x cm. Then the length is (x + 7) cm. The area equation is x(x + 7) = 30 ⇒ x² + 7x – 30 = 0. Factorise: (x + 10)(x – 3) = 0, so x = –10 or x = 3. Since a length cannot be negative, the width is 3 cm and the length is 10 cm.

设宽为 x cm，则长为 (x + 7) cm。面积方程为 x(x + 7) = 30 ⇒ x² + 7x – 30 = 0。因式分解得 (x + 10)(x – 3) = 0，所以 x = –10 或 x = 3。因为长度不能为负，所以宽为 3 cm，长为 10 cm。

Projectile motion problems also use quadratics. For example, the height h metres of a ball after t seconds is h = 20t – 5t². When does the ball hit the ground? Set h = 0 ⇒ 5t(4 – t) = 0 ⇒ t = 0 or t = 4. It hits the ground after 4 seconds.

抛体运动问题也使用二次方程。例如，一个球在 t 秒后的高度 h 米满足 h = 20t – 5t²。球何时落地？令 h = 0 ⇒ 5t(4 – t) = 0 ⇒ t = 0 或 t = 4。它在 4 秒后落地。

11. Common Mistakes and Tips | 常见错误与技巧

Mistake 1: Forgetting that a ≠ 0 in a quadratic equation. Always check the coefficient of x² before solving.

错误 1：忘记二次方程中 a ≠ 0。解题前务必检查 x² 的系数。

Mistake 2: When factorising, sign errors. Double‑check by expanding your factors.

错误 2：因式分解时符号出错。通过展开因式来检验。

Mistake 3: Forgetting to take both positive and negative square roots when using completing the square or the quadratic formula.

错误 3：在使用配方法或二次公式时忘记取正负两个平方根。

Mistake 4: Discarding a negative root without a valid reason, for example when solving for length, time, or other physical quantities.

错误 4：在求解长度、时间等物理量时，无充分理由舍弃负根。

Tip: Always rearrange the equation to the standard form ax² + bx + c = 0 before using any method. If the equation looks too hard to factorise, check the discriminant first; if it is a perfect square, factorisation is possible.

技巧：使用任何方法前，先将方程整理为标准形式 ax² + bx + c = 0。若方程难以直接因式分解，先计算判别式；若其为完全平方数，则必定可以因式分解。

12. Practice Questions | 练习与总结

1. Solve x² – 7x + 12 = 0. (Answer: x = 3 or 4)

1. 解 x² – 7x + 12 = 0。（答案：x = 3 或 4）

2. Solve 2x² + 5x – 3 = 0 by factorisation. (Answer: x = ½ or –3)

2. 用因式分解法解 2x² + 5x – 3 = 0。（答案：x = ½ 或 –3）

3. Solve x² + 6x + 1 = 0 by completing the square. (Answer: x = –3 ± 2√2)

3. 用配方法解 x² + 6x + 1 = 0。（答案：x = –3 ± 2√2）

4. Find the value of p so that the equation px² – 8x + p = 0 has equal roots. (Answer: p = ±4, because discriminant 64 – 4p² = 0)

4. 求 p 的值，使方程 px² – 8x + p = 0 有相等的根。（答案：p = ±4，因为判别式 64 – 4p² = 0）

5. The product of two consecutive positive integers is 156. Find them. (Answer: 12 and 13, from x(x + 1) = 156)

5. 两个连续正整数的乘积为 156，求这两个数。（答案：12 和 13，由 x(x + 1) = 156 解得）

In summary, choose the most efficient method: factorise when possible, otherwise use the quadratic formula or completing the square. Understand the discriminant to anticipate the nature of the roots. With regular practice, quadratic equations become trivial.

总结：选择最有效的方法：能因式分解就分解，否则用二次公式或配方法。理解判别式以预判根的性质。通过定期练习，二次方程将不再困难。

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Solving Quadratic Equations | 解二次方程