Mastering Quadratic Equations | 掌握二次方程

📚 Mastering Quadratic Equations | 掌握二次方程

Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in almost every paper, whether as direct solving questions, graph-sketching tasks, or word problems. Understanding the different methods of solving quadratics — factorisation, completing the square, and the quadratic formula — is essential for achieving top marks.

二次方程是 IGCSE 数学中最重要的内容之一。几乎每份试卷都会出现二次方程，无论是直接求解、绘制函数图像，还是应用题。掌握因式分解法、配方法和求根公式这三种解法，是取得高分的关键。

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

A quadratic equation is any equation of the form ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0. The reason a cannot be zero is simple: if a = 0, the equation becomes linear (bx + c = 0), and it no longer has a squared term.

二次方程是指形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。a 不能为零的原因很简单：如果 a = 0，方程就变成了一次方程（bx + c = 0），不再含有平方项。

The highest power of x in a quadratic equation is 2, which means a quadratic equation has at most two solutions. These solutions are also called ‘roots’ of the equation.

二次方程中 x 的最高次数是 2，这意味着二次方程最多有两个解。这些解也称为方程的”根”。

ax² + bx + c = 0 (a ≠ 0)

For example, 2x² + 5x − 3 = 0 is a quadratic equation with a = 2, b = 5 and c = −3. Meanwhile, x² − 9 = 0 is also quadratic, with b = 0.

例如，2x² + 5x − 3 = 0 是一个二次方程，其中 a = 2，b = 5，c = −3。而 x² − 9 = 0 也是二次方程，此时 b = 0。

2. Solving by Factorisation | 因式分解法

Factorisation is often the quickest method for solving quadratics, but it only works when the equation has rational roots. The underlying principle is the zero product property: if the product of two factors is zero, then at least one of the factors must be zero.

因式分解通常是解二次方程最快的方法，但仅当方程有有理数根时才适用。其核心原理是零乘积性质：如果两个因式的乘积为零，那么至少有一个因式必须为零。

Step-by-step process | 步骤详解

Step 1: Rearrange the equation so that one side equals zero.

第一步：整理方程，使一边等于零。

Step 2: Factorise the quadratic expression into the form (px + q)(rx + s).

第二步：将二次表达式因式分解为 (px + q)(rx + s) 的形式。

Step 3: Set each factor equal to zero and solve the resulting linear equations.

第三步：令每个因式等于零，并解所得的一次方程。

Consider the equation x² − 5x + 6 = 0. We look for two numbers that multiply to 6 and add to −5. These numbers are −2 and −3. Therefore the equation factorises as (x − 2)(x − 3) = 0. Setting each factor to zero gives x = 2 or x = 3.

考虑方程 x² − 5x + 6 = 0。我们需要找到两个数，它们相乘得 6，相加得 −5。这两个数是 −2 和 −3。因此方程因式分解为 (x − 2)(x − 3) = 0。令每个因式为零，得到 x = 2 或 x = 3。

For equations where the coefficient of x² is not 1, such as 2x² + 7x + 3 = 0, we use a slightly different approach. Multiply a and c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7: these are 1 and 6. Rewrite the middle term: 2x² + x + 6x + 3 = 0, then factor by grouping: x(2x + 1) + 3(2x + 1) = 0, giving (2x + 1)(x + 3) = 0. Thus x = −½ or x = −3.

对于 x² 系数不为 1 的方程，如 2x² + 7x + 3 = 0，我们采用略有不同的方法。先将 a 和 c 相乘：2 × 3 = 6。找到两个数相乘得 6、相加得 7：这两个数是 1 和 6。改写中间项：2x² + x + 6x + 3 = 0，然后分组因式分解：x(2x + 1) + 3(2x + 1) = 0，得到 (2x + 1)(x + 3) = 0。因此 x = −½ 或 x = −3。

3. Solving by Completing the Square | 配方法

Completing the square is a powerful technique that works for every quadratic equation. It involves rewriting ax² + bx + c in the form a(x + p)² + q. This method is particularly useful when factorisation is not possible, and it also helps when finding the turning point of a quadratic graph.

配方法是一种强大的技巧，适用于所有二次方程。它将 ax² + bx + c 改写为 a(x + p)² + q 的形式。当因式分解行不通时，这种方法特别有用，同时也有助于找到二次函数图像的顶点。

For a simple case where a = 1, take x² + 6x + 2 = 0. Halve the coefficient of x: 6 ÷ 2 = 3. Write (x + 3)² and subtract 3² = 9 to compensate. This gives (x + 3)² − 9 + 2 = 0, which simplifies to (x + 3)² − 7 = 0. Adding 7 to both sides and taking square roots gives x + 3 = ±√7, so x = −3 ± √7.

对于 a = 1 的简单情况，以 x² + 6x + 2 = 0 为例。将 x 的系数除以 2：6 ÷ 2 = 3。写成 (x + 3)²，并减去 3² = 9 以保持等式平衡。得到 (x + 3)² − 9 + 2 = 0，化简为 (x + 3)² − 7 = 0。两边同时加 7 并开平方：x + 3 = ±√7，所以 x = −3 ± √7。

When a ≠ 1, first factor out a from the first two terms. For example, 2x² − 8x + 5 = 0 becomes 2(x² − 4x) + 5 = 0. Complete the square inside the brackets: x² − 4x = (x − 2)² − 4. Substitute back: 2[(x − 2)² − 4] + 5 = 0, which simplifies to 2(x − 2)² − 8 + 5 = 0, giving 2(x − 2)² − 3 = 0. Then solve: (x − 2)² = 3/2, so x = 2 ± √(3/2).

当 a ≠ 1 时，先将 a 从前两项中提出。例如，2x² − 8x + 5 = 0 可写成 2(x² − 4x) + 5 = 0。在括号内配方：x² − 4x = (x − 2)² − 4。代回：2[(x − 2)² − 4] + 5 = 0，化简为 2(x − 2)² − 8 + 5 = 0，即 2(x − 2)² − 3 = 0。然后求解：(x − 2)² = 3/2，所以 x = 2 ± √(3/2)。

4. Solving by the Quadratic Formula | 求根公式法

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

求根公式是解任意二次方程的通用方法。它通过对一般形式 ax² + bx + c = 0 配方推导得出。公式如下：

x = (−b ± √(b² − 4ac)) ÷ 2a

To use the formula, simply identify the values of a, b and c from the equation, substitute them into the formula, and simplify. This method always works, even when the roots are irrational or complex.

使用公式时，只需从方程中确定 a、b、c 的值，代入公式并化简即可。这种方法永远有效，即使根是无理数或复数。

Let’s solve 3x² − 5x − 2 = 0 using the formula. Here, a = 3, b = −5 and c = −2. Substitute into the formula: x = (5 ± √((−5)² − 4 × 3 × (−2))) ÷ (2 × 3) = (5 ± √(25 + 24)) ÷ 6 = (5 ± √49) ÷ 6 = (5 ± 7) ÷ 6. This gives x = (5 + 7) ÷ 6 = 2, or x = (5 − 7) ÷ 6 = −⅓.

我们用公式法解 3x² − 5x − 2 = 0。这里 a = 3，b = −5，c = −2。代入公式：x = (5 ± √((−5)² − 4 × 3 × (−2))) ÷ (2 × 3) = (5 ± √(25 + 24)) ÷ 6 = (5 ± √49) ÷ 6 = (5 ± 7) ÷ 6。得到 x = (5 + 7) ÷ 6 = 2，或 x = (5 − 7) ÷ 6 = −⅓。

Always remember to check your final answers by substituting them back into the original equation. If both sides equal zero, your solutions are correct.

务必记得将最终答案代回原方程进行验算。如果等式两边均为零，则说明你的解是正确的。

5. The Discriminant | 判别式

The expression b² − 4ac inside the quadratic formula is called the discriminant, often denoted by Δ. The discriminant tells us the nature of the roots without solving the equation completely.

求根公式中的表达式 b² − 4ac 称为判别式，常用 Δ 表示。判别式可以告诉我们方程的根的性质，而无需完整求解。

If b² − 4ac > 0, the equation has two distinct real roots.
如果 b² − 4ac > 0，方程有两个不等实根。
If b² − 4ac = 0, the equation has exactly one repeated real root (also called a double root).
如果 b² − 4ac = 0，方程有唯一的重根（也称为二重根）。
If b² − 4ac < 0, the equation has no real roots — the roots are complex.
如果 b² − 4ac < 0，方程没有实根——根为复数。

For example, consider x² + 2x + 5 = 0. The discriminant is 2² − 4 × 1 × 5 = 4 − 20 = −16, which is negative. Therefore this equation has no real solutions. You can verify this by sketching the graph — the parabola never crosses the x-axis.

例如，考虑方程 x² + 2x + 5 = 0。判别式为 2² − 4 × 1 × 5 = 4 − 20 = −16，是负数。因此该方程没有实数解。你可以通过画图验证——抛物线永远不会与 x 轴相交。

Another common exam question asks you to find the value of k such that a quadratic has equal roots. For instance, if kx² + 4x + 2 = 0 has a double root, set the discriminant to zero: 4² − 4 × k × 2 = 0, giving 16 − 8k = 0, so k = 2.

另一种常见考试题型是求 k 的值，使得二次方程有等根。例如，若 kx² + 4x + 2 = 0 有重根，则令判别式为零：4² − 4 × k × 2 = 0，即 16 − 8k = 0，解得 k = 2。

6. The Sum and Product of Roots | 根的和与积

For a quadratic equation ax² + bx + c = 0, there are elegant relationships between the roots and the coefficients. If the roots are α (alpha) and β (beta), then:

对于二次方程 ax² + bx + c = 0，根与系数之间存在优美的关系。若方程的根为 α（阿尔法）和 β（贝塔），则：

Sum of roots: α + β = −b/a

Product of roots: α × β = c/a

These relationships are extremely useful in problems where you need to find the value of expressions involving the roots without solving the equation itself. For example, for the equation 2x² − 6x + 3 = 0, the sum of the roots is −(−6) ÷ 2 = 3, and the product is 3 ÷ 2 = 1.5.

这些关系在需要求根的对称表达式值时非常有用，无需真正解方程。例如，对于方程 2x² − 6x + 3 = 0，根的和为 −(−6) ÷ 2 = 3，根的积为 3 ÷ 2 = 1.5。

You can also use these relationships to construct a quadratic equation from its roots. If the roots are p and q, then the equation is x² − (p + q)x + pq = 0.

你也可以利用这些关系从已知的根构造二次方程。如果根为 p 和 q，则方程为 x² − (p + q)x + pq = 0。

7. Sketching Quadratic Graphs | 绘制二次函数图像

The graph of a quadratic function y = ax² + bx + c is a parabola. The sign of a determines the direction: if a > 0, the parabola opens upwards (U-shape); if a < 0, it opens downwards (n-shape).

二次函数 y = ax² + bx + c 的图像是抛物线。a 的符号决定开口方向：若 a > 0，抛物线开口向上；若 a < 0，开口向下。

To sketch a quadratic graph accurately, you need three key features:

要准确绘制二次函数图像，你需要三个关键特征：

The roots: where the curve crosses the x-axis, found by setting y = 0.
根：曲线与 x 轴的交点，令 y = 0 求得。
The y-intercept: where the curve crosses the y-axis, found by setting x = 0. This is simply the value c.
y 轴截距：曲线与 y 轴的交点，令 x = 0 求得，即常数项 c 的值。
The vertex (turning point): the minimum or maximum point of the parabola.
顶点（极值点）：抛物线的最低点或最高点。

For a quadratic written in completed square form y = a(x + p)² + q, the vertex is at (−p, q). For example, y = 2(x − 1)² + 3 has its vertex at (1, 3). The line of symmetry is the vertical line x = −p.

对于配方形式的二次函数 y = a(x + p)² + q，顶点在 (−p, q)。例如，y = 2(x − 1)² + 3 的顶点在 (1, 3)。对称轴是竖直线 x = −p。

If completing the square is not performed, you can find the x-coordinate of the vertex using the formula x = −b ÷ (2a), then substitute back to find the y-coordinate.

如果未进行配方，可以用公式 x = −b ÷ (2a) 求得顶点的 x 坐标，然后代回原式求 y 坐标。

8. Quadratic Inequalities | 二次不等式

Quadratic inequalities are a natural extension of quadratic equations. The principle is similar — first solve the corresponding equation to find critical values, then use a number line to determine where the inequality holds.

二次不等式是二次方程的自然延伸。解法原理类似——先解对应的方程找到临界值，然后利用数轴确定不等式成立的范围。

Consider x² − x − 6 > 0. First solve x² − x − 6 = 0, which factorises to (x − 3)(x + 2) = 0, giving x = 3 or x = −2. These are the critical values. Now test intervals: for x < −2, (x − 3)(x + 2) is positive; for −2 < x < 3, the product is negative; for x > 3, the product is positive again. Therefore the solution is x < −2 or x > 3.

考虑 x² − x − 6 > 0。先解 x² − x − 6 = 0，因式分解为 (x − 3)(x + 2) = 0，得到 x = 3 或 x = −2。这些是临界值。然后测试区间：当 x < −2 时，(x − 3)(x + 2) 为正；当 −2 < x < 3 时，乘积为负；当 x > 3 时，乘积再次为正。因此解为 x < −2 或 x > 3。

Be careful with the direction of the inequality: ‘greater than zero’ yields two separate regions, while ‘less than zero’ yields the region between the two critical values. For x² − x − 6 < 0, the solution would be −2 < x < 3.

注意不等式方向的不同：”>0″给出两个分离的区间，而”<0″给出两个临界值之间的区间。对于 x² − x − 6 < 0，解为 −2 < x < 3。

9. Applications of Quadratic Equations | 二次方程的应用

Quadratic equations appear frequently in real-world contexts. The most common applications in IGCSE exams include projectile motion, area problems, and economic models.

二次方程经常出现在实际应用中。IGCSE 考试中最常见的应用包括抛体运动、面积问题和经济学模型。

Area problem example | 面积问题示例

A rectangular garden has length 3 metres more than its width. The area of the garden is 70 m². Find the dimensions. Let the width be x metres; then the length is x + 3 metres. The area equation is x(x + 3) = 70, which expands to x² + 3x − 70 = 0. Factorise: (x + 10)(x − 7) = 0, giving x = 7 or x = −10. Since width cannot be negative, the width is 7 m and the length is 10 m.

一个长方形花园的长比宽多 3 米。花园面积为 70 平方米。求长方形的尺寸。设宽为 x 米，则长为 x + 3 米。面积方程为 x(x + 3) = 70，展开得 x² + 3x − 70 = 0。因式分解：(x + 10)(x − 7) = 0，得到 x = 7 或 x = −10。由于宽度不能为负，所以宽为 7 米，长为 10 米。

Projectile motion example | 抛体运动示例

A ball is thrown upwards with its height h (in metres) given by h = 20t − 5t², where t is time in seconds. To find when the ball hits the ground, set h = 0: 20t − 5t² = 0, which factorises to 5t(4 − t) = 0. Thus t = 0 or t = 4. The ball hits the ground after 4 seconds.

一个球被向上抛出，其高度 h（米）由 h = 20t − 5t² 给出，其中 t 为时间（秒）。求球何时落地：令 h = 0，即 20t − 5t² = 0，因式分解为 5t(4 − t) = 0。因此 t = 0 或 t = 4。球在 4 秒后落地。

In all application problems, pay careful attention to whether each solution is valid in the given context. A negative length, a negative time, or any value that does not make physical sense should be rejected.

在所有的应用题中，务必仔细检查每个解在具体情境中是否有效。负长度、负时间或任何不符合物理意义的数值都应舍去。

10. Common Mistakes and Exam Tips | 常见错误与应试技巧

Many students lose marks in quadratic equation questions due to avoidable errors. Here are the most frequent pitfalls and how to avoid them.

许多学生在二次方程题目中因为可避免的错误而丢分。以下是最常见的陷阱及避免方法。

Common Mistake \| 常见错误	Correction \| 正确做法
Forgetting to rearrange to ax² + bx + c = 0 before factorising	Always move all terms to one side first
忘记在因式分解前将方程整理为 ax² + bx + c = 0 的形式	始终先将所有项移到等号一侧
Losing a solution by dividing both sides by x	Factor out x instead — never divide by a variable
两边同时除以 x 而丢失一个解	应提出公因式 x——切勿除以未知数
Sign errors when substituting negative values into the quadratic formula	Use brackets carefully: (−5)² ≠ −5²
将负数代入求根公式时出现符号错误	谨慎使用括号：(−5)² ≠ −5²
Forgetting to reject invalid solutions in word problems	Check each root against the physical context
在应用题中忘记舍去无效解	将每个根与实际情境进行核对

Here are three essential exam tips for quadratic equations:

以下是关于二次方程的三条重要考试建议：

Choose the right method: Check if the quadratic factorises easily by inspection. If not, use the quadratic formula. Completing the square is best when you also need the turning point of the graph.
选择合适的方法：先观察二次式是否能轻松因式分解。若不能，使用求根公式。当你还需要图像的顶点时，配方法是最佳选择。
Show your working: Even if you make an arithmetic slip, you can earn method marks. Write down each substitution clearly.
展示计算过程：即使出现计算失误，你仍能获得方法分。请清晰写出每一步的代入过程。
Practise with non-calculator papers: Factorisation and completing the square require algebraic fluency. Many IGCSE papers allow calculators, but the algebraic manipulation is always tested regardless.
多练无计算器的试卷：因式分解和配方需要熟练的代数运算能力。虽然许多 IGCSE 试卷允许使用计算器，但代数变形能力始终是考察重点。

11. Worked Exam-Style Problem | 典型考题精讲

Let us work through a complete exam-style question to consolidate everything we have learned.

让我们完整地做一道典型考题，以巩固所学的全部内容。

Question: The equation x² + kx + 9 = 0 has a repeated root.

题目：已知方程 x² + kx + 9 = 0 有重根。

(a) Find the possible values of k.

(a) 求 k 的可能值。

(b) For each value of k, solve the equation.

(b) 对于每个 k 值，解这个方程。

Solution (a): For a repeated root, the discriminant must be zero. Here a = 1, b = k, c = 9, so Δ = k² − 4 × 1 × 9 = k² − 36. Setting Δ = 0 gives k² − 36 = 0, so k² = 36, and therefore k = 6 or k = −6.

解答 (a)：对于重根，判别式必须为零。这里 a = 1，b = k，c = 9，所以 Δ = k² − 4 × 1 × 9 = k² − 36。令 Δ = 0，得 k² − 36 = 0，即 k² = 36，因此 k = 6 或 k = −6。

Solution (b): When k = 6, the equation becomes x² + 6x + 9 = 0, which factorises as (x + 3)² = 0. So x = −3 (repeated). When k = −6, the equation becomes x² − 6x + 9 = 0, factorising as (x − 3)² = 0. So x = 3 (repeated).

解答 (b)：当 k = 6 时，方程变为 x² + 6x + 9 = 0，因式分解为 (x + 3)² = 0，所以 x = −3（重根）。当 k = −6 时，方程变为 x² − 6x + 9 = 0，因式分解为 (x − 3)² = 0，所以 x = 3（重根）。

This type of question combines the discriminant concept with factorisation, and it is a very popular exam question format. Remember that the phrase ‘repeated root’ or ‘equal roots’ is your cue to set b² − 4ac = 0.

这类题目将判别式概念与因式分解相结合，是考试中非常常见的题型。请记住，”重根”或”等根”这类关键词提示我们要令 b² − 4ac = 0。

12. Final Summary | 最终总结

Quadratic equations form a fundamental part of IGCSE Mathematics. Throughout this article, we have covered the standard form ax² + bx + c = 0, three solution methods — factorisation, completing the square, and the quadratic formula — as well as the discriminant, the sum and product of roots, graph sketching, inequalities, and real-world applications.

二次方程是 IGCSE 数学的基础内容。在本文中，我们涵盖了标准形式 ax² + bx + c = 0、三种解法——因式分解法、配方法和求根公式法——以及判别式、根的和与积、图像绘制、不等式和实际应用。

To succeed in exams, memorise the quadratic formula, understand when to use each method, and always verify your answers. With consistent practice, quadratic equations will become one of the most reliable scoring areas in your IGCSE Mathematics paper.

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Mastering Quadratic Equations | 掌握二次方程