Mastering Quadratic Equations | 掌握二次方程

📚 Mastering Quadratic Equations | 掌握二次方程

Quadratic equations appear in almost every IGCSE Mathematics paper, whether as direct solving questions, graph questions, or word problems. Understanding this topic is essential for achieving top grades.

二次方程几乎出现在每一份 IGCSE 数学试卷中，无论是直接求解题、函数图像题还是应用题。掌握这一专题是取得高分的关键。

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

A quadratic equation is a polynomial equation of degree 2. This means the highest power of the variable x is 2. The general form is written in three terms: a quadratic term, a linear term, and a constant term.

二次方程是最高次数为 2 的多项式方程，即变量 x 的最高次幂为 2。其一般形式包含三项：二次项、一次项和常数项。

a x² + b x + c = 0, where a ≠ 0

Here, a is called the leading coefficient, b is the coefficient of the linear term, and c is the constant. For example, in the equation 2x² − 7x + 3 = 0, we have a = 2, b = −7, c = 3.

其中 a 称为首项系数，b 是一次项系数，c 是常数项。例如在方程 2x² − 7x + 3 = 0 中，a = 2，b = −7，c = 3。

The solutions of a quadratic equation are called the roots or solutions. A quadratic equation can have at most two real roots.

二次方程的解称为根或解。一个二次方程最多有两个实数根。

2. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic expression can be written as a product of two linear factors. The method relies on the zero product property: if AB = 0, then either A = 0 or B = 0.

当二次表达式可以写成两个一次因式的乘积时，因式分解法是最快捷的方法。该方法依赖零积性质：如果 AB = 0，则 A = 0 或 B = 0。

To factorise x² + b x + c, we look for two numbers that multiply to give c and add to give b. For example, factorise x² − 5x + 6 = 0.

对 x² + b x + c 进行因式分解时，我们需要找两个数，它们的乘积为 c，和为 b。例如，因式分解 x² − 5x + 6 = 0。

The two numbers are −2 and −3 because (−2) × (−3) = 6 and (−2) + (−3) = −5.
两个数是 −2 和 −3，因为 (−2) × (−3) = 6，且 (−2) + (−3) = −5。
Therefore, x² − 5x + 6 = (x − 2)(x − 3) = 0.
因此，x² − 5x + 6 = (x − 2)(x − 3) = 0。
Setting each bracket to zero gives x = 2 or x = 3.
令每个括号等于零，得 x = 2 或 x = 3。

For equations where a is greater than 1, we use the method of factors of a × c. Consider 2x² − 7x + 3 = 0. Here, a × c = 6. We need two numbers whose product is 6 and whose sum is −7. These numbers are −6 and −1.

当 a 大于 1 时，我们使用 a × c 的因数分解法。考虑 2x² − 7x + 3 = 0，此时 a × c = 6。我们需要两个乘积为 6、和为 −7 的数，即 −6 和 −1。

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

Thus x = ½ or x = 3. Always expand your factorised answer to check that it matches the original expression.

因此 x = ½ 或 x = 3。务必展开因式分解的结果，以检查是否与原表达式一致。

3. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic expression in the form (x + p)² + q. This method is useful when factorisation is difficult and when finding the turning point of a graph.

配方法将二次表达式改写为 (x + p)² + q 的形式。当因式分解困难或需要求图像顶点时，此方法非常有效。

To complete the square for x² + b x + c, we take half of b, square it, and then adjust the constant. Consider x² + 6x + 2 = 0.

对于 x² + b x + c 配方时，取 b 的一半、将其平方，然后调整常数项。考虑 x² + 6x + 2 = 0。

Half of 6 is 3, and 3² = 9, so x² + 6x = (x + 3)² − 9.
6 的一半是 3，3² = 9，所以 x² + 6x = (x + 3)² − 9。
Substitute back: (x + 3)² − 9 + 2 = 0, so (x + 3)² − 7 = 0.
代回原式：(x + 3)² − 9 + 2 = 0，即 (x + 3)² − 7 = 0。
Hence (x + 3)² = 7, so x + 3 = ±√7.
因此 (x + 3)² = 7，所以 x + 3 = ±√7。
Finally, x = −3 ± √7.
最终得到 x = −3 ± √7。

When a does not equal 1, first factor out a from the first two terms before completing the square. For example, 2x² + 8x + 5 = 0 becomes 2(x² + 4x) + 5 = 0.

当 a 不等于 1 时，先从前两项中提出 a，再进行配方。例如 2x² + 8x + 5 = 0 可改写为 2(x² + 4x) + 5 = 0。

2[(x + 2)² − 4] + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3 = 0

The advantage of this form is that the vertex of the parabola is immediately visible: it is (−2, −3).

这种形式的优势在于抛物线的顶点一目了然：即 (−2, −3)。

4. Solving by the Quadratic Formula | 公式法

The quadratic formula works for every quadratic equation, even when factorisation is impossible. It is derived from completing the square and must be memorised carefully.

公式法适用于所有二次方程，即使无法因式分解也能求解。它由配方法推导而来，需要牢记。

x = (−b ± √(b² − 4ac)) / 2a

To apply it, substitute a, b and c from the equation into the formula. Use the example 3x² − 4x − 2 = 0.

应用公式时，将方程中的 a、b、c 代入公式。以 3x² − 4x − 2 = 0 为例。

Here a = 3, b = −4, c = −2.
这里 a = 3，b = −4，c = −2。
Calculate the discriminant: b² − 4ac = (−4)² − 4 × 3 × (−2) = 16 + 24 = 40.
计算判别式：b² − 4ac = (−4)² − 4 × 3 × (−2) = 16 + 24 = 40。
Then x = (4 ± √40) / 6 = (4 ± 2√10) / 6 = (2 ± √10) / 3.
所以 x = (4 ± √40) / 6 = (4 ± 2√10) / 6 = (2 ± √10) / 3。

This gives two exact answers: x = (2 + √10) / 3 and x = (2 − √10) / 3. On your calculator, these approximate to 1.72 and −0.39, useful for checking.

由此得到两个精确答案：x = (2 + √10) / 3 和 x = (2 − √10) / 3。在计算器上约等于 1.72 和 −0.39，可用于验算。

5. The Discriminant | 判别式

The expression b² − 4ac inside the square root is called the discriminant, usually denoted by Δ. It reveals the nature of the roots without solving the entire equation.

根号内的表达式 b² − 4ac 称为判别式，通常用 Δ 表示。它无需完整求解即可判断根的性质。

Δ = b² − 4ac

Discriminant 判别式	Nature of Roots 根的性质
Δ > 0	Two distinct real roots 两个不同的实数根
Δ = 0	One repeated root (equal roots) 一个二重根（等根）
Δ < 0	No real roots (complex roots) 无实数根（复数根）

If the discriminant is a perfect square, the quadratic will factorise neatly. If it is positive but not a perfect square, the roots will involve surds.

若判别式是一个完全平方数，则二次方程可以整齐地因式分解；若为正数但不是完全平方数，根将包含无理数（根式）。

If Δ is negative, we say the equation has no real roots; the graph of the equation never touches the x-axis.

若 Δ 为负数，则称方程无实数根；此时对应图像与 x 轴无交点。

6. Graphical Interpretation | 图像意义

The graph of y = a x² + b x + c is a parabola. The value of a determines the orientation: if a > 0, the parabola opens upward in a U-shape; if a < 0, it opens downward in an n-shape.

y = a x² + b x + c 的图像是抛物线。a 的值决定开口方向：当 a > 0 时，抛物线向上开口，呈 U 形；当 a < 0 时，向下开口，呈 n 形。

The roots of the equation are the x-coordinates where the parabola crosses the x-axis. This explains why no real roots means the curve never crosses the x-axis.

方程的根是抛物线与 x 轴交点的横坐标。这也解释了为什么无实数根意味着曲线与 x 轴没有交点。

If the equation has a repeated root, the parabola touches the x-axis at exactly one point, which is the turning point of the curve.

若方程有重根，则抛物线仅在一点与 x 轴相切，这一点正是曲线的顶点。

The axis of symmetry is the vertical line through the midpoint of the roots. Its equation is x = −b / (2a). This is also the x-coordinate of the turning point.

抛物线的对称轴是经过两个根中点的竖直直线，其方程为 x = −b / (2a)，同时也是顶点的横坐标。

Turning Point 顶点: ( −b/(2a), c − b²/(4a) )

7. Solving Word Problems | 应用题解法

Quadratic equations in exams are often embedded in real-world problems, such as area, projectile motion, and consecutive number problems. The key is to translate the English or Chinese words into an algebraic equation.

考试中的二次方程常融入实际问题，如面积、抛体运动和连续整数问题。关键在于将文字表述转化为代数方程。

Example: The length of a rectangle is 3 cm longer than its width, and its area is 40 cm². Find the width.

例：一个长方形的长比宽多 3 cm，面积为 40 cm²。求宽。

Let the width be w cm; then the length is (w + 3) cm.
设宽为 w cm，则长为 (w + 3) cm。
Area: w(w + 3) = 40, so w² + 3w − 40 = 0.
面积：w(w + 3) = 40，即 w² + 3w − 40 = 0。
Factorise: (w + 8)(w − 5) = 0, so w = 5 or w = −8.
因式分解：(w + 8)(w − 5) = 0，所以 w = 5 或 w = −8。
Since a length cannot be negative, the width is 5 cm and the length is 8 cm.
由于长度不能为负，宽为 5 cm，长为 8 cm。

Always remember to reject negative or physically impossible roots in context. Write your final answer with units and a full sentence.

务必舍去在实际情况中不合理的负根，最终答案要带有单位并以完整句子作答。

8. Common Mistakes | 常见错误

Many students lose marks on quadratic equations due to small but repeated errors. Being aware of these pitfalls will immediately improve your accuracy.

许多学生在二次方程上失分，往往是因为一些反复出现的小错误。提前了解这些陷阱可以立刻提高准确率。

Forgetting to rearrange the equation to the form a x² + b x + c = 0 before solving.
忘记先将方程整理为 a x² + b x + c = 0 的形式再求解。
Mistakenly cancelling x from both sides of an equation such as x² = x, which loses the root x = 0.
在 x² = x 这类方程中错误地两边约去 x，导致丢失 x = 0 这个根。
Making sign errors when substituting negative values into the quadratic formula.
将负数代入求根公式时出现符号错误。
Stopping after one solution when two distinct roots exist.
求出第一个解后停止，忽略第二个不同的根。
Not checking answers by substitution back into the original equation.
没有将答案代回原方程进行验算。

To avoid these mistakes, slow down, write all steps clearly, and always verify your final values. Half a mark saved per carelessness adds up significantly.

为避免这些错误，请放慢速度、清晰写出每一步，并始终验证最终结果。每次细心换来的半分，积累起来非常可观。

9. Choosing the Best Method | 选择最优解法

Different quadratics are best solved using different methods. Selecting the most efficient approach saves time and reduces errors in examinations.

不同的二次方程适合不同的解法。在考试中选择最高效的方法可以节省时间并减少错误。

Condition 条件	Recommended Method 推荐方法
a = 1 and c has simple factors a = 1 且 c 的因数简单	Factorisation 因式分解法
Equation cannot be factorised 方程无法因式分解	Quadratic formula 公式法
Need the turning point or vertex 需要顶点坐标	Completing the square 配方法
Need approximate answers quickly 需要快速求近似值	Calculator or formula 计算器或公式法
Checking the nature of roots only 只需判断根的性质	Discriminant 判别式

In examination contexts, the question wording often gives a clue: ‘factorise’ suggests factorisation, ‘solve using the formula’ explicitly requests the formula, and ‘hence’ links parts together.

在考试中，题目的措辞常有提示：’factorise’ 提示用因式分解法，’solve using the formula’ 明确要求用公式法，而 ‘hence’ 则暗示前后小题需要衔接。

10. Exam Practice Questions | 真题练习

Below are two typical examination-style questions. Attempt them fully before looking at the outlined solutions.

下面两道是典型的考试风格练习题。请先完整作答，再参考解答思路。

Question 1: Solve 4x² − 9 = 0 by factorisation.

题目 1：用因式分解法解方程 4x² − 9 = 0。

This is a difference of two squares: (2x)² − 3².
这是平方差公式：(2x)² − 3²。
So (2x − 3)(2x + 3) = 0, giving x = 3/2 or x = −3/2.
因此 (2x − 3)(2x + 3) = 0，得 x = 3/2 或 x = −3/2。

Question 2: The sum of two consecutive positive integers is 41, and the product of their squares is 14400. Find the integers. (Hint: let them be n and n+1.)

题目 2：两个连续正整数的和为 41，它们平方的乘积为 14400。求这两个整数。（提示：设它们为 n 和 n+1。）

Let the integers be n and n + 1. Their sum gives 2n + 1 = 41, so n = 20.
设两个整数为 n 和 n + 1。由和得 2n + 1 = 41，所以 n = 20。
Check the product of squares: 20² × 21² = 400 × 441 = 176400 — this does not match the stated 14400, so the hint root check fails.
验算平方乘积：20² × 21² = 400 × 441 = 176400 — 与题目给出的 14400 不符，因此应改用乘积条件列方程。
Alternative: let n(n+1) = 120 (since √14400 = 120), so n² + n − 120 = 0.
换一种思路：令 n(n+1) = 120（因为 √14400 = 120），即 n² + n − 120 = 0。
Factorise: (n + 12)(n − 10) = 0, so n = 10 (positive), giving 10 and 11.
因式分解：(n + 12)(n − 10) = 0，取正数 n = 10，得到 10 和 11。

Practising this style of multi-step reasoning is the best preparation for Paper 2 and Paper 4 questions.

练习这种多步思考题型，是应对 Paper 2 和 Paper 4 的最佳准备。

With consistent practice of factorisation, completing the square, the quadratic formula, and the discriminant, you will build both confidence and speed in solving quadratic equations. Remember to show every step clearly, because method marks are just as important as final answers in IGCSE Mathematics.

通过坚持练习因式分解、配方法、公式法和判别式，你将在解二次方程时建立起信心和速度。切记清晰写出每一步，因为在 IGCSE 数学考试中，过程分与最后答案同样重要。

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Mastering Quadratic Equations | 掌握二次方程