122: Quadratic Equations | 122：二次方程

📚 122: Quadratic Equations | 122：二次方程

A quadratic equation is a polynomial equation of degree two, which means the highest power of the variable is squared. In the IGCSE Edexcel Maths syllabus, you will learn to solve these equations by factorising, completing the square, using the quadratic formula and interpreting graphs. Understanding quadratics is essential for tackling more advanced algebra topics and real-world problems involving area, projectile motion and optimisation.

二次方程是一个二次多项式方程，即变量的最高次幂是平方。在 IGCSE 爱德思数学大纲中，你将学习通过因式分解、配方法、求根公式和图像解读来求解这类方程。掌握二次方程对于解决更高级的代数主题以及涉及面积、抛体运动和最优化的实际问题至关重要。

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

abola. If a is positive, the parabola opens upwards; if a is negative, it opens downwards. The solutions to the equation correspond to the x-intercepts of the graph, where the curve crosses the x‑axis.

任何可整理为标准形式 ax² + bx + c = 0 的方程都称为二次方程，其中 a、b 和 c 是常数，且 a ≠ 0。二次函数 y = ax² + bx + c 的图像是一条抛物线。若 a 为正，抛物线开口向上；若 a 为负，则开口向下。方程的解对应着图像与 x 轴交点的横坐标，即曲线穿越 x 轴的位置。

2. Standard Form and Key Terminology | 标准形式与关键术语

called the leading coefficient, b is the linear coefficient and c is the constant term. A ‘ hese parts when factorising, applying the quadratic formula or analysing the discriminant.

在选择解法之前，务必先将方程整理成一边为零的形式。标准形式为 ax² + bx + c = 0。系数 a 称为二次项系数，b 为一次项系数，c 为常数项。若 a = 1，则称为首一二次式。在进行因式分解、运用求根公式或分析判别式时，你需要能识别这些组成部分。

3. Solving by Factorising (When a = 1) | 因式分解法求解（当 a = 1 时）

+ 4) = 0 because 3 × 4 = 12 and 3 + 4 = 7. Set each factor equal to zero: x + 3 = 0 → x = –3, and x + 4 = 0 → x = –4. Always check by substituting the values back into the original equation.

如果二次式首项系数为 1 且可分解，寻找两个数，使其乘积等于 c、和等于 b。例如，x² + 7x + 12 = 0 可分解为 (x + 3)(x + 4) = 0，因为 3 × 4 = 12 且 3 + 4 = 7。令每个因式等于零：x + 3 = 0 → x = –3，x + 4 = 0 → x = –4。务必将求得的根代入原方程进行验算。

4. Factorising When the Leading Coefficient Is Not 1 | 二次项系数不为 1 时的因式分解

When a ≠ 1, use the ‘ two terms and the last two: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0. Solutions are x = –½ and x = –3. This technique appears frequently on Edexcel IGCSE papers.

当 a ≠ 1 时，使用“ac 法”：将 a 与 c 相乘，然后寻找乘积等于 ac、和等于 b 的两个数。利用这两个数将一次项拆分为两项，再通过分组进行因式分解。例如 2x² + 7x + 3 = 0：ac = 6。乘积为 6 且和为 7 的数是 6 和 1。重写为 2x² + 6x + x + 3 = 0，前两项与后两项分别组合：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0。解为 x = –½ 和 x = –3。这一技巧在爱德思 IGCSE 试卷中频繁出现。

5. Solving by Completing the Square | 配方法求解

ng all terms by a if necessary. Then move the constant to the other side. Add the square of half the coefficient of x to both sides. For example, x² + 6x – + 9 = 16, so (x + 3)² = 16. Taking square roots gives x + 3 = ±4, leading to x = 1 or x = – factorisation is not obvious and it appears in the Edexcel syllabus as an assessed skill.

配方法将二次式转化为一个完全平方三项式加上常数的形式。从 ax² + bx + c = 0 出发，如有必要先通过除以 a 使首项系数为 1。然后将常数项移到方程另一边。在两边同时加上 x 项系数一半的平方。例如 x² + 6x – 7 = 0 变为 x² + 6x = 7。6 的一半是 3，3² = 9。两边加 9：x² + 6x + 9 = 16，于是 (x + 3)² = 16。开平方得 x + 3 = ±4，从而 x = 1 或 x = –7。当因式分解不明显时，配方法尤其有用，且在爱德思大纲中明确列为考查技能。

6. Using the Quadratic Formula | 使用求根公式

om completing the square. It states that for ax² + bx + c = 0, the solutions are given by:

x = [–b ± √b² – 4ac] / (2a)

Substitute the values of a, b and c carefully, paying attention to negative signs. For 3x² – 2x – 5 = 0, a = 3, b = –2, c = –5. Then x = [–(–2) ± √((–2)² – 4×3×(–5))] / (2×3) = [2 ± √4 + 60] / 6 = [2 ± √64] / 6 = [2 ± 8] / 6, giving x = 10/6 = 5/3 and x = –6/6 = –1. The quadratic formula is especially handy when factors are not easy to spot.

求根公式适用于所有二次方程，并且直接由配方法推导得出。公式为：对于 ax² + bx + c = 0，解由下式给出：

x = [–b ± √b² – 4ac] / (2a)

使用公式前确保方程为标准形式。代入 a、b 和 c 的值时要仔细，注意负号。对于 3x² – 2x – 5 = 0，a = 3、b = –2、c = –5。那么 x = [–(–2) ± √((–2)² – 4×3×(–5))] / (2×3) = [2 ± √4 + 60] / 6 = [2 ± √64] / 6 = [2 ± 8] / 6，解得 x = 10/6 = 5/3 和 x = –6/6 = –1。当不容易看出因式时，求根公式尤其方便。

7. Understanding the Discriminant | 理解判别式

discriminant is the expression under the square root sign in the quadratic formula: Δ = b² – 4ac. It tells you the nature of the roots without solving the equation fully.

判别式是求根公式中平方根符号内的表达式：Δ = b² – 4ac。它能在不完整求解方程的情况下告诉你根的性质。

广告 / Sponsored

Discriminant Δ	Nature of roots	判别式的值	根的性质
Δ > 0	Two distinct real roots	大于 0	两个不相等的实根
Δ = 0	One repeated real root	等于 0	一个重实根
Δ < 0	No real roots	小于 0	没有实根

or to find the values of k for which a quadratic has equal roots. For equal roots, set b² – 4ac = 0 and solve for k.

IGCSE 试题通常会要求你判断给定二次方程实根的个数，或求使二次方程具有等根的 k 值。对于等根的情况，令 b² – 4ac = 0 并求解 k。

8. Graphical Solutions and Sketching | 图像解法与草图绘制

The solutions of ax² + bx + c = 0 are the x‑coordinates of the points where the parabola y = ax² + bx + c intersects the x‑ inimum value of the function. Completing the square helps to find the vertex form: y = a(x – h)² + k, where (h, k) is the vertex. For instance, from (x + 3)² – 16 = 0, the vertex of y = (x + 3)² – 16 is (–3, – ooth, symmetrical curve through the vertex and intercepts helps you visualise the problem.

ax² + bx + c = 0 的解就是抛物线 y = ax² + bx + c 与 x 轴交点的横坐标。你可以利用图像估计根并检验代数解法。抛物线的顶点对应函数的最大值或最小值。配方法有助于找到顶点式：y = a(x – h)² + k，其中 (h, k) 为顶点。例如，由 (x + 3)² – 16 = 0 可知 y = (x + 3)² – 16 的顶点为 (–3, –16)。过顶点和截距画出光滑对称的曲线有助于形象化理解问题。

9. Solving Word Problems Leading to Quadratics | 二次方程应用题

Many real‑life situations in the IGCSE syllabus give rise to quadratic equations. Typical examples include finding the dimensions of a rectangle given its area and perimeter, calculating the time after which an object hits the ground in projectile motion, or working with consecutive numbers. The key is to define your variable clearly, form an equation in standard form and then choose the most efficient solving method. Always reject solutions that do not make sense in context, such as negative lengths or negative time.

爱德思 IGCSE 大纲中许多实际问题都会归结为二次方程。典型例子包括已知矩形面积和周长求其尺寸、计算抛射运动中物体落地的时间，或处理连续整数问题。关键是要清晰地定义变量，建立标准形式的方程，然后选择最高效的求解方法。务必舍弃在情境中没有意义的解，如负长度或负时间。

10. Common Mistakes and How to Avoid Them | 常见错误与如何避免

ore factorising. Students often try to factor an expression like x² + 5x + 6 = 12 directly — instead, bring the 12 over to get x² + 5x – mistake is losing the ± when taking square roots; remember that x² = 9 gives x = 3 or x = – cially with negative b. Write down a, b and c with their signs on scrap paper to avoid mis‑substitutions. Finally, do not divide by x without considering the possibility that x = 0 — you may lose a solution.

最常见的错误之一是在因式分解前忘记将方程化为零。学生经常会直接对 x² + 5x + 6 = 12 进行因式尝试——正确的做法是把 12 移项得到 x² + 5x – 6 = 0。另一个错误是开平方时遗漏 ± 号；记住 x² = 9 的解是 x = 3 或 x = –3。使用求根公式时，符号错误也很常见，尤其是当 b 为负时。在草稿纸上将 a、b 和 c 连同符号一并列出，可以避免代入错误。此外，不要在未考虑 x = 0 的前提下贸然除以 x，否则可能丢失一个根。

11. Practice Example with Step‑by‑Step Solution | 分步练习示例

Solve 2x² – 5x – 3 = 0 using two different methods, and then check the roots graphically.

用两种不同的方法解方程 2x² – 5x – 3 = 0，然后用图像检验根。

Method 1: Factorising a = 2, b = –5, c = –3, ac = –6. We need two numbers multiplying to –6 and adding to –5: –6 and 1. Rewrite: 2x² – 6x + x – 3 = 0. Group: 2x(x – 3) + 1(x – 3) = (2x + 1)(x – 3) = 0. Solutions: x = –½ or x = 3.

方法一：因式分解 a = 2、b = –5、c = –3，ac = –6。需要两个数，其积为 –6、和为 –5：–6 和 1。重写为 2x² – 6x + x – 3 = 0。分组：2x(x – 3) + 1(x – 3) = (2x + 1)(x – 3) = 0。解：x = –½ 或 x = 3。

Method 2: Quadratic Formula Substitute into x = [–(–5) ± √((–5)² – 4×2×(–3))] / (2×2) = [5 ± √25 + 24] / 4 = [5 ± √49] / 4 = [5 ± 7] / 4. So x = 12/4 = 3 or x = –2/4 = –½. The two methods agree.

方法二：求根公式 代入 x = [–(–5) ± √((–5)² – 4×2×(–3))] / (2×2) = [5 ± √25 + 24] / 4 = [5 ± √49] / 4 = [5 ± 7] / 4。所以 x = 12/4 = 3 或 x = –2/4 = –½。两种方法结果一致。

To see this graphically, plot y = 2x² – 5x – 3. The parabola crosses the x‑axis at x = – ex lies midway between the roots at x = 1.25, and substituting gives the minimum value y = –6.125. This graphical check reassures you that the algebraic solutions are correct.

从图像上看，画出 y = 2x² – 5x – 3。该抛物线在 x = –0.5 和 x = 3 处穿过 x 轴。顶点位于两实根的中点 x = 1.25 处，代入可得最小值 y = –6.125。这一图形检验可以让你确信代数解法是正确的。

12. Summary and Final Tips for the IGCSE Exam | IGCSE 考试总结与终极建议

Quadratic equations can appear in many guises on the Edexcel IGCSE paper: stand‑ ch method fits best: factorising is quickest when it works, completing the square is excellent for finding vertices, and the quadratic formula always works. Never forget to consider the discriminant when asked about the number of solutions. Show all your working clearly, and double‑ cs will give you confidence across a wide range of topics in algebra, geometry and problem‑solving.

二次方程在爱德思 IGCSE 试卷中会以多种形式出现：独立的方程求解、涉及判别式的问题、用配方法求最值点，或者作为图像或数列大题中的一部分。始终先整理为 ax² + bx + c = 0 的形式。选择最合适的方法：因式分解最快（当适用时），配方法在求顶点时特别出色，而求根公式则是万能的。当被问及解的个数时，千万别忘了考虑判别式。清晰展示全部解题步骤，并将答案代入原方程进行复核。掌握二次方程将使你在代数、几何和问题解决等众多主题中充满信心。

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