📚 Solving Quadratic Equations | 解二次方程
Quadratic equations form a central part of the Edexcel IGCSE Mathematics syllabus. Mastering them requires a clear understanding of algebraic manipulation, factorisation, and the properties of graphs. This revision guide will walk you through every essential method, from factorisation to the quadratic formula, with worked examples and common pitfalls to avoid.
二次方程是 Edexcel IGCSE 数学教学大纲的核心内容。熟练掌握二次方程需要清晰理解代数运算、因式分解以及图像性质。本复习指南将带你逐一掌握所有关键方法，从因式分解到求根公式，并配有典型例题和需要避免的常见错误。
1. What is a Quadratic Equation? | 什么是二次方程？
, b, and c are constants, and a ≠ 0. If a = 0, the equation becomes linear, not quadratic.
二次方程是次数为 2 的多项式方程，即变量 x 的最高幂次为 2。标准一般形式写作 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。若 a = 0，方程退化为一次方程，而非二次方程。
equation may have zero, one, or two real solutions, depending on the value of the discriminant (covered later).
例如，2x² - 3x + 5 = 0 是二次方程，而 x + 4 = 0 是一次方程。二次方程可能有零个、一个或两个实数解，具体取决于判别式的值（后面会介绍）。
2. Expanding and Factorising | 展开与因式分解
Before solving, you must be comfortable expanding brackets. The product (px + q)(rx + s) expands to prx² + (ps + qr)x + qs. Factorisation is the reverse process: writing a quadratic expression as a product of two linear factors.
在求解之前，你必须熟练展开括号。乘积 (px + q)(rx + s) 展开得到 prx² + (ps + qr)x + qs。因式分解是相反的过程：将二次表达式写成两个一次因式的乘积。
c and add to b. For example, x² + 5x + 6 = (x + 2)(x + 3) because 2 × 3 = 6 and 2 + 3 = 5.
对于二次项 x² + bx + c，找到两个数，它们的乘积为 c，和为 b。例如，x² + 5x + 6 = (x + 2)(x + 3)，因为 2 × 3 = 6 且 2 + 3 = 5。
), then rewrite and factorise: 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3).
当 x² 的系数不为 1 时，例如 2x² + 7x + 3，可以使用十字相乘法。将 a 和 c 相乘（2 × 3 = 6），找到两个数乘积为 6 且和为 7（即 1 和 6），然后改写并因式分解：2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)。
3. Solving by Factorisation | 用因式分解求解
= 0. Hence x = m or x = n. This is often the quickest method when factorisation is simple.
如果二次方程可以写成 (x - m)(x - n) = 0，那么根据零乘积性质，要么 x - m = 0，要么 x - n = 0。因此 x = m 或 x = n。当因式分解简单时，这通常是最快的方法。
Example: Solve x² - 4x - 5 = 0. Factorise as (x - 5)(x + 1) = 0. So x = 5 or x = -1.
例：求解 x² - 4x - 5 = 0。因式分解得 (x - 5)(x + 1) = 0。所以 x = 5 或 x = -1。
e, x² = 3x + 10 becomes x² - 3x - 10 = 0, then (x - 5)(x + 2) = 0, giving x = 5 or x = -2.
在因式分解前，务必将方程整理成 ax² + bx + c = 0 的形式。例如，x² = 3x + 10 变为 x² - 3x - 10 = 0，然后 (x - 5)(x + 2) = 0，得到 x = 5 或 x = -2。
4. The Quadratic Formula | 二次求根公式
c formula provides a universal solution. For ax² + bx + c = 0, the solutions are given by:
当因式分解困难或无法进行时，二次求根公式提供了一种通用解法。对于 ax² + bx + c = 0，解由下式给出：
x = (-b ± √b² - 4ac) / (2a)
e formula, and simplify. The symbol ± means you calculate both the sum and the difference.
你必须记住这个公式。使用时，先确定 a、b、c，代入公式并化简。符号 ± 表示要分别计算加和减两种情形。
Example: Solve 2x² - 3x - 2 = 0. Here a = 2, b = -3, c = -2. Substituting gives x = (3 ± √9 + 16) / 4 = (3 ± 5) / 4. So x = 2 or x = -0.5.
例：求解 2x² - 3x - 2 = 0。这里 a = 2，b = -3，c = -2。代入得 x = (3 ± √9 + 16) / 4 = (3 ± 5) / 4。所以 x = 2 或 x = -0.5。
Be careful with negative signs. Always write the formula neatly, and ensure b² is positive even when b is negative.
小心负号。务必整齐地写出公式，并确保 b² 即使是负数也变为正数。
5. Completing the Square | 配方法
Another algebraic method involves rewriting the quadratic in the form (x + p)² + q. This is called completing the square. It is especially useful for finding turning points and solving equations without factorising.
另一种代数方法是将二次式改写成 (x + p)² + q 的形式，这称为配方法。它在求转折点和不解因式而解方程时特别有用。
f of b, square it, and adjust. For example: x² + 6x + 2 = (x + 3)² - 9 + 2 = (x + 3)² - 7.
对于 x² + bx + c，取 b 的一半，平方，然后调整。例如：x² + 6x + 2 = (x + 3)² - 9 + 2 = (x + 3)² - 7。
To solve (x + 3)² - 7 = 0, first add 7: (x + 3)² = 7. Then take square roots: x + 3 = ±√7, so x = -3 ± √7.
要解 (x + 3)² - 7 = 0，首先加 7： (x + 3)² = 7。然后取平方根：x + 3 = ±√7，所以 x = -3 ± √7。
bracket. For example, 2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² - 4] + 5 = 2(x + 2)² - 3.
当 x² 的系数不为 1 时，先提取 a，然后对括号内配方。例如，2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² - 4] + 5 = 2(x + 2)² - 3。
6. The Discriminant | 判别式
The expression b² - 4ac inside the quadratic formula is called the discriminant, often denoted by Δ. It determines the number of real roots of a quadratic equation.
二次求根公式中的表达式 b² - 4ac 称为判别式，通常用 Δ 表示。它决定了二次方程实数根的个数。
| Δ = b² - 4ac | Number of real roots |
| Δ > 0 | Two distinct real roots |
| Δ = 0 | One repeated real root |
| Δ < 0 | No real roots |
If Δ is a perfect square, the equation factorises neatly; otherwise the roots are irrational. This is useful for quickly checking whether to attempt factorisation.
如果 Δ 是完全平方数，方程可以整齐因式分解；否则根为无理数。这可用于快速判断是否尝试因式分解。
xactly one root, x = 2. For x² + 2x + 5 = 0, Δ = 4 - 20 = -16 < 0, so no real roots exist.
例：对于 x² - 4x + 4 = 0，Δ = 16 - 16 = 0，所以恰好有一个根 x = 2。对于 x² + 2x + 5 = 0，Δ = 4 - 20 = -16 < 0，因此没有实数根。
7. Graphs of Quadratic Functions | 二次函数图像
The graph of y = ax² + bx + c is a parabola. If a > 0, it opens upwards (a 'smile'); if a < 0, it opens downwards (a 'frown'). The roots of the equation correspond to the x-intercepts of the graph.
函数 y = ax² + bx + c 的图像是抛物线。若 a > 0，开口向上（形似微笑）；若 a < 0，开口向下（形似皱眉）。方程的根对应图像与 x 轴的交点。
where (h, k) is the vertex. The x-coordinate of the vertex is also given by x = -b / (2a).
转折点（顶点）可通过配方求得：y = a(x - h)² + k，其中 (h, k) 是顶点。顶点的 x 坐标也由 x = -b / (2a) 给出。
vertex is (3, -4). The roots are x = 1 and x = 5, found by factorising (x - 1)(x - 5) = 0.
例：对于 y = x² - 6x + 5，配方得 y = (x - 3)² - 4，所以顶点为 (3, -4)。根为 x = 1 和 x = 5，由因式分解 (x - 1)(x - 5) = 0 求得。
if the curve crosses the x-axis twice, Δ > 0; touching once, Δ = 0; never crossing, Δ < 0.
图像有助于直观理解判别式：若曲线与 x 轴相交两次，则 Δ > 0；相切一次，则 Δ = 0；永不相交，则 Δ < 0。
8. Solving Equations Reducible to Quadratics | 可化为二次方程的方程
ielding u² - 5u + 6 = 0, which factorises as (u - 2)(u - 3) = 0, so u = 2 or 3, then x = ±√2 or ±√3.
有些方程初看不是二次方程，但可以转化。例如，四次方程 x⁴ - 5x² + 6 = 0 可通过代换 u = x² 求解，得到 u² - 5u + 6 = 0，因式分解为 (u - 2)(u - 3) = 0，所以 u = 2 或 3，进而 x = ±√2 或 ±√3。
2 + 3/x = 4. Multiply by 2x: x² + 6 = 8x, giving x² - 8x + 6 = 0. Solve using the formula.
类似地，涉及分数的方程可以通过乘以公分母转化为二次方程。例如，x/2 + 3/x = 4。乘以 2x：x² + 6 = 8x，得到 x² - 8x + 6 = 0。用公式求解。
Be aware of extraneous solutions, especially when multiplying by a variable expression: check that the original denominators are not zero.
注意增根，尤其是当乘以含变量表达式时：务必验证原分母不为零。
9. Applications of Quadratic Equations | 二次方程的应用
Quadratic equations appear in physics (projectile motion), geometry (area problems), and economics (profit maximisation). The IGCSE exam often presents worded problems that require you to form and solve a quadratic equation.
二次方程出现在物理（抛体运动）、几何（面积问题）和经济学（利润最大化）中。IGCSE 考试常以文字应用题形式要求你建立并求解二次方程。
orising: (x + 10)(x - 7) = 0, so x = -10 (discard) or x = 7. Width = 7 cm, length = 10 cm.
例：长方形的长比宽多 3 cm，面积为 70 cm²。设宽为 x cm。则长为 x + 3，所以 x(x + 3) = 70，得到 x² + 3x - 70 = 0。因式分解：(x + 10)(x - 7) = 0，所以 x = -10（舍去）或 x = 7。宽为 7 cm，长为 10 cm。
Always interpret the solutions in the context of the problem. Negative answers may be mathematically valid but physically impossible.
务必结合问题背景解释解。负数答案在数学上有效，但在物理上可能不成立。
10. Common Mistakes and How to Avoid Them | 常见错误及避免方法
Not rearranging to standard form. Always set the equation to ax² + bx + c = 0 before factorising or applying the formula.
未整理成标准形式。在因式分解或套公式前，务必将方程化为 ax² + bx + c = 0。
Sign errors in the quadratic formula. Pay special attention to -b and b². For example, if b = -5, then -b = 5 and b² = 25.
二次公式中的符号错误。特别注意 -b 和 b²。例如，若 b = -5，则 -b = 5 且 b² = 25。
Incorrect factorisation leading to wrong roots. Check by expanding your factors back to the original expression.
因式分解错误导致错误根。通过将因式展开回原式来验证。
Forgetting the ± when taking square roots. If x² = 9, then x = ±3, not just x = 3.
开平方时忘记 ±。若 x² = 9，则 x = ±3，而不仅仅是 x = 3。
Dividing by a variable that could be zero. Never divide both sides by x unless you know x ≠ 0; you may lose a root.
除以可能为零的变量。除非已知 x ≠ 0，否则切勿两边同除以 x，否则可能丢失一个根。
11. Practice Questions | 练习题
Try these questions on your own before checking the answers. They cover all the methods discussed.
请先独立完成以下题目，再核对答案。这些题目覆盖了上述所有方法。
Solve x² - 7x + 12 = 0.
Solve 2x² + 5x - 3 = 0.
Solve x² + 4x - 1 = 0 by completing the square.
Find the values of k for which the equation kx² + 4x + 2 = 0 has two distinct real roots.
The product of two consecutive positive odd integers is 143. Form and solve a quadratic equation to find the integers.
Answers: (1) x = 3 or 4. (2) x = 0.5 or -3. (3) x = -2 + √5 or -2 - √5 + 2n - 143 = 0, (n + 13)(n - 11) = 0, giving n = 11 (discard -13), integers are 11 and 13.
答案：(1) x = 3 或 4。(2) x = 0.5 或 -3。(3) x = -2 + √5 或 -2 - √5。(4) k < 2 且 k ≠ 0（因为 a ≠ 0）。(5) 设整数为 n 和 n + 2；n(n + 2) = 143，所以 n² + 2n - 143 = 0，(n + 13)(n - 11) = 0，得 n = 11（舍去 -13），整数为 11 和 13。
12. Summary | 总结
Quadratic equations are a major topic in Edexcel IGCSE Mathematics. You now know how to solve them by factorisation, by using the quadratic formula, and by completing the square. You also understand the role of the discriminant and how to sketch quadratic graphs. Regular practice with these methods will help you recognise which technique to use in any situation.
二次方程是 Edexcel IGCSE 数学中的重要专题。你现在掌握了因式分解法、二次求根公式法以及配方法来解二次方程。你也了解了判别式的作用以及如何绘制二次函数草图。通过定期练习这些方法，你将能够在任何情况下识别该使用哪种技巧。
Remember to always check your solutions by substituting back into the original equation. With solid revision, quadratic equations will become one of the most reliable parts of your exam.
记住，务必通过代回原方程来检验你的解。经过扎实的复习，二次方程将成为你考试中最有把握的部分之一。
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