📚 Mastering Quadratic Equations | 掌握二次方程
In this article, we will explore the essential concepts of quadratic equations for the IGCSE Mathematics syllabus. You will learn how to solve quadratics by factorisation, use the quadratic formula, complete the square, and understand the significance of the discriminant.
在本文中，我们将学习 IGCSE 数学大纲中二次方程的核心概念。你将掌握因式分解法、公式法、配方法求解二次方程，并理解判别式的意义。
1. General Form of a Quadratic Equation | 二次方程的一般形式
A quadratic equation in one variable is an equation that can be written in the form ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0. The name “quadratic” comes from the Latin word “quadratus” meaning square, because the highest power of the variable is 2.
一元二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。 “Quadratic” 源自拉丁语 “quadratus”，意为”平方”，因为变量的最高次数为 2。
The coefficient a is called the leading coefficient. If a = 0, the equation becomes linear, not quadratic. The constants b and c can be zero; for example, 2x² = 0 and 2x² − 3x = 0 are still quadratic equations.
系数 a 称为首项系数。若 a = 0，方程变为一次方程。b 和 c 可以为零，例如 2x² = 0 和 2x² − 3x = 0 仍是一元二次方程。
In IGCSE, you will often need to rearrange equations into this standard form before solving. For instance, x² = 3x − 2 becomes x² − 3x + 2 = 0.
在 IGCSE 考试中，你通常需要先将方程重新整理为标准形式。例如，x² = 3x − 2 可化为 x² − 3x + 2 = 0。
2. Factorising Quadratic Expressions | 二次表达式的因式分解
Factorising a quadratic expression means writing it as a product of two linear factors. For a simple quadratic x² + bx + c, we look for two numbers that multiply to give c and add to give b.
因式分解二次表达式是指将其写成两个一次因式的乘积。对于简单的二次式 x² + bx + c，我们寻找两个数，使其乘积为 c，和为 b。
For example, x² + 5x + 6 = (x + 2)(x + 3) because 2 × 3 = 6 and 2 + 3 = 5.
例如，x² + 5x + 6 = (x + 2)(x + 3)，因为 2 × 3 = 6，且 2 + 3 = 5。
When the coefficient of x² is not 1, the process becomes slightly more involved. Consider 2x² + 7x + 3. We use the “ac method”: multiply a and c (2 × 3 = 6), then find two numbers that multiply to 6 and add to 7. These are 6 and 1. Rewrite the middle term: 2x² + 6x + x + 3, then factor by grouping: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
当 x² 的系数不等于 1 时，因式分解稍复杂。以 2x² + 7x + 3 为例：先计算 a × c (2 × 3 = 6)，再找到两个数，其乘积为 6，和为 7，即 6 和 1。将中间项改写为 2x² + 6x + x + 3，然后分组提取公因式：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。
3. Solving by Factorisation | 用因式分解法求解
Once a quadratic is factorised, solving the equation is straightforward. If (x − p)(x − q) = 0, then either x − p = 0 or x − q = 0, giving x = p or x = q. This uses the zero-product property.
当二次式完成因式分解后，求解方程就变得直接。若 (x − p)(x − q) = 0，则 x − p = 0 或 x − q = 0，因此 x = p 或 x = q。这利用了零乘积性质。
For example, solve x² − 7x + 10 = 0. Factorise first: (x − 2)(x − 5) = 0. Therefore x = 2 or x = 5.
例如，解方程 x² − 7x + 10 = 0。先因式分解：(x − 2)(x − 5) = 0，所以 x = 2 或 x = 5。
Before factorising, always check that the equation is in standard form. A common mistake is to factorise the expression before rearranging, which can lead to incorrect solutions.
因式分解前，务必确认方程已化为标准形式。常见错误是在整理前就分解表达式，导致答案错误。
4. The Quadratic Formula | 二次求根公式
Some quadratic equations cannot be factorised easily using integers. In such cases, we use the quadratic formula, which gives the solutions for ax² + bx + c = 0 as:
有些二次方程难以用整数因式分解。此时我们使用二次求根公式，它给出了 ax² + bx + c = 0 的解：
x = (−b ± √(b² − 4ac)) / 2a
This formula works for any quadratic equation, provided a ≠ 0. It is particularly useful when the roots are irrational or complex (though complex roots are not required at IGCSE, the formula still gives the discriminant guidance).
这个公式适用于任何一元二次方程（a ≠ 0）。当根为无理数或非整数时尤为有用。
Let’s solve 2x² − 4x − 6 = 0 using the quadratic formula. Here a = 2, b = −4, c = −6. Substitute: x = (−(−4) ± √((−4)² − 4×2×(−6))) / (2×2) = (4 ± √(16 + 48)) / 4 = (4 ± √64) / 4 = (4 ± 8) / 4, giving x = 3 or x = −1.
我们用公式法解 2x² − 4x − 6 = 0。这里 a = 2，b = −4，c = −6。代入得：x = (−(−4) ± √((−4)² − 4×2×(−6))) / (2×2) = (4 ± √(16 + 48)) / 4 = (4 ± √64) / 4 = (4 ± 8) / 4，得到 x = 3 或 x = −1。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This is useful for identifying the vertex of a parabola and solving equations when factorisation is inconvenient.
配方法是将二次式化为 a(x + p)² + q 的形式。这有助于找出抛物线的顶点，也便于解方程。
For a simple quadratic x² + bx + c, we add and subtract (b/2)². For example, x² + 6x + 2 becomes (x + 3)² − 9 + 2 = (x + 3)² − 7.
对于简单的二次式 x² + bx + c，我们加减 (b/2)²。例如，x² + 6x + 2 可化为 (x + 3)² − 9 + 2 = (x + 3)² − 7。
To solve an equation by completing the square, isolate the squared term. For instance, (x + 3)² = 7, then x + 3 = ±√7, so x = −3 ± √7.
用配方法解方程时，先隔离平方项。例如 (x + 3)² = 7，则 x + 3 = ±√7，所以 x = −3 ± √7。
When the coefficient a is not 1, factor it out first. For 2x² − 8x + 5 = 0, write 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3.
当 a 不是 1 时，先提取 a。例如 2x² − 8x + 5 = 0 可化为 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3。
6. The Discriminant and the Nature of Roots | 判别式与根的性质
The discriminant, denoted by Δ, is the part of the quadratic formula under the square root: Δ = b² − 4ac. Its value tells us the nature of the roots without fully solving the equation.
判别式用 Δ 表示，是二次求根公式中平方根下的部分：Δ = b² − 4ac。其值无需解方程即可判断根的性质。
If Δ > 0, the equation has two distinct real roots. If Δ = 0, there is exactly one repeated real root. If Δ < 0, there are no real roots (the roots are complex).
若 Δ > 0，方程有两个不相等的实数根；若 Δ = 0，有一个重根；若 Δ < 0，没有实数根（根为复数）。
For example, for 3x² − 2x + 1 = 0, Δ = (−2)² − 4×3×1 = 4 − 12 = −8 < 0, so the equation has no real solutions.
例如，对于 3x² − 2x + 1 = 0，Δ = (−2)² − 4×3×1 = 4 − 12 = −8 < 0，因此方程无实数解。
7. Graphs of Quadratic Functions | 二次函数的图像
The graph of a quadratic function y = ax² + bx + c is a parabola. If a > 0, the parabola opens upwards; if a < 0, it opens downwards.
二次函数 y = ax² + bx + c 的图像是抛物线。若 a > 0，抛物线开口向上；若 a < 0，开口向下。
The vertex (turning point) can be found by completing the square. For y = a(x − h)² + k, the vertex is at (h, k). The axis of symmetry is the vertical line x = h.
抛物线的顶点可通过配方法求得。对于 y = a(x − h)² + k，顶点为 (h, k)，对称轴为 x = h。
The y-intercept is found by setting x = 0, giving y = c. The x-intercepts are the real roots of the equation ax² + bx + c = 0. The discriminant tells us whether the graph crosses the x-axis, touches it, or never meets it.
令 x = 0 得到 y 截距，即 y = c。x 截距是方程 ax² + bx + c = 0 的实数根。判别式决定了抛物线与 x 轴相交、相切还是无交点。
8. Solving Quadratic Inequalities | 解二次不等式
Quadratic inequalities, such as x² − x − 6 > 0, are solved by first finding the boundary values from the equation, then testing intervals.
二次不等式（如 x² − x − 6 > 0）先通过对应方程求出边界值，再在区间内进行测试。
Factorise: (x − 3)(x + 2) = 0, so the critical values are x = −2 and x = 3. These split the number line into three intervals: x < −2, −2 < x < 3, and x > 3. Testing a point in each interval shows the inequality is satisfied when x < −2 or x > 3.
因式分解得 (x − 3)(x + 2) = 0，临界值为 x = −2 和 x = 3。这两个点将数轴分为三段：x < −2、−2 < x < 3、x > 3。每段取点测试，可知不等式的解为 x < −2 或 x > 3。
If the inequality is x² − x − 6 ≤ 0, the solution becomes −2 ≤ x ≤ 3.
若不等式为 x² − x − 6 ≤ 0，则解为 −2 ≤ x ≤ 3。
9. Quadratic Functions and Transformations | 二次函数的变换
In IGCSE, you may need to determine the equation of a parabola after translations and reflections. The completed square form y = a(x − h)² + k shows the transformation from y = x².
在 IGCSE 中，你可能需要根据平移和翻转变换来确定抛物线方程。配方法形式 y = a(x − h)² + k 显示了从 y = x² 的变换关系。
The graph y = a(x − h)² + k is a translation of y = x² by h units horizontally (right if h > 0) and k units vertically (up if k > 0). The coefficient a represents a vertical stretch if |a| > 1, or a compression if 0 < |a| < 1; if a < 0, it is reflected in the x-axis.
图像 y = a(x − h)² + k 是 y = x² 沿水平方向平移 h 个单位（h > 0 向右）、沿垂直方向平移 k 个单位（k > 0 向上）的结果。系数 a 表示纵向伸缩：|a| > 1 时拉伸，0 < |a| < 1 时压缩；若 a < 0，则关于 x 轴对称。
For example, y = 2(x + 1)² − 3 has a vertex at (−1, −3), opens upwards, and is narrower than y = x².
例如，y = 2(x + 1)² − 3 的顶点为 (−1, −3)，开口向上，且比 y = x² 更窄。
10. Problem-Solving with Quadratic Equations | 二次方程的应用题
Many real-world problems, such as projectile motion, area optimization, and revenue calculations, lead to quadratic equations. You must translate the words into a mathematical statement, solve it, and check the answer against the context.
许多实际问题（如抛体运动、面积优化、收入计算）都会导出二次方程。你需要将文字转化为数学表达式、求解，并结合实际情境检验答案。
For instance, the area of a rectangle is 30 cm² and its length is 2 cm more than its width. Let the width be x cm. Then x(x + 2) = 30, so x² + 2x − 30 = 0. Using the quadratic formula gives x = −1 ± √31. Since a width cannot be negative, x = −1 + √31 ≈ 4.57 cm.
例如，一个矩形的面积为 30 cm²，长比宽多 2 cm。设宽为 x cm，则 x(x + 2) = 30，即 x² + 2x − 30 = 0。用求根公式得 x = −1 ± √31。因为宽度不能为负，所以 x = −1 + √31 ≈ 4.57 cm。
Always reject negative or impossible solutions in such context, and give the final answer to an appropriate degree of accuracy.
在应用题中要舍去负解或不合理的解，并按合适精度给出最终答案。
11. Common Mistakes and Exam Tips | 常见错误与考试要点
A frequent error is forgetting to make the coefficient a positive before factorising or using the formula. If the equation is −x² + 4x − 3 = 0, multiply throughout by −1 to get x² − 4x + 3 = 0, then solve.
常见错误之一是忘记先将首项系数化为正。如果方程是 −x² + 4x − 3 = 0，应两边乘以 −1 得到 x² − 4x + 3 = 0，然后再求解。
Another common mistake is misapplying the zero-product property. Remember that (x − 3)(x + 2) = 0 means x = 3 or x = −2, not x = −3 or x = 2.
另一个常见错误是误用零乘积性质。请注意 (x − 3)(x + 2) = 0 意味着 x = 3 或 x = −2，而不是 x = −3 或 x = 2。
In the quadratic formula, be careful with signs. For −b, if b is negative, then −b is positive. Also, compute b² − 4ac underneath the radical first, and never separate the ± incorrectly.
使用求根公式时注意符号。如果 b 为负，则 −b 为正。另外，先计算根号下的 b² − 4ac，避免错误展开 ±。
When solving, always substitute your answers back into the original equation to verify. This takes only seconds and often prevents careless marks lost.
解完后，务必将答案代回原方程检验。这一步骤仅需几秒钟，却常能避免不必要的失分。
12. Practice Questions and Summary | 练习与总结
Let’s review the key techniques with a final set of short exercises. Solve each equation and state the method used.
让我们用一组简短的练习来复习核心技巧。解下列方程，并说明所用的方法。
1. x² + 3x − 10 = 0 (factorisation)
2. 4x² − 12x + 9 = 0 (quadratic formula or factorisation)
3. x² + 4x − 1 = 0 (completing the square)
4. 3x² = 5x + 2 (rearrange first, then solve)
5. x² − 2x + 5 = 0 (discriminant test)
1. x² + 3x − 10 = 0（因式分解法）
2. 4x² − 12x + 9 = 0（公式法或因式分解法）
3. x² + 4x − 1 = 0（配方法）
4. 3x² = 5x + 2（先整理，再求解）
5. x² − 2x + 5 = 0（判别式检验）
Answers: 1. x = 2 or x = −5; 2. x = 1.5 (repeated); 3. x = −2 ± √5; 4. x = 2 or x = −1/3; 5. No real roots (Δ = −16).
答案：1. x = 2 或 x = −5；2. x = 1.5（重根）；3. x = −2 ± √5；4. x = 2 或 x = −1/3；5. 无实数根（Δ = −16）。
In this article, we have covered the general form, factorisation, the quadratic formula, completing the square, the discriminant, graphs, inequalities, transformations, and practical applications. Master these methods, and you will be well prepared for any quadratic equation problem in your IGCSE exam.
在本文中，我们涵盖了一般形式、因式分解、求根公式、配方法、判别式、图像、不等式、变换和实际应用。熟练掌握这些方法，你就能从容应对 IGCSE 考试中的所有二次方程问题。
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