📚 Quadratic Equations: Factorising, Formula & Completing the Square | 二次方程：因式分解、公式法与配方法
Quadratic equations are one of the most important topics in the Edexcel IGCSE Mathematics syllabus. They appear in both non-calculator and calculator papers, and strong command of factorising, completing the square, and the quadratic formula is essential for exam success.
二次方程是 Edexcel IGCSE 数学大纲中最重要的主题之一。它同时出现在非计算器试卷和计算器试卷中，熟练掌握因式分解、配方法和二次公式对考试成功至关重要。
1. What Is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants, and a ≠ 0. The highest power of the variable x is 2, which is why it is called ‘quadratic’ (from the Latin ‘quadratus’, meaning square).
二次方程是可以写成 ax² + bx + c = 0 形式的任意方程，其中 a、b、c 是常数，且 a ≠ 0。变量 x 的最高次数是 2，因此称为“二次”（源自拉丁语“quadratus”，意为“平方”）。
Linear equations such as 2x + 3 = 0 have only one solution, but quadratic equations can have two, one or even no real solutions. In the next sections we will explore three systematic methods for solving them.
像 2x + 3 = 0 这样的线性方程只有一个解，而二次方程可能有两个、一个甚至没有实数解。在接下来的小节中，我们将探讨三种系统的解法。
2. The Standard Form | 标准形式
Before solving any quadratic equation, rewrite it so that all non-zero terms are on one side and the other side is zero. The order should usually be the x² term first, then the x term, then the constant term.
在求解任何二次方程之前，先将所有非零项移到等号一侧，使另一侧为零。通常按 x² 项、x 项、常数项的顺序排列。
For example, 3x² + 5x = 2 should be rearranged as 3x² + 5x − 2 = 0. Expanding brackets and collecting like terms are also common first steps.
例如，3x² + 5x = 2 应改写为 3x² + 5x − 2 = 0。展开括号和合并同类项也是常见的操作。
ax² + bx + c = 0, where a ≠ 0
ax² + bx + c = 0，其中 a ≠ 0
3. Solving by Factorising | 因式分解法
Factorising is often the quickest method when the quadratic factorises neatly over the integers. To solve by factorising, write the left-hand side as a product of two linear expressions, then set each factor equal to zero.
当二次式能整齐地分解为整数因式时，因式分解法通常是最快的。通过因式分解求解时，将左侧写成两个一次因式的乘积，然后令每个因式等于零。
For example, solve x² + 7x + 12 = 0. We look for two numbers that multiply to 12 and add to 7: 3 and 4. Hence (x + 3)(x + 4) = 0, so x = −3 or x = −4.
例如，解 x² + 7x + 12 = 0。寻找两个数，乘积为 12 且和为 7：即 3 和 4。因此 (x + 3)(x + 4) = 0，所以 x = −3 或 x = −4。
When the coefficient of x² is not 1, the method becomes harder. For 2x² + 5x − 3 = 0, we find factors of 2 × (−3) = −6 that add to 5: 6 and −1. Split the middle term: 2x² + 6x − x − 3, then factor by grouping: 2x(x + 3) − 1(x + 3) = (2x − 1)(x + 3). Therefore x = ½ or x = −3.
当 x² 的系数不为 1 时，因式分解变得更难。对于 2x² + 5x − 3 = 0，我们找到 2 × (−3) = −6 的两个因数，其和为 5：即 6 和 −1。拆分中间项：2x² + 6x − x − 3，然后分组因式分解：2x(x + 3) − 1(x + 3) = (2x − 1)(x + 3)。因此 x = ½ 或 x = −3。
4. Solving by Completing the Square | 配方法
Completing the square transforms x² + bx + c into the form (x + p)² + q. This method works for every quadratic, including those that do not factorise.
配方法将 x² + bx + c 变形为 (x + p)² + q 的形式。这种方法适用于所有二次方程，包括无法因式分解的方程。
Take the coefficient of x, halve it, and use that number inside the bracket. For x² + 6x + 5 = 0, half of 6 is 3, so (x + 3)² = x² + 6x + 9. Thus x² + 6x + 5 = (x + 3)² − 4.
取 x 的系数，将其除以 2，然后在括号内使用该数。对于 x² + 6x + 5 = 0，6 的一半是 3，所以 (x + 3)² = x² + 6x + 9。因此 x² + 6x + 5 = (x + 3)² − 4。
Now solve (x + 3)² − 4 = 0. Add 4 to both sides: (x + 3)² = 4. Then take the square root: x + 3 = ±2. So x = −1 or x = −5.
现在解 (x + 3)² − 4 = 0。两边加 4：(x + 3)² = 4。然后开平方：x + 3 = ±2。所以 x = −1 或 x = −5。
If the coefficient of x² is not 1, first factor it out before completing the square. For example, 2x² + 8x − 10 = 0 becomes 2(x² + 4x − 5) = 0, then complete the square inside the brackets.
如果 x² 的系数不为 1，先提取该系数再进行配方。例如，2x² + 8x − 10 = 0 变成 2(x² + 4x − 5) = 0，然后在括号内配方。
5. The Quadratic Formula | 二次公式
The quadratic formula is the most general algebraic method. It is derived by completing the square on ax² + bx + c = 0 and gives the solutions directly.
二次公式是最通用的代数方法。它通过对 ax² + bx + c = 0 配方推导得出，并直接给出解。
x = [−b ± √(b² − 4ac)] / (2a)
x = [−b ± √(b² − 4ac)] / (2a)
Use this formula whenever factorising is difficult or impossible. Substitute the values of a, b and c carefully, and simplify the square root first if possible.
当因式分解困难或不可能时，使用这个公式。仔细代入 a、b、c 的值，并尽可能先化简根号。
For 2x² − 3x − 2 = 0, here a = 2, b = −3, c = −2. Plugging in gives x = [3 ± √(9 + 16)] / 4 = [3 ± 5] / 4, so x = 2 or x = −½.
对于 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 = −½。
6. The Discriminant | 判别式
The expression b² − 4ac in the quadratic formula is called the discriminant. It tells us the number and type of roots without solving the equation.
二次公式中的表达式 b² − 4ac 称为判别式。它无需解方程就能告诉我们根的个数和类型。
- If b² − 4ac > 0, there are two real and distinct roots.
- 如果 b² − 4ac > 0，有两个不同实数根。
- If b² − 4ac = 0, there is exactly one real root (a repeated root).
- 如果 b² − 4ac = 0，有一个实数根（重根）。
- If b² − 4ac < 0, there are no real roots.
- 如果 b² − 4ac < 0，没有实数根。
For example, x² + 4x + 5 = 0 has discriminant 16 − 20 = −4, so it has no real roots. Meanwhile 4x² − 12x + 9 = 0 has discriminant 144 − 144 = 0, so it has one repeated root, x = 1.5.
例如，x² + 4x + 5 = 0 的判别式为 16 − 20 = −4，因此没有实数根。而 4x² − 12x + 9 = 0 的判别式为 144 − 144 = 0，因此有一个重根 x = 1.5。
7. Roots and the Parabola | 根与抛物线
The graph of y = ax² + bx + c is a parabola. The real roots are the x-coordinates where the curve crosses the x-axis.
y = ax² + bx + c 的图像是抛物线。实数根是曲线与 x 轴交点的 x 坐标。
If the parabola opens upwards (a > 0), it has a minimum point; if it opens downwards (a < 0), it has a maximum point. The x-coordinate of the vertex is given by −b / (2a).
如果抛物线开口向上（a > 0），它有最小值点；如果开口向下（a < 0），它有最大值点。顶点的 x 坐标为 −b / (2a)。
This gives a quick check: if the discriminant is zero, the vertex lies exactly on the x-axis. If the discriminant is negative, the parabola does not cross the x-axis at all.
这提供了一个快速检验：如果判别式为零，顶点正好在 x 轴上；如果判别式为负，抛物线完全不与 x 轴相交。
8. Applications in Real Life | 实际应用
Quadratic equations appear in physics, business, and geometry. For example, the height of a ball thrown upward is described by h(t) = −5t² + 20t + 2, and you can find when it hits the ground by setting h(t) = 0.
二次方程出现在物理、商业和几何中。例如，上抛小球的高度由 h(t) = −5t² + 20t + 2 描述，令 h(t) = 0 即可求出落地时间。
In business, profit and revenue models sometimes involve quadratic formulas, and the break-even points occur when profit equals zero. In geometry, finding the dimensions of a rectangle with a given area and perimeter also leads to a quadratic equation.
在商业中，利润和收入模型有时涉及二次公式，当利润等于零时即为盈亏平衡点。在几何中，已知面积和周长求矩形边长也会得到二次方程。
Always interpret the solutions in the context of the problem. Negative lengths or times are usually rejected, even if they satisfy the equation algebraically.
始终要结合问题背景解释解。通常会舍去负数长度或时间，即使它们在代数上满足方程。
9. Common Mistakes to Avoid | 常见错误避免
One common mistake is forgetting to set the equation to zero before factorising or using the formula. The formula only works when the equation is in the form ax² + bx + c = 0.
一个常见错误是忘记在因式分解或使用公式前将方程化为零。公式只在方程为 ax² + bx + c = 0 时才有效。
Another error is sign errors when substituting negative values into the quadratic formula. Write down the formula, substitute carefully, and then simplify step by step.
另一个错误是在代入负数到二次公式时出现符号错误。写出公式，仔细代入，然后逐步化简。
When factorising, check your factors by expanding them again. A common trap is to forget that (x + p)² = x² + 2px + p², so the middle term is doubled.
因式分解时，通过再次展开来检查因式。一个常见陷阱是忘记 (x + p)² = x² + 2px + p²，即中间项要乘以 2。
10. Summary | 总结
To succeed with quadratic equations, follow this strategy: first rewrite in standard form, then factorise if possible. If factorising is difficult, use the quadratic formula or complete the square.
要成功解决二次方程，请遵循以下策略：首先改写为标准形式，然后尽可能因式分解。如果因式分解困难，使用二次公式或配方法。
Always check your answers by substituting back into the original equation. With consistent practice, these methods become automatic, and you will gain full confidence for the Edexcel IGCSE exam.
务必通过代入原方程来检查答案。通过持续练习，这些方法会变得熟练，你将对 Edexcel IGCSE 考试充满信心。
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