📚 Solving Quadratic Equations by Factorising, Completing the Square, and the Quadratic Formula | 解二次方程：因式分解、配方法与公式法
Mathematics syllabus, students must be able to solve such equations using factorisation, the quadratic formula, and completing the square, as well as interpret their solutions graphically.
二次方程是指任何可以写成 ax² + bx + c = 0 形式的方程，其中 a ≠ 0。在 Edexcel IGCSE 数学考纲中，学生必须能够使用因式分解、二次公式和配方法求解此类方程，并能从图像上解释其解的意义。
1. The General Form and Key Concepts | 一般形式与核心概念
The values of a, b, and c may be integers, fractions, or decimals, but a must not be zero.
在尝试解二次方程之前，务必将其整理成标准形式 ax² + bx + c = 0。a、b、c 的值可以是整数、分数或小数，但 a 不能为零。
² + bx + c, the roots are the x-coordinates where the curve crosses or touches the x-axis.
二次方程的解也称为方程的根。在 y = ax² + bx + c 的图像上，根是曲线与 x 轴相交或相切处的 x 坐标。
ax² + bx + c = 0, a ≠ 0
2. Solving by Factorisation | 因式分解法
Factorisation is usually the fastest method when the quadratic has simple integer roots. To factorise x² + px + q, find two numbers whose product is q and whose sum is p.
当二次方程具有简单的整数根时，因式分解通常是最快的方法。要因式分解 x² + px + q，需要找到两个数，它们的积为 q，和为 p。
-5. These are -2 and -3, so the equation becomes (x - 2)(x - 3) = 0. Then x = 2 or x = 3.
例如，解 x² - 5x + 6 = 0，需要找到两个数，它们相乘得 6，相加得 -5。这两个数是 -2 和 -3，因此方程变为 (x - 2)(x - 3) = 0。于是 x = 2 或 x = 3。
of x² is not 1, such as 2x² + 5x - 3 = 0, you may need to factorise by grouping or use the 'ac' by grouping: 2x(x + 3) - 1(x + 3) = 0, giving (2x - 1)(x + 3) = 0. Hence x = ½ or x = -3.
当 x² 的系数不为 1 时，例如 2x² + 5x - 3 = 0，你可能需要使用分组因式分解或 "ac" 法。这里 a = 2，c = -3，所以 ac = -6。找到两个数，积为 -6，和为 5：6 和 -1。重写中间项：2x² + 6x - x - 3 = 0，然后分组因式分解：2x(x + 3) - 1(x + 3) = 0，得到 (2x - 1)(x + 3) = 0。故 x = ½ 或 x = -3。
3. Solving by Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form p(x + q)² + r. This method works for any quadratic, including those that do not factorise with integers.
配方法将二次表达式改写为 p(x + q)² + r 的形式。这种方法适用于任何二次方程，包括那些不能用整数因式分解的方程。
e it, and adjust the constant. For example, x² + 6x + 4 = (x + 3)² - 9 + 4 = (x + 3)² - 5.
要对 x² + bx + c 配方法，取 x 系数的一半，平方，然后调整常数项。例如，x² + 6x + 4 = (x + 3)² - 9 + 4 = (x + 3)² - 5。
² + 6x + 4 = 0, set (x + 3)² - 5 = 0, so (x + 3)² = 5. Taking square roots gives x + 3 = ±√5, so x = -3 ± √5.
要解 x² + 6x + 4 = 0，令 (x + 3)² - 5 = 0，因此 (x + 3)² = 5。取平方根得 x + 3 = ±√5，所以 x = -3 ± √5。
out first: ax² + bx + c = a(x² + (b/a)x) + c, then complete the square inside the bracket.
当 x² 的系数为 a 时，先提取 a：ax² + bx + c = a(x² + (b/a)x) + c，然后在括号内配方法。
4. The Quadratic Formula | 二次公式
l method that solves any quadratic equation. For ax² + bx + c = 0, the roots are given by:
二次公式是求解任何二次方程的通用方法。对于 ax² + bx + c = 0，根由下式给出：
x = (−b ± √b² − 4ac) / (2a)
This formula is derived from completing the square and is crucial for questions that give a non-factorisable quadratic. Always substitute a, b, and c carefully, noting the signs.
该公式由配方法推导而来，对于不可因式分解的二次方程至关重要。代入 a、b、c 时务必小心，注意符号。
b = 3, c = -2. The discriminant b² - 4ac = 9 - 4(2)(-2) = 9 + 16 = 25. Therefore x = (−3 ± √25) / 4 = (−3 ± 5) / 4, giving x = ½ or x = -2.
例如，解 2x² + 3x - 2 = 0。这里 a = 2，b = 3，c = -2。判别式 b² - 4ac = 9 - 4(2)(-2) = 9 + 16 = 25。因此 x = (−3 ± √25) / 4 = (−3 ± 5) / 4，得到 x = ½ 或 x = -2。
5. The Discriminant and the Number of Roots | 判别式与根的个数
The expression b² - 4ac is called the discriminant. It tells us how many real roots a quadratic equation has without solving it:
表达式 b² - 4ac 称为判别式。无需解方程，它就能告诉我们二次方程有多少个实数根：
- If b² - 4ac > 0, there are two distinct real roots.
- If b² - 4ac = 0, there is exactly one real root (a repeated root).
- If b² - 4ac < 0, there are no real roots (the roots are complex).
如果 b² - 4ac > 0，有两个不同的实数根。
如果 b² - 4ac = 0，恰好有一个实数根（重根）。
如果 b² - 4ac < 0，没有实数根（根为复数）。
Graphically, a positive discriminant means the parabola crosses the x-axis twice; a zero discriminant means it touches the x-axis at one point; a negative discriminant means it never meets the x-axis.
从图像上看，判别式为正意味着抛物线与 x 轴相交两次；判别式为零意味着抛物线在一点处与 x 轴相切；判别式为负意味着抛物线不与 x 轴相交。
6. Solving Quadratic Equations by Factorising: Worked Examples | 因式分解法解方程的例题
+ 8 = 0. Group: x(3x - 4) - 2(3x - 4) = 0, so (x - 2)(3x - 4) = 0. Hence x = 2 or x = 4/3.
考虑方程 3x² - 10x + 8 = 0。我们使用 ac 法：a = 3，c = 8，所以 ac = 24。找到两个数，积为 24，和为 -10：-4 和 -6。拆开中间项：3x² - 4x - 6x + 8 = 0。分组：x(3x - 4) - 2(3x - 4) = 0，因此 (x - 2)(3x - 4) = 0。从而 x = 2 或 x = 4/3。
is already factorised. For instance, (x + 5)(x - 2) = 0 immediately gives x = -5 or x = 2.
始终注意方程是否已经因式分解。例如，(x + 5)(x - 2) = 0 直接给出 x = -5 或 x = 2。
7. Completing the Square: Worked Example with Fraction Digits | 配方法的分数例题
² - 49/4 + 8/4 = (x - 7/2)² - 41/4. Set equal to zero: (x - 7/2)² = 41/4. Then x - 7/2 = ±√41/4 = ±√41 / 2. Thus x = (7 ± √41) / 2.
用配方法解 x² - 7x + 2 = 0。将 -7 除以 2 得 -7/2。平方得 49/4。改写：x² - 7x + 2 = (x - 7/2)² - 49/4 + 2 = (x - 7/2)² - 49/4 + 8/4 = (x - 7/2)² - 41/4。令其等于零：(x - 7/2)² = 41/4。于是 x - 7/2 = ±√41/4 = ±√41 / 2。因此 x = (7 ± √41) / 2。
This method is especially useful when the coefficient of x is odd, because the expression stays tidy.
当 x 的系数为奇数时，这种方法尤其有用，因为表达式保持整洁。
8. The Quadratic Formula: Worked Examples | 二次公式的例题
la. Here a = 5, b = 4, c = -1. Discriminant = 16 - 4(5)(-1) = 16 + 20 = 36. Then x = (−4 ± √36) / 10 = (−4 ± 6) / 10. So x = 2/10 = 1/5 or x = -10/10 = -1.
使用二次公式解 5x² + 4x - 1 = 0。这里 a = 5，b = 4，c = -1。判别式 = 16 - 4(5)(-1) = 16 + 20 = 36。于是 x = (−4 ± √36) / 10 = (−4 ± 6) / 10。所以 x = 2/10 = 1/5 或 x = -10/10 = -1。
If the discriminant is not a perfect square, leave the answer in surd form, e.g. x = (3 ± √5) / 2.
尽可能简化根式。如果判别式不是完全平方数，将答案保留为根式形式，例如 x = (3 ± √5) / 2。
9. Graphical Interpretation and Solving via the Curve | 图像意义与通过图像求解
x + c = 0. If the curve does not intersect the x-axis, the equation has no real solutions.
y = ax² + bx + c 的图像与 x 轴交点的 x 坐标正是 ax² + bx + c = 0 的实数根。若曲线不与 x 轴相交，则方程无实数解。
Sometimes you may be given a graph and asked to estimate the roots. Reading the intersection points to one decimal place is often sufficient in exam questions.
有时你可能会得到一张图并要求估算根。在考试题中，将交点读数精确到一位小数通常就足够了。
orizontal line y = k and reading the x-coordinates of the intersections with the parabola.
你还可以利用图像解 ax² + bx + c = k 形式的方程：画出水平线 y = k，读取它与抛物线的交点的 x 坐标。
10. Choosing the Best Method | 选择最佳方法
In examinations, speed and accuracy matter. Follow this general guide:
在考试中，速度和准确性都很重要。遵循以下一般指南：
- First try factorisation — if the quadratic has simple integer or fractional roots, this is quickest.
- If factorisation is not obvious, use the quadratic formula — it always works.
- Use completing the square when the question explicitly asks for it, or when you need the turning point of the parabola.
首先尝试因式分解——如果二次方程有简单的整数或分数根，这是最快的方法。
如果因式分解不明显，使用二次公式——它总是有效。
当题目明确要求配方法，或者你需要抛物线的顶点时，使用配方法。
Remember: every quadratic equation can be solved by the formula, and completing the square also always works, but factorisation only works when the roots are rational.
记住：每个二次方程都可以用公式法求解，配方法也总是有效，但因式分解只在根为有理数时有效。
11. Common Mistakes and Exam Tips | 常见错误与考试技巧
ore applying methods. For example, x² = 5x - 4 must be rewritten as x² - 5x + 4 = 0 first.
一个常见错误是忘记在应用方法之前将方程整理成 ax² + bx + c = 0 的形式。例如，x² = 5x - 4 必须先改写为 x² - 5x + 4 = 0。
ng the ± sign when taking square roots. Always write both the positive and negative roots.
另一个错误是取平方根时遗漏 ± 号。务必写出正负两个根。
When using the quadratic formula, check the value of the discriminant before calculating the square root. A negative discriminant means you should stop and state 'no real roots' unless complex numbers are allowed.
使用二次公式时，在计算平方根之前先检查判别式的值。如果判别式为负，除非允许复数，否则应停止并说明"没有实数根"。
Finally, always check your answers by substituting them back into the original equation.
最后，务必通过将答案代入原方程来检查。
12. Practice Questions with Answers | 练习与答案
Try the following questions on your own, then check the solutions below.
先亲自尝试以下问题，然后核对下方的解答。
| Question | Answer |
| x² - 9x + 18 = 0 | x = 3 or 6 |
| 2x² - 3x - 5 = 0 | x = 2.5 or -1 |
| x² + 4x + 1 = 0 (use formula) | x = -2 ± √3 |
| 4x² - 12x + 9 = 0 | x = 1.5 (repeated root) |
For question 5, complete the square for x² - 6x + 2 = 0: (x - 3)² - 7 = 0, so x = 3 ± √7.
对于第 5 题，对 x² - 6x + 2 = 0 配方法：(x - 3)² - 7 = 0，所以 x = 3 ± √7。
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