📚 Mastering Quadratic Equations: Factorisation, Completing the Square and the Quadratic Formula | 掌握二次方程：因式分解、配方法与求根公式
Quadratic equations are central to the IGCSE Mathematics course. A solid understanding of their structure and solution methods is essential for success in algebra, graph sketching and many real-world applications. This revision guide introduces the standard form of a quadratic, shows three main methods of solving it, and explains how the discriminant helps you predict the number of roots without solving the equation.
二次方程是 IGCSE 数学课程的核心内容。牢固掌握二次方程的结构与求解方法，对于代数、函数图像以及许多实际应用中的成功至关重要。本复习指南将介绍二次方程的标准形式，展示三种主要求解方法，并说明如何利用判别式在不求解的情况下判断根的数量。
1. Recognising a Quadratic Equation | 识别二次方程
A quadratic equation is any equation that can be rearranged into the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the unknown is 2, and that single feature distinguishes a quadratic from a linear equation. If a were zero, the x² term would vanish and the equation would become linear rather than quadratic.
二次方程是指可以整理成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数，且 a ≠ 0。未知数的最高次数是 2，正是这一特征把二次方程与一次方程区分开来。如果 a 等于零，x² 项就会消失，方程便退化为一次方程而不是二次方程。
Example: 3x² + 2x - 5 = 0 is quadratic, because the highest power of x is 2.
示例：3x² + 2x - 5 = 0 是二次方程，因为 x 的最高次数是 2。
Example: 5x - 7 = 0 is not quadratic, because the highest power of x is 1.
示例：5x - 7 = 0 不是二次方程，因为 x 的最高次数是 1。
9 is quadratic, and it can be rewritten as x² - 9 = 0, giving the roots x = 3 and x = -3.
示例：x² = 9 是二次方程，可改写为 x² - 9 = 0，其根为 x = 3 与 x = -3。
2. The Standard Form ax² + bx + c = 0 | 标准形式 ax² + bx + c = 0
to obtain the standard form ax² + bx + c = 0. This standard form makes factorisation, completing the square and the quadratic formula easier to apply, and it prevents sign errors later on.
在应用任何求解方法之前，应先将方程整理为所有项都在等号一侧、另一侧为零。各项按 x 的降幂排列，即得到标准形式 ax² + bx + c = 0。这个标准形式会使因式分解、配方法和求根公式的应用更加简便，也能避免后续出现符号错误。
reliably factorise it or identify the values of a, b and c. Here a = 1, b = 3 and c = -10.
例如，x² + 3x = 10 必须先整理为 x² + 3x - 10 = 0。只有这样，你才能准确地进行因式分解或确定 a、b、c 的值。此处 a = 1，b = 3，c = -10。
nd the discriminant still work perfectly, but only if you treat the missing coefficient as zero rather than ignoring it.
注意缺项的情况：在 x² - 5 = 0 中 b = 0；在 4x² + x = 0 中 c = 0。求根公式与判别式依然完全适用，但前提是你把缺失的系数当作 0，而不是直接忽略它。
3. Solving by Factorisation | 因式分解法
that if the product of two factors is zero, then at least one of the factors must be zero. So if (px + q)(rx + s) = 0, then either px + q = 0 or rx + s = 0. Solving each linear equation gives the roots of the original quadratic.
如果一个二次式可以因式分解为两个一次括号的乘积，就可以使用零乘积性质。该性质指出：若两个因子的乘积为零，则其中至少有一个因子必须为零。因此若 (px + q)(rx + s) = 0，则 px + q = 0 或 rx + s = 0。分别解这两个一次方程，就得到原二次方程的根。
3x - 10 = 0, the two numbers are 5 and -2, because 5 × (-2) = -10 and 5 + (-2) = 3. Hence:
对于 a = 1 的 x² + bx + c，需要找到两个数，使它们的积为 c、和为 b。在 x² + 3x - 10 = 0 中，这两个数是 5 和 -2，因为 5 × (-2) = -10，且 5 + (-2) = 3。因此：
(x + 5)(x - 2) = 0 → x = -5 or x = 2
Factorisation is usually the fastest method, but it only works neatly when the roots are rational. If no pair of integers or simple fractions fits, switch to completing the square or the quadratic formula.
因式分解通常是最快的方法，但它只在根为有理数时才简洁有效。如果找不到合适的整数或简单分数组合，就应改用配方法或求根公式。
4. Factorisation Techniques and Special Cases | 因式分解技巧与特殊情况
nd factorise in pairs. For 3x² + 2x - 5 = 0, a × c = -15, and the numbers 5 and -3 give a sum of 2, so 3x² + 5x - 3x - 5 = 0, which factorises to (3x + 5)(x - 1) = 0, giving x = -5/3 or x = 1.
当 x² 的系数不为 1 时，先计算 a 与 c 的乘积，再找两个数使其积等于该乘积、和等于 b。然后拆开中间项并分组分解。对于 3x² + 2x - 5 = 0，a × c = -15，而 5 与 -3 之和为 2，于是 3x² + 5x - 3x - 5 = 0，分解得 (3x + 5)(x - 1) = 0，即 x = -5/3 或 x = 1。
Common factor first: 2x² - 8x = 0 becomes 2x(x - 4) = 0, so x = 0 or x = 4.
先提取公因式：2x² - 8x = 0 化为 2x(x - 4) = 0，故 x = 0 或 x = 4。
Difference of two squares: a² - b² = (a + b)(a - b), so x² - 49 = 0 gives x = ±7.
平方差公式：a² - b² = (a + b)(a - b)，故 x² - 49 = 0 给出 x = ±7。
e: a² ± 2ab + b² = (a ± b)², so x² - 6x + 9 = 0 gives (x - 3)² = 0, a repeated root x = 3.
完全平方：a² ± 2ab + b² = (a ± b)²，故 x² - 6x + 9 = 0 化为 (x - 3)² = 0，得到重根 x = 3。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q, which reveals the vertex of the parabola and handles irrational roots that factorisation cannot reach. Start by dividing through by a so the coefficient of x² becomes 1, then add and subtract the square of half the coefficient of x.
配方法把二次式改写为 a(x + p)² + q 的形式，既能直接读出抛物线的顶点，也能处理因式分解无法解决的无理根。首先两边同时除以 a，使 x² 的系数变为 1，然后加上并减去一次项系数一半的平方。
For a monic quadratic x² + bx + c, the identity is:
对于首项系数为 1 的二次式 x² + bx + c，恒等式为：
x² + bx + c = (x + b/2)² - (b/2)² + c
+ 3)² - 9 + 1 = (x + 3)² - 8. Setting this equal to zero gives (x + 3)² = 8, so x + 3 = ±√8 and therefore:
以 x² + 6x + 1 = 0 为例。6 的一半是 3，3² = 9，因此 x² + 6x + 1 = (x + 3)² - 9 + 1 = (x + 3)² - 8。令其等于零得 (x + 3)² = 8，故 x + 3 = ±√8，于是：
x = -3 ± 2√2 (≈ -0.172 or ≈ -5.828)
+ 3)² - 8 has a minimum of -8, reached at x = -3. This is exactly the vertex of the curve.
配方式还直接给出最小值：由于 (x + 3)² ≥ 0，式子 (x + 3)² - 8 的最小值为 -8，在 x = -3 时取得。这正是该曲线的顶点。
6. The Quadratic Formula | 求根公式
e on the general form ax² + bx + c = 0. Divide by a, complete the square, and rearrange to isolate x:
求根公式可以求解任何二次方程，无论其根是有理数、无理数还是复数。它正是对一般形式 ax² + bx + c = 0 配方推导而来的。两边除以 a，配方，再整理以解出 x：
x = (-b ± √b² - 4ac) / (2a)
d place brackets around negative values of b and c to avoid sign mistakes. For 2x² - 4x - 3 = 0, a = 2, b = -4 and c = -3, so b² - 4ac = 16 + 24 = 40 and:
符号 ± 意味着必须算出两个值，一个取加号、一个取减号。代入前请先写出公式，并把 b、c 的负值加上括号，以避免符号错误。对于 2x² - 4x - 3 = 0，a = 2，b = -4，c = -3，因此 b² - 4ac = 16 + 24 = 40，于是：
x = (4 ± √40) / 4 = (2 ± √10) / 2 (≈ 2.581 or ≈ -0.581)
Leave surd answers in exact form unless the question asks for a decimal, and give decimals to three significant figures when accuracy is specified.
除非题目要求小数，否则根式答案应保留精确形式；若题目规定了精确度，则小数答案通常保留三位有效数字。
7. The Discriminant and the Number of Roots | 判别式与根的个数
The expression under the square root, b² - 4ac, is called the discriminant and is often written as Δ. Because you cannot take the square root of a negative number in real numbers, the sign of the discriminant decides how many real roots the equation has, without any need to solve it.
根号下的表达式 b² - 4ac 称为判别式，常记作 Δ。由于在实数范围内不能对负数开平方，判别式的正负决定了方程有多少个实根，而无需真正求解。
| Δ = b² - 4ac | Number of real roots | 实根个数 | Graph | 图像 |
| Δ > 0 | Two distinct real roots | 两个不同实根 | Crosses the x-axis twice | 与 x 轴交于两点 |
| Δ = 0 | One repeated real root | 一个重根 | Touches the x-axis once | 与 x 轴相切于一点 |
| Δ < 0 | No real roots | 没有实根 | Does not meet the x-axis | 与 x 轴不相交 |
x = 2. Meanwhile x² + x + 5 = 0 has Δ = 1 - 20 = -19 < 0, so it has no real solutions.
例如，x² - 4x + 4 = 0 的 Δ = 16 - 16 = 0，故在 x = 2 处有重根。而 x² + x + 5 = 0 的 Δ = 1 - 20 = -19 < 0，因此没有实数解。
8. Choosing the Best Method | 选择最合适的方法
All three methods give the same roots, so the choice is about speed and accuracy. Factorisation is quickest when the roots are integers or simple fractions. Completing the square is the natural choice when the question asks for a vertex, a minimum value or a proof of a surd form. The quadratic formula is the universal fallback that always works.
三种方法给出的根完全相同，区别在于速度与准确度。当根为整数或简单分数时，因式分解最快。当题目要求顶点、最小值或证明根式形式时，配方法是自然选择。求根公式则是万能的后备方案，永远适用。
Factorisation: fastest, but only when roots are rational and easy to spot.
因式分解：最快，但仅在根为有理数且容易看出时适用。
Completing the square: shows the vertex and works with irrational roots.
配方法：能显示顶点，并可处理无理根。
Quadratic formula: always works, but requires careful substitution.
求根公式：永远适用，但需要小心代入计算。
A useful exam habit is to attempt factorisation first for a few seconds; if nothing obvious appears, move straight to the formula rather than losing time.
一个实用的考试习惯是：先用几秒钟尝试因式分解；如果没有明显结果，就直接改用求根公式，而不要在猜测上浪费时间。
9. Quadratics and Their Graphs | 二次方程与图像
. Its roots are exactly the x-coordinates where the curve crosses the x-axis, which is why the discriminant and the graph tell the same story. If a > 0 the parabola opens upwards; if a < 0 it opens downwards.
y = ax² + bx + c 的图像是一条抛物线。方程的根恰好就是曲线与 x 轴交点的横坐标，这也是判别式与图像给出的结论完全一致的原因。若 a > 0，抛物线开口向上；若 a < 0，开口向下。
ntercept is simply the constant c, here 1. Notice that the two roots found earlier, -3 ± 2√2, sit symmetrically on either side of the axis of symmetry.
配方把 y = x² + 6x + 1 化为 y = (x + 3)² - 8，因此顶点为 (-3, -8)，对称轴为 x = -3。y 轴截距就是常数 c，此处为 1。可以注意到，前面求得的两根 -3 ± 2√2 正好对称地分布在对称轴两侧。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
d minus signs and forgetting the ± are the classic culprits. Checking each answer by substituting it back into the original equation takes only a few seconds and catches almost every mistake.
二次方程题目的失分大多来自细小的代数失误，而不是方法理解错误。移项错误、漏掉负号、忘记 ± 都是经典问题。把答案代回原方程检验只需几秒钟，却能捕捉到几乎所有的错误。
by x: from 2x² = 8x you must not cancel x, or you lose the root x = 0. Factorise instead.
不要两边除以 x：由 2x² = 8x 不能约去 x，否则会丢失根 x = 0。应改用因式分解。
over: when rearranging x² + 3x = 10, subtract 10 from both sides, not just from the left.
整项移项：整理 x² + 3x = 10 时，两边都要减 10，而不是只减左边。
Bracket your negatives: writing (-4)² rather than -4² prevents a sign disaster in the formula.
负值加括号：写成 (-4)² 而不是 -4²，可避免求根公式中的符号灾难。
Check the discriminant first: if Δ < 0 you can state immediately that there are no real roots.
先算判别式：若 Δ < 0，可以直接说明方程没有实根。
11. Worked Exam-Style Questions | 考试式例题精讲
² - 5x + 6 = 0. The product is 6 and the sum is -5, so the numbers are -2 and -3. Hence (x
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