📚 Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与求根公式
Quadratic equations are a core topic in IGCSE Mathematics and a foundation skill for A Level. They appear in algebra, graphs, geometry and problem-solving questions. This guide explains how to solve quadratics by factorising, completing the square and using the quadratic formula, and how to interpret the roots graphically.
二次方程是 IGCSE 数学的核心主题，也是 A Level 的重要基础。它们广泛出现在代数、图像、几何测量和应用题中。本指南将讲解如何用因式分解法、配方法和求根公式求解二次方程，以及如何从图像上理解方程根的意义。
1. What Is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is a polynomial equation of degree 2. The highest power of the variable x is x², which means the equation can have up to two real solutions, called roots.
二次方程是次数为 2 的多项式方程。未知数 x 的最高次幂是 x²，因此方程最多可以有两个实数解，称为根。
The standard form is ax² + bx + c = 0 ts and a cannot be zero. If a = 0, the x² term disappears and the equation becomes linear.
标准形式为 ax² + bx + c = 0，其中 a、b、c 为常数且 a 不能为零。如果 a = 0，x² 项消失，方程就变为一次方程。
h as x² + 4x = 5. Always rearrange these into standard form before identifying a, b and c.
例如，2x² − 5x + 3 = 0 是二次方程，因为 a = 2、b = −5、c = 3。x² − 9 = 0 也属于二次方程，其中 b = 0。你还可能看到形如 x² + 4x = 5 的隐藏形式。务必先将其整理为标准形式，再确定 a、b、c 的值。
In exam papers the word 'root' means a value of x that makes the expression equal to zero. Roots are also called solutions or zeros, and on a graph they are the x-intercepts of the curve.
在考题中，'根'指使表达式等于零的 x 值。根也称为解或零点，在图像上表现为曲线与 x 轴交点的横坐标。
2. Standard Form and Key Vocabulary | 标准形式与关键术语
ives the standard form ax² + bx + c = 0, and only then can you read off a, b and c safely.
在求解任何二次方程之前，先把所有项移到一边，使另一边等于零。这样就得到标准形式 ax² + bx + c = 0，只有此时才能安全地读出 a、b、c 的值。
ing terms are written with coefficient zero, which is why x² − 9 = 0 has b = 0 and c = −9.
系数 a 是 x² 的系数，b 是 x 的系数，c 是常数项。缺失的项系数记为零，所以 x² − 9 = 0 中 b = 0、c = −9。
| Equation | Standard form | a, b, c |
| 3x² = 12 | 3x² − 12 = 0 | a = 3, b = 0, c = −12 |
| x(x + 7) = 0 | x² + 7x = 0 | a = 1, b = 7, c = 0 |
| 5 − 2x = x² | x² + 2x − 5 = 0 | a = 1, b = 2, c = −5 |
| (x + 1)(x − 4) = 6 | x² − 3x − 10 = 0 | a = 1, b = −3, c = −10 |
Notice the last row: the equation is given in factorised form, but it is not equal to zero, so you must expand and rearrange first. A factorised expression is only directly useful when the product equals zero.
注意最后一行：方程以因式形式给出，但右边不等于零，所以必须先展开并整理。只有乘积等于零时，因式形式才能直接用来求根。
3. Solving by Factorising | 用因式分解法求解
ve zero, at least one of them must be zero. So if (x − p)(x − q) = 0, then x = p or x = q.
因式分解法依据零乘积法则：若两个数相乘得零，则其中至少一个为零。因此若 (x − p)(x − q) = 0，则 x = p 或 x = q。
duct is c and whose sum is b. Then x² + bx + c = (x + p)(x + q) with pq = c and p + q = b.
当 a = 1 时，寻找两个数，使它们的乘积为 c、和为 b。于是 x² + bx + c = (x + p)(x + q)，其中 pq = c，p + q = b。
d 3 multiply to 6 and add to 5, so (x + 2)(x + 3) = 0 and the roots are x = −2 and x = −3.
例：x² + 5x + 6 = 0。2 与 3 相乘得 6、相加得 5，所以 (x + 2)(x + 3) = 0，根为 x = −2 与 x = −3。
x² − 2x − 3x + 3 = 0, then 2x(x − 1) − 3(x − 1) = 0, so (2x − 3)(x − 1) = 0 and x = 3/2 or x = 1.
当 a ≠ 1 时，需要的乘积是 ac、和仍是 b，然后拆开中间项。对于 2x² − 5x + 3 = 0，ac = 6、b = −5，取 −2 与 −3，于是 2x² − 2x − 3x + 3 = 0，再得 2x(x − 1) − 3(x − 1) = 0，即 (2x − 3)(x − 1) = 0，所以 x = 3/2 或 x = 1。
quare gives a repeated root, for example x² + 6x + 9 = 0 gives (x + 3)² = 0 and x = −3 only.
两个特殊模式值得记忆：平方差 x² − k² = (x − k)(x + k)，以及完全平方 x² + 2kx + k² = (x + k)²。完全平方给出重根，例如 x² + 6x + 9 = 0 得 (x + 3)² = 0，只有 x = −3 一个根。
If no integer pair works, the expression may still factorise with surds, but in an exam it is usually a signal to switch to the formula or to completing the square.
如果找不到整数对，表达式或许仍可用根式因式分解，但在考试中这通常提示你改用求根公式或配方法。
4. Completing the Square | 配方法
Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. This form is very useful because it reveals the minimum or maximum point of the graph immediately.
配方法把 ax² + bx + c 改写为 a(x + p)² + q 的形式。这个形式非常有用，因为它能立刻显示图像的最低点或最高点。
+ bx = (x + b/2)² − (b/2)². Halve the coefficient of x, square that half, and subtract it.
对于首项系数为 1 的二次式，使用规则：x² + bx = (x + b/2)² − (b/2)²。把 x 的系数除以 2，再平方，然后减去它。
x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8
Setting this equal to zero gives (x + 3)² = 8, so x + 3 = ±√8 = ±2√2, and therefore x = −3 + 2√2 or x = −3 − 2√2. These exact surd values are usually required unless the question asks for decimals.
令其等于零得 (x + 3)² = 8，所以 x + 3 = ±√8 = ±2√2，因此 x = −3 + 2√2 或 x = −3 − 2√2。除非题目要求小数，否则一般要保留这种精确的根式形式。
6x) + 7 = 2[(x − 3)² − 9] + 7 = 2(x − 3)² − 11. The vertex of the curve is then (3, −11).
当 a ≠ 1 时，先从前两项提出系数 a。对于 2x² − 12x + 7，写成 2(x² − 6x) + 7 = 2[(x − 3)² − 9] + 7 = 2(x − 3)² − 11，曲线顶点为 (3, −11)。
5. Deriving and Using the Quadratic Formula | 求根公式的推导与使用
bx + c = 0. Starting from the standard form and dividing by a gives x² + (b/a)x + c/a = 0.
求根公式正是对一般方程 ax² + bx + c = 0 配方得到的。从标准形式出发并除以 a，得 x² + (b/a)x + c/a = 0。
)² = (b² − 4ac) / 4a², and taking square roots of both sides leads to the familiar result:
配方后得到 (x + b/2a)² = (b² − 4ac) / 4a²，两边开平方即得到熟悉的结果：
x = (−b ± √b² − 4ac) ÷ (2a)
To use it: rearrange into standard form, write down a, b and c with their signs, substitute carefully, then evaluate the discriminant b² − 4ac before simplifying the square root.
使用方法：先整理成标准形式，写出带符号的 a、b、c，仔细代入，然后先计算判别式 b² − 4ac，再化简根号。
1.5 or x = 1. This matches the factorised answer from Section 3, which is a useful check.
对于 2x² − 5x + 3 = 0，a = 2、b = −5、c = 3，所以 b² − 4ac = 25 − 24 = 1，得 x = (5 ± 1) ÷ 4，即 x = 1.5 或 x = 1。这与第 3 节因式分解的结果一致，是一个很好的检验。
Always show the substitution line in your working. Method marks are awarded for a correct substitution even if a later arithmetic slip changes the final answer.
答题时一定要写出代入的那一行。即使之后的计算失误改变了最终答案，正确的代入仍可获得方法分。
6. The Discriminant and the Number of Roots | 判别式与根的个数
The expression b² − 4ac is called the discriminant, often written as Δ. It tells you how many real roots the equation has before you do any further work.
表达式 b² − 4ac 称为判别式，通常记作 Δ。它能在进一步计算之前告诉你方程有多少个实根。
| Discriminant | Nature of roots | Graph |
| Δ > 0 | Two distinct real roots | Crosses the x-axis twice |
| Δ = 0 | One repeated real root | Touches the x-axis at the vertex |
| Δ < 0 | No real roots | Does not meet the x-axis |
ated root x = −2. The equation x² + x + 1 = 0 has Δ = 1 − 4 = −3, so it has no real roots.
例如，x² + 4x + 4 = 0 的 Δ = 16 − 16 = 0，只有一个重根 x = −2。方程 x² + x + 1 = 0 的 Δ = 1 − 4 = −3，因此没有实根。
sks for the values of k that give equal roots. Set Δ = 0: k² − 36 = 0, so k = 6 or k = −6.
常见考题会给出未知常数，例如 x² + kx + 9 = 0，问 k 取何值时方程有等根。令 Δ = 0：k² − 36 = 0，所以 k = 6 或 k = −6。
7. Roots, Intercepts and Graphs | 根、交点与图像
. If a > 0 it opens upwards and has a minimum point; if a < 0 it opens downwards and has a maximum point.
y = ax² + bx + c 的图像是一条抛物线。若 a > 0，开口向上并有最低点；若 a < 0，开口向下并有最高点。
are exactly the x-coordinates where the curve crosses the x-axis. The y-intercept is always the point (0, c).
方程 ax² + bx + c = 0 的根正好是曲线与 x 轴交点的横坐标。y 轴截距永远是点 (0, c)。
he same information in the friendlier form y = a(x + p)² + q, where the vertex is (−p, q).
对称轴经过顶点，方程为 x = −b/2a。配方法以更直观的形式 y = a(x + p)² + q 给出同样的信息，顶点为 (−p, q)。
. The vertex is (3, −4), the axis of symmetry is x = 3, and the roots are x = 1 and x = 5.
对于 y = x² − 6x + 5，配方得 y = (x − 3)² − 4。顶点为 (3, −4)，对称轴为 x = 3，根为 x = 1 与 x = 5。
o ax² + bx + c = 0, and solve. The number of intersections is decided by the discriminant.
当题目问直线与曲线的交点时，令两个表达式相等，整理为 ax² + bx + c = 0 后求解。交点的个数由判别式决定。
8. Hidden Quadratics and Equations That Need Rearranging | 隐藏的二次方程与需要整理的方程
² + 4x − 5 = 0 before solving, which factorises to (x + 5)(x − 1) = 0, so x = −5 or x = 1.
有些方程需要整理后才成为二次方程。x² + 4x = 5 必须先写成 x² + 4x − 5 = 0，再因式分解为 (x + 5)(x − 1) = 0，所以 x = −5 或 x = 1。
then 3x² − x − 2 = 0. Remember that any value making a denominator zero must be rejected.
形如 (x + 2)/x
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