📚 Master Quadratics: A Complete IGCSE Guide | 掌握二次方程：IGCSE 数学完整指南
A quadratic equation is a polynomial equation of degree 2. In IGCSE Mathematics, you must solve these equations and interpret their graphs. They appear in algebra, coordinate geometry, and problem-solving questions across both examination papers.
二次方程是次数为 2 的多项式方程。在 IGCSE 数学中，你必须会求解这类方程并解读其图像。它们出现在代数、坐标几何以及两份试卷的应用题中。
1. Introduction to Quadratic Equations | 二次方程简介
A quadratic equation always contains an x² term and can be written in various equivalent forms. The highest power of the variable is 2, which is why it is called "quadratic" (from the Latin quadratus, meaning square).
二次方程一定包含 x² 项，并且可以写成多种等价形式。变量的最高次数为 2，因此被称为"二次"（quadratic，源自拉丁语 quadratus，意为"平方"）。
In the IGCSE syllabus, you are expected to solve quadratics by factorisation, by using the quadratic formula, and by completing the square. You should also be able to sketch the corresponding parabola and identify its key features.
在 IGCSE 考纲中，你需要掌握因式分解法、二次公式法和配方法求解二次方程，同时还要能画出对应的抛物线，并识别其关键特征。
2. The Standard Form | 标准形式
Every quadratic equation can be rearranged into the standard form, where a, b and c are real numbers and a ≠ 0.
每个二次方程都可以化为标准形式，其中 a、b、c 为实数，且 a ≠ 0。
ax² + bx + c = 0
c is the constant term. For example, in 2x² − 3x + 5 = 0, we have a = 2, b = −3 and c = 5.
系数 a 是 x² 前面的数；b 是 x 的系数；c 是常数项。例如，在 2x² − 3x + 5 = 0 中，a = 2，b = −3，c = 5。
The value of a cannot be zero, because then the equation would simply become linear (a straight line). The sign of a also determines whether the parabola opens upward or downward.
a 的值不能为零，否则方程就退化为一次方程（直线）。a 的正负还决定抛物线的开口方向。
3. Solving by Factorisation | 因式分解法求解
When a quadratic expression factorises neatly, this is usually the quickest method. You convert the sum of terms into a product of two brackets, then use the fact that if a product is zero, at least one factor must be zero.
当二次式可以顺利因式分解时，这通常是最快的方法。你把各项之和转化为两个括号之积，然后利用"乘积为零则至少一个因式为零"的性质。
Rearrange the equation into the form ax² + bx + c = 0.
先将方程整理为 ax² + bx + c = 0 的形式。
Factorise the left-hand side into two linear brackets.
将左侧因式分解为两个一次括号。(px + q)(rx + s) = 0
Set each bracket equal to zero and solve the two linear equations.
令每个括号等于零，然后分别解两个一次方程。
Example 1: Solve x² − 5x + 6 = 0.
例 1：解方程 x² − 5x + 6 = 0。
(x − 2)(x − 3) = 0 ⇒ x = 2 或 x = 3
Example 2: Solve 2x² + 5x − 3 = 0.
例 2：解方程 2x² + 5x − 3 = 0。
(2x − 1)(x + 3) = 0 ⇒ x = ½ 或 x = −3
Always check whether the constant term and the coefficient of x allow simple integer factors. If factorisation is not obvious within a few seconds, move on to the quadratic formula.
始终检查常数项和 x 的系数是否能构成简单的整数因式。如果几秒内看不出因式，就改用二次公式。
4. The Quadratic Formula | 二次公式
ys be found using the quadratic formula. This works even when factorisation is difficult or impossible.
对于任何形如 ax² + bx + c = 0 的二次方程，都可以用二次公式求出解。即使因式分解很困难甚至不可能，该方法也有效。
x = (−b ± √b² − 4ac) / 2a
means there are two possible values: one using the plus sign and one using the minus sign.
符号 ± 表示有两个可能的值：一个取加号，一个取减号。
Worked example: Solve 2x² + 3x − 5 = 0 using the formula.
例题：用公式法解方程 2x² + 3x − 5 = 0。
= 2, b = 3 and c = −5. Substitute these values carefully, keeping the negative sign of c.
这里 a = 2，b = 3，c = −5。代入时要小心，保留 c 的负号。
x = (−3 ± √(3² − 4 × 2 × (−5))) / (2 × 2) = (−3 ± √9 + 40) / 4 = (−3 ± 7) / 4
x = 1 或 x = −2.5
Write down each substitution step; examiners award method marks even if your final arithmetic is wrong.
写出每一步代入过程；即使最后计算有误，阅卷官也会给方法分。
5. Completing the Square | 配方法
Completing the square rewrites x² + bx + c in the form (x + p)² + q. This form is especially useful for finding the vertex and for solving equations without factorising.
配方法将 x² + bx + c 改写为 (x + p)² + q 的形式。这种形式特别适合求顶点，也适合不解因式直接求根。
(x + b/2)² + (c − b²/4)
Example: Write x² + 6x + 2 in completed-square form.
例：将 x² + 6x + 2 写成配方法形式。
x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7
rearrange to (x + 3)² = 7, then take the square root of both sides. Remember to include ±.
解 (x + 3)² − 7 = 0 时，先移项得 (x + 3)² = 7，然后两边开平方。记得要加 ±。
x = −3 ± √7
When the coefficient of x² is not 1, first divide the whole equation by a before completing the square.
当 x² 的系数不为 1 时，先两边除以 a，再做配方。
6. The Discriminant | 判别式
The expression b² − 4ac inside the square root is called the discriminant, denoted by Δ (delta). It tells you the nature of the roots without solving the equation.
根号内的表达式 b² − 4ac 称为判别式，记为 Δ（德尔塔）。它无需解方程就能告诉你根的性质。
Δ = b² − 4ac
If Δ > 0, the equation has two distinct real roots.
若 Δ > 0，方程有两个不同的实数根。
If Δ = 0, the equation has one repeated real root (two equal roots).
若 Δ = 0，方程有一个重根（两个相等的实数根）。
If Δ < 0, the equation has no real roots.
若 Δ < 0，方程没有实数根。
Example: tly one root x = 2. For x² + 2x + 5 = 0, Δ = 4 − 20 = −16 < 0, so there are no real roots.
例：对 x² − 5x + 6 = 0，Δ = 25 − 24 = 1 > 0，所以有两个根。对 x² − 4x + 4 = 0，Δ = 16 − 16 = 0，所以只有一个根 x = 2。对 x² + 2x + 5 = 0，Δ = 4 − 20 = −16 < 0，所以没有实数根。
7. Quadratic Graphs | 二次函数图像
always a parabola. The sign of a controls the orientation, while c gives the y-intercept.
y = ax² + bx + c 的图像永远是抛物线。a 的符号控制开口方向，c 给出 y 轴截距。
If a > 0, the parabola opens upward (∪ shape) and has a minimum point.
若 a > 0，抛物线开口向上（∪ 形），有最低点。
If a < 0, the parabola opens downward (∩ shape) and has a maximum point.
若 a < 0，抛物线开口向下（∩ 形），有最高点。
The y-intercept is at (0, c), where the graph crosses the y-axis.
y 轴截距在 (0, c)，即图像与 y 轴的交点。
cepts (if they exist) are exactly the real roots found from the equation ax² + bx + c = 0.
x 轴截距（若存在）正是方程 ax² + bx + c = 0 的实数根。
When asked to sketch a parabola, always mark the vertex, the y-intercept, and either the roots or the discriminant sign.
当要求画抛物线草图时，务必标出顶点、y 轴截距，以及根或判别式的符号。
8. Vertex and Axis of Symmetry | 顶点与对称轴
Every parabola has a vertical axis of symmetry passing through its vertex. The formula for the axis is derived from the completed-square form.
每条抛物线都有一条经过顶点的竖直对称轴。对称轴的公式可由配方法形式推导出来。
x = −b / (2a)
as y = (x − h)² + k, then the vertex is simply (h, k), and the axis of symmetry is x = h.
若把方程写成 y = (x − h)² + k，那么顶点就是 (h, k)，对称轴就是 x = h。
Example: re the vertex is (2, −3), the axis of symmetry is x = 2, and the minimum value of y is −3.
例：对 y = x² − 4x + 1，配方得 y = (x − 2)² − 3。因此顶点为 (2, −3)，对称轴为 x = 2，y 的最小值为 −3。
9. Real-World Applications | 实际应用
Quadratic equations model many real situations, including area problems, projectile motion and economic profit functions. You must translate the words into algebra before solving.
二次方程可以建模许多实际情境，包括面积问题、抛体运动和利润函数。你必须先把文字转化为代数式再求解。
Area problem: A rectangle is 4 cm longer than it is wide, and its area is 45 cm². Find its dimensions.
面积问题：一个矩形比宽长 4 cm，面积为 45 cm²。求其尺寸。
w(w + 4) = 45 ⇒ w² + 4w − 45 = 0 ⇒ (w + 9)(w − 5) = 0
e negative value because a length cannot be negative. The rectangle measures 5 cm by 9 cm.
因此 w = 5 或 w = −9。舍去负值，因为长度不能为负。矩形尺寸为 5 cm × 9
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