📚 IGCSE Maths: Quadratic Equations and Graphs | IGCSE 数学：二次方程与图像
Quadratic equations and their graphs are a central topic in IGCSE Mathematics. You need to solve quadratics by factorising, completing the square, and using the quadratic formula, and you must be able to sketch and interpret parabolas confidently.
二次方程及其图像是 IGCSE 数学的核心主题。你需要掌握因式分解法、配方法和求根公式法来解二次方程，并能够熟练绘制和解读抛物线。
1. Recognising Quadratic Expressions | 识别二次表达式
s and a ≠ 0. If a = 0, the expression becomes linear, so the condition a ≠ 0 is essential.
二次表达式的一般形式为 ax² + bx + c，其中 a、b、c 为常数且 a ≠ 0。如果 a = 0，表达式就变成一次式，因此 a ≠ 0 这一条件至关重要。
The coefficient a controls the shape and direction of the parabola, while b and c affect its position. In IGCSE questions, you may be given a quadratic in standard form or one that needs rearranging first.
系数 a 决定抛物线的形状和开口方向，b 和 c 则影响其位置。在 IGCSE 题目中，二次式可能已经写成标准形式，也可能需要先进行整理。
Standard form: ax² + bx + c = 0
2. Solving by Factorising | 因式分解法求解
it as a product of two brackets. For example, x² − 5x + 6 = 0 becomes (x − 2)(x − 3) = 0.
当二次式可以因式分解时，最快的方法是将它写成两个括号的乘积。例如 x² − 5x + 6 = 0 可化为 (x − 2)(x − 3) = 0。
roperty: if AB = 0, then A = 0 or B = 0. So x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.
然后运用零乘积性质：若 AB = 0，则 A = 0 或 B = 0。因此 x − 2 = 0 或 x − 3 = 0，解得 x = 2 或 x = 3。
Always check by substituting the solutions back into the original equation. This helps you catch sign errors and factorising mistakes.
一定要将解代回原方程检验。这可以帮助你发现符号错误和因式分解的错误。
3. The Quadratic Formula | 求根公式
be factorised easily, use the quadratic formula. For ax² + bx + c = 0, the solutions are:
如果二次式不容易因式分解，就使用求根公式。对于 ax² + bx + c = 0，其解为：
x = (−b ± √b² − 4ac) / 2a
This formula works for every quadratic equation, including those with irrational or complex roots. In IGCSE, you will normally work with real roots unless a question says otherwise.
这个公式适用于所有二次方程，包括具有无理根或复数根的方程。在 IGCSE 考试中，除非题目另有说明，通常只涉及实数根。
Write down the values of a, b and c carefully before substituting. Pay special attention to negative coefficients, because mistakes often occur when b is negative.
代入前先仔细写出 a、b、c 的值。要特别注意负系数，因为当 b 为负数时最容易出错。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x − h)² + k. This form quickly reveals the vertex of the parabola and helps solve equations when factorising is awkward.
配方法将二次式改写为 a(x − h)² + k 的形式。这种形式能直接显示抛物线的顶点，并且在不易因式分解时有助于解方程。
first take half of 6 to get 3, then write (x + 3)² − 9 + 1 = 0, which is (x + 3)² − 8 = 0.
以 x² + 6x + 1 = 0 为例，先取 6 的一半得到 3，然后写成 (x + 3)² − 9 + 1 = 0，即 (x + 3)² − 8 = 0。
Then solve: (x + 3)² = 8, so x + 3 = ±√8, giving x = −3 ± 2√2. This method is also essential for deriving the quadratic formula.
然后求解：(x + 3)² = 8，因此 x + 3 = ±√8，得到 x = −3 ± 2√2。配方法也是推导求根公式的基础。
5. Discriminant and Nature of Roots | 判别式与根的性质
− 4ac. It tells you the number and type of roots without solving the equation completely.
判别式 Δ 定义为 Δ = b² − 4ac。它可以在不完整解方程的情况下告诉你根的个数和类型。
ots. If Δ = 0, there is exactly one repeated real root. If Δ < 0, there are no real roots.
如果 Δ > 0，则有两个不同的实数根；如果 Δ = 0，则有一个重根；如果 Δ < 0，则没有实数根。
ample, x² − 4x + 4 = 0 has Δ = (−4)² − 4(1)(4) = 0, so the roots are repeated: x = 2 only.
例如，x² − 4x + 4 = 0 的判别式 Δ = (−4)² − 4(1)(4) = 0，因此根是重根：只有 x = 2。
6. Graphs of Quadratic Functions | 二次函数图像
. 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，开口向上且有最低点；若 a < 0，开口向下且有最高点。
g x = 0 gives y = c. The x-intercepts are the real roots of the equation ax² + bx + c = 0.
y 轴截距为 (0, c)，因为代入 x = 0 可得 y = c。x 轴截距就是方程 ax² + bx + c = 0 的实数根。
To sketch a parabola, label the intercepts, the vertex, and the axis of symmetry. Use smooth curves, never straight line segments.
绘制抛物线时，要标出截距、顶点和对称轴。用平滑曲线连接，绝不能用直线段。
7. Vertex and Axis of Symmetry | 顶点与对称轴
s given by x = −b / 2a. Substitute this value into the quadratic to find the y-coordinate.
顶点的 x 坐标由 x = −b / 2a 给出。将这个值代入二次式即可求出 y 坐标。
he vertical line x = −b / 2a. This line divides the parabola into two mirror-image halves.
对称轴是竖直直线 x = −b / 2a。这条线将抛物线分成左右镜像对称的两部分。
k, the vertex is simply (h, k). This is often the fastest way to locate the turning point.
如果方程是配方法形式 y = a(x − h)² + k，顶点就是 (h, k)。这通常是确定转折点最快的方法。
Vertex x-coordinate: x = −b / 2a
8. Interpreting Real-Life Problems | 实际应用解读
Quadratic functions often model projectile motion, area problems, and profit maximisation. In these contexts, the vertex gives the maximum or minimum value.
二次函数常用于建模抛体运动、面积问题和利润最大化。在这些情境中，顶点给出最大值或最小值。
ds may be h = −5t² + 20t + 1. The maximum height occurs at t = −20 / (2 × −5) = 2 seconds.
例如，一个球在 t 秒后的高度可表示为 h = −5t² + 20t + 1。最大高度出现在 t = −20 / (2 × −5) = 2 秒时。
Interpret answers in the context of the problem, and reject impossible values such as negative time.
代入 t = 2 求出最大高度：h = −5(2)² + 20(2) + 1 = 21 米。要结合题意解释答案，并舍去不合理值（如负时间）。
9. Common Mistakes | 常见错误
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