📚 Quadratic Equations, Graphs and Inequalities for IGCSE Maths | IGCSE 数学：二次方程、图像与不等式
Quadratic equations are one of the core algebraic topics in IGCSE Mathematics. They involve an x² term and appear in a wide range of problems, from factorising and using the quadratic formula to sketching parabolas and solving real-world problems.
二次方程是 IGCSE 数学的核心代数主题之一。它们含有 x² 项，出现在各类问题中，从因式分解、使用求根公式，到绘制抛物线图像和解决实际问题。
In this revision guide, you will learn how to recognise, solve and graph quadratic equations. You will also see how the discriminant controls the number of roots and how quadratic inequalities can be solved using a sketch.
在这份复习指南中，你将学习如何识别、求解和绘制二次方程图像。你还会看到判别式如何决定根的数量，以及如何利用草图解二次不等式。
1. What Is a Quadratic Equation? | 什么是二次方程？
be written in the general form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.
二次方程是次数为 2 的多项式方程。它总能写成一般形式 ax² + bx + c = 0，其中 a、b、c 是常数且 a ≠ 0。
The condition a ≠ 0 is essential. If a were zero, the x² term would disappear and the equation would become linear, not quadratic. The x² term is what gives a quadratic equation its characteristic curve, called a parabola.
条件 a ≠ 0 非常重要。如果 a 为零，x² 项就会消失，方程就变成一次方程，而不是二次方程。x² 项正是二次方程具有抛物线这一特征曲线的根本原因。
Some common examples are shown below.
以下是一些常见例子。
In x² + 5x + 6 = 0, the coefficients are a = 1, b = 5 and c = 6.
在 x² + 5x + 6 = 0 中，系数为 a = 1，b = 5，c = 6。
In 3x² − 2x + 1 = 0, the coefficients are a = 3, b = −2 and c = 1.
在 3x² − 2x + 1 = 0 中，系数为 a = 3，b = −2，c = 1。
In −x² + 4x = 0, the coefficients are a = −1, b = 4 and c = 0.
在 −x² + 4x = 0 中，系数为 a = −1，b = 4，c = 0。
2. Writing Quadratics in Standard Form | 将二次式写成标准形式
is means moving all terms onto one side of the equals sign and simplifying where possible.
在求解二次方程之前，通常最好先把方程整理成标准形式 ax² + bx + c = 0。这意味着把所有项移到等号的一边，并尽可能化简。
2x² = 8x − 6 is not yet in standard form. Move all terms to the left-hand side to obtain:
例如，2x² = 8x − 6 还不是标准形式。把所有项移到左边，得到：
2x² − 8x + 6 = 0
common factor can make later steps much easier, but you must divide every term by the same number.
你可以进一步化简：每一项都除以公因数 2，得到 x² − 4x + 3 = 0。除以公因数可以让后面的步骤简单很多，但必须每项都除以同一个数。
= 3 becomes x² + 2x = 6 after multiplying by 2, and then x² + 2x − 6 = 0 in standard form.
如果方程含有分数，可以两边乘以最小公分母来去掉分母。例如，x²/2 + x = 3 乘以 2 后变成 x² + 2x = 6，再变为标准形式 x² + 2x − 6 = 0。
3. Solving by Factorisation | 因式分解法求解
Factorising is often the fastest method when a quadratic equation has integer roots. The idea is to write ax² + bx + c as a product of two brackets and then set each bracket equal to zero.
当二次方程有整数根时，因式分解通常是最快的方法。思路是把 ax² + bx + c 写成两个括号的乘积，然后令每个括号等于零。
le quadratics where a = 1, you need two numbers that multiply to give c and add to give b.
对于 a = 1 的简单二次式，你需要找到两个数，它们相乘等于 c，相加等于 b。
Example: Solve x² − 5x + 6 = 0.
例题：解 x² − 5x + 6 = 0。
We need two numbers that multiply to +6 and add to −5. The numbers are −2 and −3.
我们需要两个数，相乘为 +6，相加为 −5。这两个数是 −2 和 −3。
x² − 5x + 6 = (x − 2)(x − 3) = 0
Now set each bracket equal to zero: x − 2 = 0 or x − 3 = 0. Therefore x = 2 or x = 3.
现在令每个括号等于零：x − 2 = 0 或 x − 3 = 0。因此 x = 2 或 x = 3。
3 = 6. The two numbers that multiply to 6 and add to 7 are 6 and 1. Split the middle term:
当 a > 1 时，你可以使用拆分中项法。对于 2x² + 7x + 3，先计算 a × c：2 × 3 = 6。两个数相乘为 6、相加为 7，它们是 6 和 1。拆分中项：
2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)
Then solve (2x + 1)(x + 3) = 0 to get x = −1/2 or x = −3.
然后解 (2x + 1)(x + 3) = 0，得到 x = −1/2 或 x = −3。
4. Solving by Completing the Square | 配方法求解
Completing the square is especially useful when a quadratic cannot be factorised easily. It rewrites ax² + bx + c in the form a(x + p)² + q, which also helps when finding the vertex of a graph.
当二次式不容易因式分解时，配方法特别有用。它把 ax² + bx + c 改写成 a(x + p)² + q 的形式，这也有助于求图像的顶点。
For a monic quadratic such as x² + 6x + 2, take half of the coefficient of x, which is 3, and square it to get 9. Then write:
对于 x² + 6x + 2 这类首项系数为 1 的二次式，取 x 项系数的一半，即 3，再平方得到 9。然后写成：
x² + 6x + 2 = (x + 3)² − 7
To solve x² + 6x + 2 = 0, set (x + 3)² − 7 = 0. Then add 7 to both sides: (x + 3)² = 7.
要解 x² + 6x + 2 = 0，令 (x + 3)² − 7 = 0。然后两边加 7：(x + 3)² = 7。
x + 3 = ±√7
x = −3 ± √7
If a is not 1, factor it out of the first two terms first. For example, 2x² + 8x + 1 becomes 2(x² + 4x) + 1, and then complete the square inside the bracket before expanding the outside factor.
如果 a 不是 1，要先把 a 从前两项中提出来。例如，2x² + 8x + 1 先写成 2(x² + 4x) + 1，然后在括号内完成配方，最后再乘以外面的系数。
5. The Quadratic Formula | 求根公式
t is given in the IGCSE exam formula sheet, but you must know how to substitute correctly.
求根公式可以解任何形式为 ax² + bx + c = 0 的二次方程。IGCSE 考试公式表中会给出该公式，但你必须知道如何正确代入。
x = (−b ± √b² − 4ac) ÷ (2a)
Always identify a, b and c carefully, including any negative signs. Then calculate the discriminant b² − 4ac before simplifying the full expression.
一定要仔细识别 a、b、c，包括负号。然后先计算判别式 b² − 4ac，再化简整个表达式。
Example: Solve 2x² + 3x − 2 = 0.
例题：解 2x² + 3x − 2 = 0。
Here a = 2, b = 3 and c = −2. Substitute into the formula:
这里 a = 2，b = 3，c = −2。代入公式：
