📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are one of the core topics in IGCSE Mathematics. Mastering them is essential for both Paper 2 and Paper 4, and they appear in a wide variety of problem-solving contexts. In this revision guide, we will walk through every step you need to know.
二次方程是 IGCSE 数学的核心主题之一。掌握二次方程对于 Paper 2 和 Paper 4 都至关重要，并且在各种应用题中频繁出现。在本复习指南中，我们将逐步讲解你所需知道的每一个步骤。
1. What is a Quadratic Equation? | 什么是二次方程
x − 3 = 0. The general form is 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。这意味着变量的最高次数是 2。一个简单的例子是 x² = 9。另一个例子是 2x² + 5x − 3 = 0。一般形式是 ax² + bx + c = 0，其中 a、b 和 c 是常数，且 a ≠ 0。
2. Standard Form | 标准形式
3x² = 2x + 5 can be rearranged to 3x² − 2x − 5 = 0 by subtracting 2x + 5 from both sides.
在解任何二次方程之前，务必将其重新整理为标准形式 ax² + bx + c = 0。例如，方程 3x² = 2x + 5 可通过从两边减去 2x + 5 整理为 3x² − 2x − 5 = 0。
Why is this important? es that if A × B = 0, then either A = 0 or B = 0. This is the foundation of factorisation.
为什么这很重要？因为所有求解方法都依赖于让一边等于零。特别地，零乘积性质指出：如果 A × B = 0，那么 A = 0 或 B = 0。这是因式分解法的基础。
3. Solving by Factorisation | 因式分解法
and add to give 5. Those numbers are 2 and 3. Therefore, x² + 5x + 6 = (x + 2)(x + 3) = 0.
当方程容易因式分解时，因式分解法通常是最快的方法。为了解 x² + 5x + 6 = 0，我们寻找两个数，它们相乘得到 6，相加得到 5。这两个数是 2 和 3。因此 x² + 5x + 6 = (x + 2)(x + 3) = 0。
+ 2 = 0 gives x = −2, and x + 3 = 0 gives x = −3. So the solutions are x = −2 and x = −3.
利用零乘积性质，我们将每个因式设为零：x + 2 = 0 得 x = −2，x + 3 = 0 得 x = −3。因此解为 x = −2 和 x = −3。
or example, 2x² + 7x + 3 = 0 can be written as (2x + 1)(x + 3) = 0. Then x = −½ or x = −3.
当 x² 的系数不是 1 时，因式分解需要更仔细。例如，2x² + 7x + 3 = 0 可以写成 (2x + 1)(x + 3) = 0。于是 x = −½ 或 x = −3。
4. Solving by Completing the Square | 配方法
r example, x² + 6x + 1 = 0 becomes (x + 3)² − 9 + 1 = 0, which simplifies to (x + 3)² = 8.
配方法将二次式转化为一个完全平方加上一个常数。基本思路是把 x² + bx + c 重写为 (x + b/2)² − (b/2)² + c。例如，x² + 6x + 1 = 0 变成 (x + 3)² − 9 + 1 = 0，即简化为 (x + 3)² = 8。
Now take square roots: x + 3 = ±√8. Since √8 = 2√2, we get x = −3 ± 2√2. These are the exact solutions.
然后开平方：x + 3 = ±√8。因为 √8 = 2√2，所以 x = −3 ± 2√2。这些是精确解。
This method is especially useful when factorisation is not obvious, but it is more time-consuming. However, it is also the key to deriving the quadratic formula.
当因式分解不明显时，此方法尤其有用，但更耗时。然而，它也是推导求根公式的关键。
5. Solving by the Quadratic Formula | 求根公式法
The quadratic formula works for every quadratic equation, even when factorisation is impossible. The formula is:
x = (−b ± √b² − 4ac) / (2a)
xample, solve 2x² − 4x − 3 = 0. Here a = 2, b = −4, c = −3. Substituting into the formula:
只要把方程写成标准形式，就可以直接使用该公式。例如，解 2x² − 4x − 3 = 0，这里 a = 2，b = −4，c = −3。代入公式：
x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2
So the solutions are x = (2 + √10)/2 and x = (2 − √10)/2. Always simplify the square root when possible.
因此解为 x = (2 + √10)/2 和 x = (2 − √10)/2。当可能时，务必化简根号。
6. The Discriminant | 判别式
The expression b² − 4ac inside the square root is called the discriminant, often denoted by Δ (Delta). It tells us how many real solutions a quadratic equation has without solving it completely.
根号内的表达式 b² − 4ac 称为判别式，通常用 Δ（德尔塔）表示。它不需要完全求解就能告诉我们一个二次方程有多少个实数解。
If Δ > 0, there are two distinct real solutions.
如果 Δ > 0，则有两个不相等的实数解。
If Δ = 0, there is exactly one real solution (a repeated root).
如果 Δ = 0，则恰好有一个实数解（重根）。
If Δ < 0, there are no real solutions (the solutions are complex numbers).
如果 Δ < 0，则没有实数解（解为复数）。
ample, for x² + 2x + 5 = 0, Δ = 2² − 4×1×5 = 4 − 20 = −16 < 0, so there are no real roots.
例如，对于 x² + 2x + 5 = 0，Δ = 2² − 4×1×5 = 4 − 20 = −16 < 0，所以没有实数根。
7. Word Problems | 应用题
appear in word problems. The key is to translate the words into an equation. For example, "The product of two consecutive positive integers is 42. Find the integers." Let the first integer be n. Then the second is n + 1. The equation is n(n + 1) = 42.
二次方程经常出现在应用题中。关键是将文字转化为方程。例如："两个连续正整数的乘积是 42，求这两个整数。"设第一个整数为 n，那么第二个是 n + 1。方程为 n(n + 1) = 42。
, so n = −7 or n = 6. Since the integers are positive, n = 6, so the integers are 6 and 7.
展开得 n² + n − 42 = 0。因式分解得 (n + 7)(n − 6) = 0，所以 n = −7 或 n = 6。由于整数为正，n = 6，因此这两个整数是 6 和 7。
Always check whether the question restricts the sign of the variable. Negative roots can be rejected depending on the context.
始终检查题目是否限制变量的符号。根据实际情况，负根可能需要舍去。
8. Graphs of Quadratic Functions | 二次函数图像
When a quadratic expression is set equal to y, we get a quadratic function. Its graph is a parabola. If a > 0, the parabola opens upwards; if a < 0, it opens downwards.
当二次表达式等于 y 时，我们得到二次函数。它的图像是抛物线。如果 a > 0，抛物线开口向上；如果 a < 0，抛物线开口向下。
of the graph. The number of x-intercepts matches the discriminant conditions: two, one, or zero.
方程 ax² + bx + c = 0 的解对应于图像与 x 轴的交点。x 轴交点的个数与判别式的情况一致：两个、一个或零个。
nt) can be found using x = −b/(2a). You can then substitute back to find the y-coordinate.
抛物线的顶点（最高点或最低点）可以用 x = −b/(2a) 求得。然后代入原式求出 y 坐标。
9. Solving Quadratics by Taking Square Roots | 用开平方法解二次方程
quick for special cases. For example, (x − 3)² = 16 gives x − 3 = ±4, so x = 7 or x = −1.
如果二次方程具有 (x + p)² = q 的形式，我们可以直接开平方求解。这种方法对特殊情况非常快捷。例如，(x − 3)² = 16 得到 x − 3 = ±4，所以 x = 7 或 x = −1。
f both sides, you must include both the positive and negative square roots. Forgetting the '±' symbol is one of the most common errors in mathematics.Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
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