📚 Mastering Quadratic Equations for IGCSE Mathematics | IGCSE 数学二次方程完全掌握
In IGCSE Mathematics, quadratic equations are one of the most important algebraic topics. They appear in both core and extended papers, and they link algebra, graphs and real-world problem solving. This article explains the main methods you need to master: factorising, completing the square and using the quadratic formula.
在 IGCSE 数学中，二次方程是代数最重要的主题之一。它们出现在核心卷和扩展卷中，并且连接代数、图像与现实问题求解。本文解释你需要掌握的主要方法：因式分解法、配方法和求根公式法。
1. What is a Quadratic Equation? | 什么是二次方程
on a ≠ 0 is essential because if a = 0, the equation becomes linear rather than quadratic.
二次方程是变量最高次数为 2 的方程。它的一般形式是 ax² + bx + c = 0，其中 a、b 和 c 是常数，且 a ≠ 0。a ≠ 0 这一条件非常重要，因为如果 a = 0，方程就变成一次方程而不是二次方程。
For example, x² + 3x = 10 is a quadratic because it can be rearranged to x² + 3x - 10 = 0.
IGCSE 题目经常给出展开式、因式分解式或整理后的二次方程。你必须能够识别它们，并在求解前将其改写为标准形式。例如，x² + 3x = 10 是二次方程，因为它可以整理为 x² + 3x - 10 = 0。
2. Standard Form and Key Terms | 标准形式与关键术语
. This makes it easier to identify the coefficients and choose the most efficient method. The quadratic term is ax², the linear term is bx, and the constant term is c.
在求解任何二次方程之前，都要将其改写为标准形式 ax² + bx + c = 0。这样可以更容易确定系数并选择最有效的方法。二次项是 ax²，一次项是 bx，常数项是 c。
-8. Identifying these values correctly is especially important for the quadratic formula.
例如，在 4x² - 7x + 3 = 0 中，a = 4，b = -7，c = 3。在 2x² - 8 = 0 中，没有一次项，所以 b = 0，a = 2，c = -8。正确识别这些值对于使用求根公式尤其重要。
3. Solving by Factorisation | 因式分解法
to 6 and add to 5. These numbers are 2 and 3, so the equation becomes (x + 2)(x + 3) = 0.
当二次方程可以写成两个一次括号的乘积时，可以使用因式分解法。对于 x² + 5x + 6 = 0，我们要找两个数，它们的乘积为 6，和为 5。这两个数是 2 和 3，因此方程变为 (x + 2)(x + 3) = 0。
the solutions x = -2 or x = -3. Always set the equation equal to zero before factorising.
然后使用零乘积性质：如果 (x + 2)(x + 3) = 0，那么 x + 2 = 0 或 x + 3 = 0。这样得到解 x = -2 或 x = -3。在因式分解之前，一定要使方程等于零。
x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0. The solutions are x = -1/2 and x = -3.
当 a ≠ 1 时，因式分解需要更加仔细。对于 2x² + 7x + 3 = 0，将中间项拆开，需要两个数，它们的乘积为 2 × 3 = 6，和为 7：这两个数是 6 和 1。因此 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0。解为 x = -1/2 和 x = -3。
4. Solving by Completing the Square | 配方法
÷ 2)² = 9 to both sides: x² + 6x + 9 = 4. The left side is a perfect square: (x + 3)² = 4.
配方法将 x² + bx + c = 0 转化为 (x + p)² = q 的形式。从 x² + 6x + 5 = 0 开始。首先移项：x² + 6x = -5。然后两边加上 (6 ÷ 2)² = 9：x² + 6x + 9 = 4。左边是完全平方：(x + 3)² = 4。
Taking square roots gives x + 3 = ±√4 5. If the coefficient of x² is not 1, divide every term by a before completing the square.
开平方得到 x + 3 = ±√4 = ±2。因此 x = -3 ± 2，所以 x = -1 或 x = -5。如果 x² 的系数不是 1，则在进行配方法之前，先让每一项除以 a。
This method is especially useful when the quadratic does not factorise neatly, and it also helps you find the vertex of a parabola. The expression (x + 3)² gives the vertex (-3, 0) in the related quadratic function.
当二次方程不能整齐地因式分解时，配方法特别有用，而且它还能帮助你找到抛物线的顶点。表达式 (x + 3)² 给出相关二次函数的顶点 (-3, 0)。
5. The Quadratic Formula | 求根公式
The quadratic formula gives the solutions of ax² + bx + c = 0 directly:
x = (-b ± √b² - 4ac) / 2a
求根公式直接给出 ax² + bx + c = 0 的解：
x = (-b ± √b² - 4ac) / 2a
and c. For 3x² - 5x - 2 = 0, we have a = 3, b = -5 and c = -2. Substituting gives x = (5 ± √25 + 24) / 6 = (5 ± √49) / 6 = (5 ± 7) / 6, so x = 2 or x = -1/3.
在代入 a、b 和 c 之前，一定要把方程整理成标准形式。对于 3x² - 5x - 2 = 0，我们有 a = 3，b = -5，c = -2。代入得到 x = (5 ± √25 + 24) / 6 = (5 ± √49) / 6 = (5 ± 7) / 6，所以 x = 2 或 x = -1/3。
The quadratic formula works for every quadratic equation, even when factorisation is difficult or impossible. It is the most reliable method for extended paper questions involving exact answers.
求根公式适用于所有二次方程，即使因式分解困难或不可能时也可以使用。它是扩展卷中涉及精确答案的题目最可靠的方法。
6. Discriminant and Nature of Roots | 判别式与根的性质
how many real solutions a quadratic has without solving it. The table below summarises the relationship:
判别式 D = b² - 4ac 可以在不解方程的情况下告诉你二次方程有多少个实数解。下表总结了这个关系：
| Discriminant D | Nature of roots | 根的性质 |
|---|---|---|
| D > 0 | Two distinct real roots | 两个不同的实数根 |
| D = 0 | One repeated real root | 一个重复的实数根 |
| D < 0 | No real roots | 没有实数根 |
e discriminant greater than zero: b² - 4ac > 0, so 16 - 4k > 0. This gives -4k > -16, and dividing by -4 reverses the inequality: k < 4.
IGCSE 考试题经常要求你求 k 的取值范围，使得 kx² + 4x + 1 = 0 有两个不同的实数根。令判别式大于零：b² - 4ac > 0，所以 16 - 4k > 0。由此得到 -4k > -16，除以 -4 时不等号反向：k < 4。
7. Graphical Interpretation | 图像解释
x² + bx + c = 0 are the x-intercepts of this graph, where the parabola crosses the x-axis.
y = ax² + bx + c 的图像是一条抛物线。方程 ax² + bx + c = 0 的解就是这条图像与 x 轴的交点，即抛物线与 x 轴相交的位置。
If the parabola touches the x-axis at one point, the equation has one repeated real root and the discriminant is zero. If the parabola stays entirely above or below the x-axis, then D < 0 and there are no real roots.
如果抛物线只与 x 轴接触于一点，方程有一个重复的实数根，判别式为零。如果抛物线始终位于 x 轴上方或下方，那么 D < 0，没有实数根。
When a > 0, the parabola opens upwards; when a < 0, it opens downwards. This shape helps you check whether the roots you find by algebra make sense graphically.
当 a > 0 时，抛物线开口向上；当 a < 0 时，抛物线开口向下。这个形状可以帮助你检查通过代数求出的根在图像上是否合理。
8. Applications and Word Problems | 应用题
Factorising gives (x + 8)(x - 5) = 0, so x = 5 because length and width must be positive.
许多 IGCSE 问题涉及面积、连续数的乘积或抛体运动。例如，一个矩形的长为 x + 3，
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