📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are among the most frequently tested topics in IGCSE Mathematics. Whether you are solving for unknown roots, sketching a parabola, or modelling a real-life situation, a secure command of quadratics is essential. This article presents the complete toolkit: standard form, three solving methods, the discriminant, graph properties, and practical applications, each illustrated with clear worked examples.
二次方程是 IGCSE 数学中考查频率最高的内容之一。无论是求未知根、绘制抛物线图像，还是建立实际生活模型，熟练掌握二次方程都至关重要。本文将为你提供完整的工具包：标准形式、三种求解方法、判别式、图像性质以及实际应用，每个部分都配有清晰的例题讲解。
1. The Standard Form | 标准形式
A quadratic equation is an equation that can be rearranged into the standard form below, where (a), (b) and (c) are real numbers and (a neq 0):
二次方程是指可以整理为如下标准形式的方程，其中 (a)、(b)、(c) 为实数，且 (a neq 0)：
ax² + bx + c = 0
The coefficient (a) is called the leading coefficient and must not be zero; if (a = 0), the equation reduces to a linear one. The coefficient (b) or the constant (c) may be zero, but (a) is always non-zero.
系数 (a) 称为首项系数，不能为零；若 (a = 0)，方程就退化成一次方程。系数 (b) 或常数项 (c) 可以为 0，但 (a) 始终不能为 0。
For example, (2x² + 3x – 5 = 0) is a quadratic equation with (a = 2), (b = 3) and (c = -5). Likewise, (x² = 9) can be rewritten as (x² – 9 = 0), where (b = 0).
例如，(2x² + 3x – 5 = 0) 是一个二次方程，其中 (a = 2)，(b = 3)，(c = -5)。同样，(x² = 9) 可以改写为 (x² – 9 = 0)，此时 (b = 0)。
2. Solving by Factorisation | 因式分解法
Factorisation is the fastest method when the quadratic expression can be written as a product of two linear factors with integer coefficients. The principle is the zero-product property: if (p times q = 0), then (p = 0) or (q = 0).
当二次表达式可以写成两个整数系数一次因式的乘积时，因式分解是最快的解法。其原理是零积性质：若 (p times q = 0)，则 (p = 0) 或 (q = 0)。
For a quadratic of the form (x² + bx + c), we look for two numbers whose product is (c) and whose sum is (b). For (ax² + bx + c) with (a neq 1), we look for two numbers whose product is (a times c) and whose sum is (b), then split the middle term.
对于形如 (x² + bx + c) 的二次式，我们寻找两个数，使其乘积为 (c)，和为 (b)。对于 (a neq 1) 的 (ax² + bx + c)，我们寻找两个数，使其乘积为 (a times c)，和为 (b)，然后拆分中间项。
Example 1: Solve (x² – 7x + 12 = 0). Two numbers with product 12 and sum -7 are -3 and -4. Hence:
例 1：解方程 (x² – 7x + 12 = 0)。乘积为 12、和为 -7 的两个数是 -3 和 -4。因此：
(x – 3)(x – 4) = 0，所以 x = 3 或 x = 4。
Example 2: Solve (x² + x – 6 = 0). Two numbers with product -6 and sum 1 are 3 and -2. Thus:
例 2：解方程 (x² + x – 6 = 0)。乘积为 -6、和为 1 的两个数是 3 和 -2。因此：
(x + 3)(x – 2) = 0，所以 x = -3 或 x = 2。
When the coefficient of (x²) is not 1, for instance (2x² + 5x – 3 = 0), multiply (a) and (c) to get (2 times (-3) = -6). Find two numbers with product -6 and sum 5, namely 6 and -1. Split the middle term:
当 (x²) 的系数不为 1 时，例如 (2x² + 5x – 3 = 0)，先计算 (a times c = 2 times (-3) = -6)。找到乘积为 -6、和为 5 的两个数，即 6 和 -1。然后拆分中间项：
2x² + 6x – x – 3 = 0 → 2x(x + 3) – 1(x + 3) = 0 → (x + 3)(2x – 1) = 0 → x = -3 或 x = ½
3. Solving by Completing the Square | 配方法
Completing the square rewrites a quadratic as a perfect square plus a constant. This method is especially useful for solving equations that do not factorise, and it directly reveals the turning point of the graph.
配方法将二次式改写成一个完全平方加上一个常数。这种方法特别适合解不能因式分解的方程，并且可以直接揭示图像的顶点坐标。
For a monic quadratic (x² + bx + c), we use the identity:
对于首项系数为 1 的二次式 (x² + bx + c)，我们使用恒等式：
x² + bx + c = (x + b/2)² – (b/2)² + c
Example: Solve (x² + 6x + 5 = 0). Here (b = 6), so (b/2 = 3). Therefore:
例：解方程 (x² + 6x + 5 = 0)。这里 (b = 6)，所以 (b/2 = 3)。于是：
(x + 3)² – 9 + 5 = 0 → (x + 3)² – 4 = 0 → (x + 3)² = 4
Taking square roots on both sides gives (x + 3 = ±2), so (x = -1) or (x = -5).
两边开平方得 (x + 3 = ±2)，因此 (x = -1) 或 (x = -5)。
If the coefficient of (x²) is not 1, first factor out that coefficient from the (x²) and (x) terms, then complete the square inside the brackets.
若 (x²) 的系数不为 1，应先将该系数从 (x²) 项和 (x) 项中提出，再在括号内配方。
4. Solving by the Quadratic Formula | 公式法
The quadratic formula is the most general method and always works, even when factorisation fails. For (ax² + bx + c = 0), the solutions are given by:
二次公式是最通用的方法，即使因式分解行不通也始终有效。对于 (ax² + bx + c = 0)，解由下式给出：
x = (-b ± √(b² – 4ac)) / 2a
Example: Solve (2x² + 3x – 5 = 0). Here (a = 2), (b = 3), (c = -5). Substitute into the formula:
例：解方程 (2x² + 3x – 5 = 0)。这里 (a = 2)，(b = 3)，(c = -5)。代入公式：
x = (-3 ± √(3² – 4 × 2 × (-5))) / (2 × 2) = (-3 ± √(9 + 40)) / 4 = (-3 ± √49) / 4 = (-3 ± 7) / 4
Thus (x = (-3 + 7)/4 = 1), or (x = (-3 – 7)/4 = -10/4 = -2.5).
因此 (x = (-3 + 7)/4 = 1)，或 (x = (-3 – 7)/4 = -10/4 = -2.5)。
Always simplify the expression fully, and write the square root in simplified surd form when the discriminant is not a perfect square. On the IGCSE exam, answers may be required either as exact surds or as decimals rounded to the required degree of accuracy.
务必对结果进行彻底化简；当判别式不是完全平方数时，应将根号写成最简根式形式。在 IGCSE 考试中，答案可能要求以精确根式或保留指定小数位数的形式给出。
5. The Discriminant | 判别式
The expression (b² – 4ac), known as the discriminant and often denoted by (Delta), determines the nature of the roots of a quadratic equation without solving it fully.
式子 (b² – 4ac) 称为判别式，通常记作 (Delta)。通过判别式可以在不解方程的情况下判断根的性質。
There are three cases:
共有三种情形：
-
If (Delta > 0), the equation has two distinct real roots. The graph crosses the x-axis at two points.
若 (Delta > 0)，方程有两个不相等的实数根。图像与 x 轴有两个交点。
-
If (Delta = 0), the equation has exactly one repeated real root. The graph touches the x-axis at one point, the vertex.
若 (Delta = 0)，方程有一个重根（两个相等的实数根）。图像与 x 轴相切于一点，即顶点处。
-
If (Delta < 0), the equation has no real roots. The graph does not intersect the x-axis at all.
若 (Delta < 0)，方程没有实数根。图像与 x 轴没有交点。
| 判别式 Δ | 根的性質 | 图像特征 |
| Δ > 0 | 两个不相等的实数根 | 与 x 轴有两个交点 |
| Δ = 0 | 一个重根 | 与 x 轴相切于顶点 |
| Δ < 0 | 无实数根 | 与 x 轴无交点 |
Example: The equation (x² – 4x + k = 0) is required to have two distinct roots. Find the range of (k).
例：若方程 (x² – 4x + k = 0) 要有两个不相等的实根，求 (k) 的取值范围。
Δ = (-4)² – 4 × 1 × k = 16 – 4k > 0 → k < 4
6. Graphs of Quadratic Functions | 二次函数图像
The graph of (y = ax² + bx + c) is a curve called a parabola. The sign of (a) determines its orientation: when (a > 0), the parabola opens upward like a cup, and the vertex is a minimum point; when (a < 0), it opens downward like a cap, and the vertex is a maximum point.
函数 (y = ax² + bx + c) 的图像是一条称为抛物线的曲线。(a) 的正负决定了开口方向：当 (a > 0) 时，抛物线开口向上，像一个杯子，顶点是最低点；当 (a < 0) 时，开口向下，像一个帽子，顶点是最高点。
Key features of the graph include:
图像的关键特征包括：
-
The x-intercepts, if any, are the real roots of (ax² + bx + c = 0).
x 轴截距（如果存在）就是方程 (ax² + bx + c = 0) 的实数根。
-
The y-intercept is always the constant term (c), because setting (x = 0) gives (y = c).
y 轴截距始终等于常数项 (c)，因为令 (x = 0) 时 (y = c)。
-
The axis of symmetry is the vertical line (x = -b/(2a)), which passes through the vertex.
对称轴是垂直直线 (x = -b/(2a))，它经过顶点。
When sketching a parabola, always label the vertex, the intercepts, and the axis of symmetry clearly. These three features are enough to produce an accurate sketch.
绘制抛物线草图时，一定要清楚地标出顶点、截距和对称轴。这三点特征足以画出准确的草图。
7. Axis of Symmetry and Vertex | 对称轴与顶点
The axis of symmetry of the parabola (y = ax² + bx + c) is given by the formula below. The vertex lies on this axis.
抛物线 (y = ax² + bx + c) 的对称轴由下面的公式给出，顶点位于对称轴上。
x = -b / (2a)
The x-coordinate of the vertex is (x = -b/(2a)). To find the y-coordinate, substitute this value back into the original equation. Alternatively, completing the square rewrites the function in the form (y = a(x – h)² + k), where ((h, k)) is the vertex.
顶点的 x 坐标为 (x = -b/(2a))。要求 y 坐标，只需将该值代回原方程。另一种方法是通过配方法将函数写成 (y = a(x – h)² + k) 的形式，其中 ((h, k)) 就是顶点。
Example: Find the vertex of (y = x² – 6x + 11). Complete the square:
例：求抛物线 (y = x² – 6x + 11) 的顶点。使用配方法：
y = (x – 3)² – 9 + 11 = (x – 3)² + 2
Hence the vertex is ((3, 2)) and the axis of symmetry is (x = 3).
因此顶点为 ((3, 2))，对称轴为 (x = 3)。
8. Sum and Product of Roots | 根的和与积
For a quadratic equation (ax² + bx + c = 0) with roots (alpha) and (beta), the sum and product of the roots can be read directly from the coefficients:
对于以 (alpha) 和 (beta) 为根的二次方程 (ax² + bx + c = 0)，根的和与积可以直接从系数中读出：
α + β = -b/a，αβ = c/a
This result is derived from the factorised form (a(x – alpha)(x – beta) = 0). Expanding gives (ax² – a(alpha + beta)x + aalphabeta = 0), and comparing coefficients with (ax² + bx + c = 0) yields the two relations.
这个结论由因式分解形式 (a(x – alpha)(x – beta) = 0) 推导而来。展开得 (ax² – a(alpha + beta)x + aalphabeta = 0)，与 (ax² + bx + c = 0) 对照系数即可得到上述两个关系式。
Example: The roots of (3x² – 6x + 2 = 0) are (alpha) and (beta). Find (alpha + beta) and (alphabeta).
例：已知方程 (3x² – 6x + 2 = 0) 的两根为 (alpha) 和 (beta)，求 (alpha + beta) 与 (alphabeta)。
α + β = -(-6)/3 = 2，αβ = 2/3
This property is extremely useful in solving problems involving symmetric functions of the roots, such as (alpha² + beta²), without finding the roots explicitly.
这一性质在解决涉及根对称函数（如 (alpha² + beta²)）的问题时十分有用，无需显式求出每个根。
9. Forming Equations from Given Roots | 由已知根建立方程
If the roots of a quadratic equation are known, we can reconstruct the equation. For a monic quadratic (leading coefficient 1), the formula is:
若已知二次方程的根，我们可以重建该方程。对于首项系数为 1 的二次方程，公式为：
x² – (α + β)x + αβ = 0
Example: Find the quadratic equation whose roots are 2 and -5.
例：求以 2 和 -5 为根的二次方程。
α + β = 2 + (-5) = -3，αβ = 2 × (-5) = -10 → x² + 3x – 10 = 0
If the equation is required to have an integer leading coefficient other than 1, multiply both sides by that coefficient. This technique appears frequently in both Section A and Section B questions.
如果题目要求首项系数为 1 以外的整数，则可将方程两边同时乘以该系数。这一技巧在考试的第一部分和第二部分题目中经常出现。
10. Word Problems and Applications | 实际应用题
Quadratic equations arise naturally in geometry, physics, and number theory. The key to solving word problems is to translate the given conditions into a quadratic equation, solve it, and then check that the solutions make sense in the context.
二次方程在几何、物理和数论中自然产生。解答应用题的关键是将题目条件转化为二次方程，求解后再检验答案是否符合实际情境。
Example 1 (Geometry): A rectangle is 3 cm longer than it is wide. Its area is 40 cm². Find its dimensions.
例 1（几何）：一个长方形的长比宽多 3 cm，面积为 40 cm²。求它的长和宽。
Let the width be (x) cm. Then the length is ((x + 3)) cm. Since area = length × width:
设宽为 (x) cm，则长为 ((x + 3)) cm。因为面积 = 长 × 宽：
x(x
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