📚 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 ≠ 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 ≠ 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 × 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 × 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 ≠ 1, we look for two numbers whose product is a × 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 ≠ 1 的 ax² + bx + c，我们寻找两个数，使其乘积为 a × 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 × (-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 × c = 2 × (-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 Δ, 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 Δ。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 Δ > 0, the equation has two distinct real roots. The graph crosses the x-axis at two points.
若 Δ > 0，方程有两个不相等的实数根。图像与 x 轴有两个交点。
If Δ = 0, the equation has exactly one repeated real root. The graph touches the x-axis at one point, the vertex.
若 Δ = 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 Δ < 0, the equation has no real roots. The graph does not intersect the x-axis at all.
若 Δ < 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 α and β, the sum and product of the roots can be read directly from the coefficients:
对于以 α 和 β 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 - α)(x - β) = 0. Expanding gives ax² - a(α + β)x + aαβ = 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 - α)(x - β) = 0 推导而来。展开得 ax² - a(α + β)x + aαβ = 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 α and β. Find α + β and αβ.
例：已知方程 3x² - 6x + 2 = 0 的两根为 α 和 β，求 α + β 与 αβ。
α + β = -(-6)/3 = 2，αβ = 2/3
This property is extremely useful in solving problems involving symmetric functions of the roots, such as α² + β², without finding the roots explicitly.
这一性质在解决涉及根对称函数（如 α² + β²）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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