📚 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 \(ax² – a(\alpha + \beta)x + a\alpha\beta = 0\), 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 \(ax² – a(\alpha + \beta)x + a\alpha\beta = 0\)，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 \(\alpha\beta\).
例：已知方程 \(3x² – 6x + 2 = 0\) 的两根为 \(\alpha\) 和 \(\beta\)，求 \(\alpha + \beta\) 与 \(\alpha\beta\)。
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
Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导