📚 IGCSE Maths: Quadratic Equations and Graphs | IGCSE 数学：二次方程与图像
Quadratic equations and their graphs are a central part of the IGCSE Mathematics syllabus. This article explains the key methods for solving quadratics, drawing and interpreting their graphs, and avoiding common mistakes in examinations. It is designed for students who want a clear, structured revision guide with worked reasoning and exam-focused tips.
二次方程及其图像是 IGCSE 数学大纲中的核心内容。本文讲解求解二次方程的主要方法、绘制并理解二次函数图像，以及如何在考试中避免常见错误。本文适合希望获得清晰、结构化复习指南和考试技巧的学生。
1. Recognising Quadratic Expressions | 识别二次表达式
ssion becomes linear instead of quadratic. The term ax² is called the quadratic term, bx is the linear term, and c is the constant term.
二次表达式的一般形式为 ax² + bx + c，其中 a、b、c 为常数且 a ≠ 0。a ≠ 0 这个条件至关重要，因为如果 a = 0，表达式就变成了一次表达式，而不是二次表达式。ax² 称为二次项，bx 称为一次项，c 称为常数项。
For example, 2x² + 3x − 5 is quadratic because the highest power of x is 2. However, 4x + 7 is not quadratic because there is no x² term. In an IGCSE question, you may be asked to identify whether an expression is quadratic or to arrange it into the standard form.
例如，2x² + 3x − 5 是二次式，因为 x 的最高次数为 2。而 4x + 7 不是二次式，因为没有 x² 项。在 IGCSE 考题中，你可能需要判断一个表达式是否为二次式，或者将它整理成标准形式。
2. Solving by Factorising | 因式分解法
actorise the left-hand side, and then use the fact that if a × b = 0, then a = 0 or b = 0.
因式分解通常是求解二次方程最快的方法，适用于表达式可以写成两个一次因式乘积的情况。标准步骤是先将方程移项使右边等于 0，再对左边进行因式分解，然后利用“如果 a × b = 0，则 a = 0 或 b = 0”这一性质。
x² + 5x + 6 = 0 → (x + 2)(x + 3) = 0 → x = −2 or x = −3
In this example, the two numbers 2 and 3 multiply to give the constant term 6 and add to give the coefficient 5. Always check your factors by expanding them back to the original expression before solving.
在这个例子中，数字 2 和 3 相乘得到常数项 6，相加得到一次项系数 5。求解前，一定要通过展开因式来检验是否与原式一致。
Sometimes the coefficient of x² is not 1. In that case, you may need to use a trial method or split the middle term. For example, 2x² + 7x + 3 can be factorised as (2x + 1)(x + 3).
有时 x² 的系数不是 1。此时你可能需要使用尝试法或拆分中间项。例如，2x² + 7x + 3 可以因式分解为 (2x + 1)(x + 3)。
3. Solving by Completing the Square | 配方法
Completing the square is a useful method when a quadratic cannot be factorised easily, and it also helps in finding the turning point of a graph. The idea is to rewrite ax² + bx + c in the form a(x + p)² + q, where p and q are constants.
配方法适用于二次方程不易因式分解的情况，同时也有助于求图像顶点。其思路是把 ax² + bx + c 改写为 a(x + p)² + q 的形式，其中 p 和 q 为常数。
(x + 3)² − 7. To solve x² + 6x + 2 = 0, set (x + 3)² − 7 = 0 and then isolate the square.
对于首项系数为 1 的简单二次式，可使用恒等式 x² + bx = (x + b/2)² − (b/2)²。例如，x² + 6x + 2 可写为 (x + 3)² − 9 + 2 = (x + 3)² − 7。要求解 x² + 6x + 2 = 0，就令 (x + 3)² − 7 = 0，然后解出平方项。
(x + 3)² = 7 → x + 3 = ±√7 → x = −3 ± √7
This method also reveals the vertex of the parabola. The graph of x² + 6x + 2 has a minimum point at (−3, −7) because the expression is written as (x + 3)² − 7.
配方法还能直接揭示抛物线的顶点。由于 x² + 6x + 2 已写成 (x + 3)² − 7，其图像的最低点为 (−3, −7)。
4. The Quadratic Formula | 求根公式
ng is difficult or the roots are irrational. The formula is derived from completing the square and is given below.
求根公式可以求解任何形如 ax² + bx + c = 0 的二次方程，包括难以因式分解或根为无理数的情况。该公式由配方法推导而来，表示如下。
x = (−b ± √b² − 4ac) / (2a)
To use the formula correctly, first rearrange the equation into standard form and identify the values of a, b and c. Substitute these values carefully, especially when they are negative, and then simplify the numerator and denominator step by step.
要正确使用公式，应先把方程整理为标准形式，并确定 a、b 和 c 的值。代入时要特别注意负数，然后逐步化简分子和分母。
mple, solve 3x² − 4x − 2 = 0. Here a = 3, b = −4 and c = −2, so the formula gives x = (4 ± √16 + 24) / 6 = (4 ± √40) / 6 = (4 ± 2√10) / 6 = (2 ± √10) / 3.
例如，求解 3x² − 4x − 2 = 0。这里 a = 3，b = −4，c = −2，代入公式得 x = (4 ± √16 + 24) / 6 = (4 ± √40) / 6 = (4 ± 2√10) / 6 = (2 ± √10) / 3。
The quadratic formula is particularly powerful because it works universally. In an exam, if a question says 'give your answer correct to 2 decimal places', you can use a calculator after simplifying the square root.
求根公式非常强大，因为它普遍适用。在考试中，如果题目要求“答案精确到 2 位小数”，你可以在化简根号后用计算器求出近似值。
5. Discriminant and Nature of Roots | 判别式与根的性质
ls you how many real roots a quadratic equation has without actually solving the equation.
判别式是求根公式中根号内的部分：D = b² − 4ac。它可以在不实际求解方程的情况下，判断二次方程有多少个实数根。
d real root, meaning the graph touches the x-axis at exactly one point. If D < 0, it has no real roots, so the graph does not cross or touch the x-axis at all.
如果 D > 0，方程有两个不同的实数根；如果 D = 0，方程有一个重根，即图像与 x 轴恰好相切于一点；如果 D < 0，方程没有实数根，因此图像与 x 轴既不相交也不相切。
oots. In contrast, x² − 4x + 4 = 0 has D = 16 − 16 = 0, so it has one repeated root x = 2.
例如，x² + 2x + 3 = 0 的判别式 D = 4 − 12 = −8，所以它没有实数根。相反，x² − 4x + 4 = 0 的判别式 D = 16 − 16 = 0，所以它有一个重根 x = 2。
A common IGCSE question asks you to find the range of values of k for which a quadratic has two distinct real roots. To answer it, set up the discriminant and solve the resulting inequality.
IGCSE 常见题型是求使二次方程有两个不同实数根的 k 值范围。解答方法是写出判别式，然后解所得不等式。
x² + kx + 4 = 0 → D = k² − 16 > 0 → k < −4 or k > 4
6. Sketching Quadratic Graphs | 绘制二次函数图像
x² + bx + c is a parabola. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.
二次函数 y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上并有一个最低点；如果 a < 0，抛物线开口向下并有一个最高点。
ng x = 0, giving the point (0, c). The x-intercepts are found by solving ax² + bx + c = 0.
要准确绘制二次函数图像，应找出 y 轴截距、x 轴截距（如果存在）以及顶点。y 轴截距令 x = 0 求得，即点 (0, c)。x 轴截距通过解方程 ax² + bx + c = 0 求得。
ercepts are (1, 0) and (3, 0). The turning point is (2, −1). Plotting these points gives a clear U-shaped curve opening upwards.
例如，绘制 y = x² − 4x + 3 的图像。y 轴截距为 (0, 3)。解 x² − 4x + 3 = 0 得 (x − 1)(x − 3) = 0，所以 x 轴截距为 (1, 0) 和 (3, 0)。顶点为 (2, −1)。描出这些点后即可得到一条开口向上的清晰 U 形曲线。
Always label the intercepts and turning point on your sketch, and make sure the shape is smooth rather than made of straight line segments. In an exam, a rough sketch with correct key features is usually enough for full marks.
绘图时务必在图上标出截距和顶点，并确保曲线平滑，而不是由直线段拼成。在考试中，一张标有关键特征的正确草图通常足以获得满分。
7. Turning Point and Axis of Symmetry | 顶点与对称轴
he quadratic function y = ax² + bx + c, the axis of symmetry has the equation x = −b/(2a).
每条抛物线都有一条对称轴，即经过其顶点的竖直直线。对于二次函数 y = ax² + bx + c，对称轴方程为 x = −b/(2a)。
The turning point is the point where the graph changes direction. Its x-coordinate is −b/(2a), and its y-coordinate is found by substituting this x-value back into the original equation. If a > 0, the turning point is a minimum; if a < 0, it is a maximum.
顶点是图像改变方向的点。其 x 坐标为 −b/(2a)，y 坐标可通过将该 x 值代回原方程求得。如果 a > 0，顶点为最低点；如果 a < 0，顶点为最高点。
= −2. Substituting x = −2 gives y = 2(4) − 16 + 5 = −3, so the turning point is (−2, −3).
例如，对于 y = 2x² + 8x + 5，对称轴为 x = −8/(2×2) = −2。将 x = −2 代入得 y = 2(4) − 16 + 5 = −3，因此顶点为 (−2, −3)。
of symmetry is x = −p. This is why completing the square is so useful for graph sketching.
如果二次函数已经写成配方法形式 a(x + p)² + q，那么顶点就是 (−p, q)，对称轴为 x = −p。这就是配方法在绘图时非常实用的原因。
8. Solving Graphically | 图像法求解
x² + bx + c crosses the x-axis. These x-values are the real solutions of ax² + bx + c = 0.
二次方程也可以通过图像法求解，即找出抛物线 y = ax² + bx + c 与 x 轴交点的 x 坐标。这些 x 值就是方程 ax² + bx + c = 0 的实数解。
If the graph just touches the x-axis at one point, the equation has one repeated root. If the graph does not meet the x-axis, the equation has no real roots. This visual method helps you understand the connection between algebraic solutions and graph behaviour.
如果图像与 x 轴恰好相切于一点，则方程有一个重根；如果图像与 x 轴没有交点，则方程没有实数根。这种直观方法有助于理解代数解与图像行为之间的联系。
the x-values where the curve crosses the x-axis. Here the solutions are x = −1 and x = 3.
在 IGCSE 考试中，题目可能给你一张图像，要求利用它解方程，如 x² − 2x − 3 = 0。你只需读出曲线与 x 轴交点的 x 值。此例中解为 x = −1 和 x = 3。
You can also solve simultaneous equations graphically, for example finding where a straight line intersects a parabola. The x-coordinates of their intersection points give the solutions to the pair of equations.
你还可以用图像法解联立方程，例如找出直线与抛物线的交点。交点横坐标就是联立方程的解。
9. Real-Life Applications | 实际应用
area of 40 m², then x(x + 3) = 40, which leads to the quadratic equation x² + 3x − 40 = 0.
二次方程出现在许多实际情境中，例如面积问题、抛体运动以及经济学。例如，一个长方形花园的长为 x + 3 米，宽为 x 米，面积为 40 平方米，则 x(x + 3) = 40，由此得到二次方程 x² + 3x − 40 = 0。
ot be negative, the only valid solution is x = 5 metres, making the dimensions 5 m by 8 m.
通过因式分解解该方程得 (x + 8)(x − 5) = 0，因此 x = 5 或 x = −8。由于长度不能为负，唯一有效的解是 x = 5 米，因此尺寸为 5 米和 8 米。
initial height. Setting h = 0 and solving for t tells you when the object hits the ground.
抛体问题常用方程 h = −5t² + vt + h₀，其中 h 为高度，t 为时间，v 为初始竖直速度，h₀ 为初始高度。令 h = 0 并求解 t，即可得出物体落地的时间。
In such questions, always interpret the mathematical solutions in the context of the problem and reject values that do not make sense, such as negative times or negative lengths.
在这类问题中，务必结合题意解释数学解，并舍去不合理的值，例如负时间或负长度。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
² + 3x = 4 by writing x(x + 3) = 4 directly; you must first rewrite it as x² + 3x − 4 = 0.
最常见的错误之一是在因式分解前忘记把方程移项使右边等于 0。你不能直接把 x² + 3x = 4 写成 x(x + 3) = 4；必须先将它改写为 x² + 3x − 4 = 0。
Another frequent error is mishandling negative signs when using the quadratic formula. Write down the values of a, b and c clearly, and use brackets when substituting, especially for negative coefficients.
另一个常见错误是在使用求根公式时处理负号不当。应清楚地写出 a、b、c 的值，并在代入时使用括号，尤其是当系数为负数时。
Students also sometimes give answers without checking whether they satisfy the original equation. After solving, substitute your values back into the equation to catch arithmetic mistakes. For graph questions, make sure the curve has the correct direction of opening and a smooth turning point.
学生有时给出答案后不检验是否满足原方程。解出后应把值代回原方程以发现计算错误。对于图像题，要确保曲线开口方向正确，顶点处平滑。
When a question asks for exact answers, leave surds and fractions rather than rounding too early. If it asks for decimal approximations, round only at the final step to the accuracy requested.
当题目要求精确答案时，应保留根式和分数，不要过早取近似值。如果要求用小数表示，则只在最后一步按要求精度四舍五入。
Finally, remember that the discriminant can quickly tell you whether a real solution exists before you spend time solving. This is a powerful checking tool in both algebraic and graphical questions.
最后，记住判别式可以在你花时间求解之前快速判断是否存在实数解。在代数和图像题中，这都是一个强大的检验工具。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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