Mastering Quadratic Equations | 掌握二次方程

📚 Mastering Quadratic Equations | 掌握二次方程

Quadratic equations and their graphs are a cornerstone of the IGCSE Mathematics syllabus. Whether you are solving an equation, sketching a parabola, or interpreting a real-world model, a deep understanding of quadratics is essential for success in both Paper 2 and Paper 4.

二次方程及其图像是 IGCSE 数学课程的核心内容。无论是解方程、画抛物线草图，还是解释现实世界中的数学模型，深入理解二次方程都是你在 Paper 2 和 Paper 4 中取得成功的必要条件。

1. General Form | 一般形式

A quadratic expression or equation is defined by the highest power of the variable being 2. The standard general form is written as:

二次表达式或方程的定义是变量的最高次数为 2。其标准一般形式可写为：

ax² + bx + c = 0, where a ≠ 0

Here, \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. The value of \(a\) must never be zero; if \(a = 0\), the equation becomes linear, not quadratic.

这里的 \(a\)、\(b\)、\(c\) 均为常数，\(x\) 为变量。其中 \(a\) 的值绝不能为零；如果 \(a = 0\)，方程将变为线性方程，而非二次方程。

For example, \(2x² + 3x – 5 = 0\) is a quadratic equation in general form. In this case, \(a = 2\), \(b = 3\), and \(c = -5\).

例如，\(2x² + 3x – 5 = 0\) 就是一个一般形式的二次方程。这里 \(a = 2\)，\(b = 3\)，\(c = -5\)。

2. Solving by Factorisation | 因式分解法

Factorisation is often the quickest method when the quadratic has simple integer roots. The goal is to express the equation as a product of two binomials.

当二次方程具有简单的整数根时，因式分解法通常是最快捷的方法。其目标是将方程表示为两个二项式的乘积。

If ax² + bx + c = (px + q)(rx + s) = 0, then x = -q/p or x = -s/r

For example, solve \(x² – 5x + 6 = 0\). We look for two numbers that multiply to 6 and add to -5. These numbers are -2 and -3. Thus:

例如，解 \(x² – 5x + 6 = 0\)。我们寻找两个数，它们相乘得 6，相加得 -5。这两个数是 -2 和 -3。因此：

(x – 2)(x – 3) = 0 ⇒ x = 2 or x = 3

Always expand your factorised form to verify your answer. Factorisation becomes more challenging when \(a ≠ 1\), but the underlying principle remains the same: find two numbers whose product is \(ac\) and whose sum is \(b\).

务必展开因式分解后的形式来验证答案。当 \(a ≠ 1\) 时，因式分解会更具挑战性，但其基本原理相同：寻找两个数，使它们的乘积为 \(ac\)，和为 \(b\)。

3. Solving by Completing the Square | 配方法

Completing the square transforms a quadratic into a perfect square plus a constant. This method is especially useful when the quadratic cannot be factorised.

配方法将二次方程转化为一个完全平方加上一个常数。当二次方程无法因式分解时，此方法尤其有用。

x² + bx = (x + b/2)² – (b/2)²

For example, solve \(x² + 6x + 7 = 0\). First, complete the square for \(x² + 6x\):

例如，解 \(x² + 6x + 7 = 0\)。首先对 \(x² + 6x\) 配方：

(x + 3)² – 9 + 7 = 0 ⇒ (x + 3)² – 2 = 0

Then rearrange: \((x + 3)² = 2\). Taking square roots gives \(x + 3 = ±√2\), so \(x = -3 ± √2\).

然后重新整理：\((x + 3)² = 2\)。两边开平方得 \(x + 3 = ±√2\)，所以 \(x = -3 ± √2\)。

This method also directly reveals the turning point of the graph, which we will discuss later.

这种方法还能直接揭示图像的顶点坐标，我们将在后面讨论。

4. Solving by the Quadratic Formula | 公式法

The quadratic formula is a universal method that works for every quadratic equation, regardless of whether it factorises. It is particularly useful when the roots are irrational or complex.

二次公式是适用于所有二次方程的通用方法，无论其是否可以因式分解。当根为无理数或复数时，它尤其有用。

x = (-b ± √(b² – 4ac)) / (2a)

For example, solve \(3x² – 4x – 2 = 0\). Here \(a = 3\), \(b = -4\), \(c = -2\). Substitute into the formula:

例如，解 \(3x² – 4x – 2 = 0\)。这里 \(a = 3\)，\(b = -4\)，\(c = -2\)。代入公式：

x = (4 ± √(16 + 24)) / 6 = (4 ± √40) / 6 = (4 ± 2√10) / 6

Simplify to \(x = (2 ± √10) / 3\). Always check whether the calculator is allowed; in a non-calculator paper, leave your answer in exact surd form.

化简得 \(x = (2 ± √10) / 3\)。务必注意是否允许使用计算器；在不允许使用计算器的试卷中，请将答案保留为精确的根式形式。

5. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root: \(b² – 4ac\). It tells us the nature of the roots without fully solving the equation.

判别式是二次公式中根号内的部分：\(b² – 4ac\)。它无需完全解方程就能告诉我们根的性质。

If \(b² – 4ac > 0\), there are two distinct real roots.
If \(b² – 4ac = 0\), there is exactly one real root (a repeated root).
If \(b² – 4ac < 0\), there are no real roots (two complex roots).

若 \(b² – 4ac > 0\)，则有两个不同的实数根。
若 \(b² – 4ac = 0\)，则有一个实数根（重根）。
若 \(b² – 4ac < 0\)，则没有实数根（有两个复数根）。

For example, the equation \(2x² – 4x + 3 = 0\) has discriminant \( (-4)² – 4×2×3 = 16 – 24 = -8 < 0\), so it has no real roots.

例如，方程 \(2x² – 4x + 3 = 0\) 的判别式为 \( (-4)² – 4×2×3 = 16 – 24 = -8 < 0\)，因此它没有实数根。

6. Quadratic Graphs | 二次函数图像

The graph of a quadratic function \(y = ax² + bx + c\) is a smooth, symmetric curve called a parabola. The shape depends on the sign of \(a\).

二次函数 \(y = ax² + bx + c\) 的图像是一条平滑、对称的曲线，称为抛物线。其形状取决于 \(a\) 的符号。

If \(a > 0\), the parabola opens upwards (u-shaped) and has a minimum point.
If \(a < 0\), the parabola opens downwards (n-shaped) and has a maximum point.

若 \(a > 0\)，抛物线开口向上（U 形），有最小值点。
若 \(a < 0\)，抛物线开口向下（n 形），有最大值点。

The larger the absolute value of \(a\), the narrower the parabola. For \(a = 0\), the graph degenerates into a straight line.

\(a\) 的绝对值越大，抛物线越窄。当 \(a = 0\) 时，图像退化为一条直线。

7. Turning Point and Axis of Symmetry | 顶点与对称轴

Every parabola has a vertical line of symmetry that passes through the turning point (vertex). The x-coordinate of the vertex is given by:

每条抛物线都有一条垂直对称轴，它经过顶点（转向点）。顶点的 x 坐标由以下公式给出：

x = -b / (2a)

To find the y-coordinate of the vertex, substitute this \(x\)-value back into the original function.

要求顶点的 y 坐标，将此 \(x\) 值代回原函数即可。

Alternatively, if the quadratic is written in completed square form \(y = a(x – h)² + k\), then the vertex is simply \((h, k)\).

或者，如果二次函数写成配方法形式 \(y = a(x – h)² + k\)，则顶点坐标直接为 \((h, k)\)。

For example, for \(y = 2x² – 8x + 5\), \(x = 8/(2×2) = 2\). Substituting \(x = 2\) gives \(y = 2(2)² – 8(2) + 5 = 8 – 16 + 5 = -3\). So the vertex is \((2, -3)\).

例如，对于 \(y = 2x² – 8x + 5\)，\(x = 8/(2×2) = 2\)。代入 \(x = 2\) 得 \(y = 2(2)² – 8(2) + 5 = 8 – 16 + 5 = -3\)。所以顶点为 \((2, -3)\)。

8. Roots and Intercepts | 根与截距

The roots of a quadratic equation \(ax² + bx + c = 0\) are the x-values where the graph crosses the x-axis. These correspond to the points \((r₁, 0)\) and \((r₂, 0)\).

二次方程 \(ax² + bx + c = 0\) 的根是图像与 x 轴交点的 x 坐标。这些交点对应于点 \((r₁, 0)\) 和 \((r₂, 0)\)。

The y-intercept is always at \((0, c)\), because setting \(x = 0\) gives \(y = c\). For example, in \(y = x² – 3x + 2\), the y-intercept is \((0, 2)\).

y 截距总是在 \((0, c)\)，因为令 \(x = 0\) 得 \(y = c\)。例如，在 \(y = x² – 3x + 2\) 中，y 截距为 \((0, 2)\)。

If the discriminant is positive, there are two distinct x-intercepts. If it is zero, the parabola touches the x-axis at exactly one point (the vertex). If negative, the graph never touches the x-axis.

如果判别式为正，则有两个不同的 x 截距。如果为零，抛物线与 x 轴恰好相切于一点（即顶点）。如果为负，则图像与 x 轴无交点。

9. Sketching Quadratics | 画二次函数草图

To sketch a quadratic graph efficiently, use the following step-by-step approach:

要高效地画出二次函数草图，请遵循以下步骤：

Determine the direction of the parabola from the sign of \(a\).
Find the y-intercept by substituting \(x = 0\).
Find the roots by solving the equation \(ax² + bx + c = 0\).
Find the vertex using \(x = -b/(2a)\) and substitute back.
Plot the points and draw a smooth curve through them.

根据 \(a\) 的符号确定抛物线的开口方向。
令 \(x = 0\) 求 y 截距。
解方程 \(ax² + bx + c = 0\) 求根。
使用 \(x = -b/(2a)\) 并代回求顶点坐标。
描点并通过这些点画一条平滑曲线。

For example, sketch \(y = -x² + 4x – 3\). Here \(a = -1\), so the parabola opens downwards. The y-intercept is \((0, -3)\). Solving \(-x² + 4x – 3 = 0\) gives \(x = 1\) and \(x = 3\). The vertex has \(x = -4/(2×-1) = 2\), and \(y = -4 + 8 – 3 = 1\). So the vertex is \((2, 1)\).

例如，画出 \(y = -x² + 4x – 3\) 的草图。这里 \(a = -1\)，所以抛物线开口向下。y 截距为 \((0, -3)\)。解 \(-x² + 4x – 3 = 0\) 得 \(x = 1\) 和 \(x = 3\)。顶点的 \(x = -4/(2×-1) = 2\)，\(y = -4 + 8 – 3 = 1\)。因此顶点为 \((2, 1)\)。

10. Applications in Word Problems | 应用题

Quadratic equations frequently appear in real-world contexts such as projectile motion, area problems, and optimisation problems. The key is to translate the words into a quadratic equation correctly.

二次方程经常出现在现实情境中，例如抛体运动、面积问题和最优化问题。关键是将文字正确地转化为二次方程。

Consider a rectangle whose length is 3 metres more than its width. If the area is 40 square metres, let the width be \(x\). Then the length is \(x + 3\), and the area equation is \(x(x + 3) = 40\).

考虑一个长方形，其长比宽多 3 米。如果面积为 40 平方米，设宽为 \(x\)。则长为 \(x + 3\)，面积方程为 \(x(x + 3) = 40\)。

x² + 3x – 40 = 0

Factorising gives \((x + 8)(x – 5) = 0\), so \(x = -8\) or \(x = 5\). Since a width cannot be negative, the width is 5 metres and the length is 8 metres.

因式分解得 \((x + 8)(x – 5) = 0\)，所以 \(x = -8\) 或 \(x = 5\)。因为宽度不可能为负数，所以宽度为 5 米，长度为 8 米。

Always check the feasibility of your solutions within the context of the problem. Discard any root that does not make physical sense.

始终检查你的解在问题情境中是否合理。丢弃任何没有实际意义的根。

11. Common Mistakes | 常见错误

Even strong students often lose marks on quadratic questions due to avoidable errors. Here are some mistakes to watch out for:

即使是优秀的学生，也常常因为一些可以避免的错误而在二次函数题目上丢分。以下是一些需要警惕的错误：

Forgetting to rearrange the equation into the form \(ax² + bx + c = 0\) before solving.
Losing negative signs when substituting into the quadratic formula.
Misinterpreting the discriminant: “no real roots” is not the same as “no roots” in the context of complex numbers.
Drawing the parabola incorrectly when \(a < 0\) – the vertex must be a maximum.
Using the wrong x-coordinate for the vertex: remember \(x = -b/(2a)\), not \(b/(2a)\).

在求解前忘记将方程整理为 \(ax² + bx + c = 0\) 的形式。
代入二次公式时丢失负号。
误解读判别式：在复数背景下，“没有实数根”不等于“没有根”。
当 \(a < 0\) 时画错抛物线——顶点必须是最大值点。
使用错误的顶点 x 坐标：记住是 \(x = -b/(2a)\)，而不是 \(b/(2a)\)。

By being mindful of these pitfalls, you can significantly improve your accuracy in exams.

通过注意这些陷阱，你可以显著提高考试中的准确性。

12. Practice Questions | 练习题

To consolidate your understanding, try the following questions before checking the answers:

为了巩固你的理解，请在查看答案之前尝试以下问题：

Solve \(2x² – 9x + 4 = 0\) by factorisation.
Use completing the square to solve \(x² + 8x + 2 = 0\). Give your answer in surd form.
Find the discriminant of \(x² – 6x + 9\). What does this tell you about the roots?
Sketch the graph of \(y = x² – 2x – 3\), labelling the intercepts and the turning point.
A company makes a profit \(P(x) = -2x² + 12x – 7\), where \(x\) is the number of items sold in hundreds. Find the maximum profit and the number of items sold at that profit.

用因式分解法解 \(2x² – 9x + 4 = 0\)。
用配方法解 \(x² + 8x + 2 = 0\)，答案以根式表示。
求 \(x² – 6x + 9\) 的判别式。这告诉你什么关于根的信息？
画出 \(y = x² – 2x – 3\) 的草图，标出截距和顶点坐标。
一家公司的利润为 \(P(x) = -2x² + 12x – 7\)，其中 \(x\) 是以百件计的销售数量。求最大利润及此时售出的件数。

Answers: 1) \(x = 1/2\) or \(x = 4\). 2) \(x = -4 ± √14\). 3) Discriminant = 0, so there is one repeated root. 4) Vertex \((1, -4)\), roots \(x = -1\) and \(x = 3\), y-intercept \((0, -3)\). 5) Maximum profit is 11,000 (when \(x = 3\), so 300 items).

答案：1) \(x = 1/2\) 或 \(x = 4\)。2) \(x = -4 ± √14\)。3) 判别式为 0，所以有一个重根。4) 顶点 \((1, -4)\)，根 \(x = -1\) 和 \(x = 3\)，y 截距 \((0, -3)\)。5) 最大利润为 11,000（此时 \(x = 3\)，即 300 件）。

If you completed all these correctly, you have a strong foundation for quadratic equations in your IGCSE exam.

如果你全部答对，说明你已经为 IGCSE 考试中的二次方程打下了坚实的基础。

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Mastering Quadratic Equations | 掌握二次方程