Quadratic Functions and Equations | 二次函数与方程

📚 Quadratic Functions and Equations | 二次函数与方程

A quadratic equation is an equation of the form ax² + bx + c = 0, where a ≠ 0. It appears frequently in the IGCSE Edexcel Mathematics syllabus, and mastering it is essential for higher-level algebra. This guide covers solving, graphing, and analysing quadratics step by step.

二次方程是形如 ax² + bx + c = 0 的方程，其中 a ≠ 0。它在 IGCSE Edexcel 数学大纲中频繁出现，掌握它对于更高阶的代数学习至关重要。本指南将逐步讲解二次方程的求解、绘图与分析。

1. Standard Form and Terminology | 标准形式与术语

Any quadratic expression can be rearranged into the standard form ax² + bx + c = 0. Here, a is the coefficient of x², b is the coefficient of x, and c is the constant term. For example, 2x² − 5x + 3 = 0 has a = 2, b = −5, c = 3.

任何二次表达式都可以整理成标准形式 ax² + bx + c = 0。其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。例如，2x² − 5x + 3 = 0 中 a = 2，b = −5，c = 3。

If the equation is not given in standard form, expand and simplify before solving. For instance, (x − 4)(x + 2) = 7 must first be expanded to x² − 2x − 8 = 7, then rewritten as x² − 2x − 15 = 0.

如果方程未以标准形式给出，则需要先展开并化简。例如，(x − 4)(x + 2) = 7 必须先展开为 x² − 2x − 8 = 7，再改写为 x² − 2x − 15 = 0。

2. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when the quadratic has simple integer roots. The idea is to write ax² + bx + c as a product of two linear factors, then use the rule that if the product is zero, at least one factor must be zero.

当二次式具有简单的整数根时，因式分解通常是最快的方法。其思路是将 ax² + bx + c 写成两个一次因式的乘积，然后利用“若乘积为零，则至少有一个因式为零”的规则。

For example, solve x² − 5x + 6 = 0. We look for two numbers that multiply to 6 and add to −5. These are −2 and −3. Therefore (x − 2)(x − 3) = 0, so x = 2 or x = 3.

例如，解 x² − 5x + 6 = 0。我们需要找到两个数，它们相乘为 6，相加为 −5。这两个数是 −2 和 −3。因此 (x − 2)(x − 3) = 0，所以 x = 2 或 x = 3。

When the coefficient a is not 1, factorisation requires more care. For 2x² + 7x + 3, we look for factors of 2 × 3 = 6 that add to 7, which are 6 and 1. Then split the middle term: 2x² + 6x + x + 3, and factor by grouping: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

当系数 a 不为 1 时，因式分解需要更小心。对于 2x² + 7x + 3，我们寻找 2 × 3 = 6 的因数，它们相加为 7，即 6 和 1。然后拆分中间项：2x² + 6x + x + 3，再分组因式分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)。

(2x + 1)(x + 3) = 0 ⇒ x = −1/2 or x = −3

(2x + 1)(x + 3) = 0 ⇒ x = −1/2 或 x = −3

3. Completing the Square | 配方法

Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. This is useful for solving equations and for identifying the vertex of a parabola. For a = 1, halve the coefficient of x to find p.

配方法将 ax² + bx + c 改写为 a(x + p)² + q 的形式。这既有助于解方程，也有助于确定抛物线的顶点。当 a = 1 时，将 x 的系数减半即可找到 p。

Take x² + 6x − 2 = 0. The coefficient of x is 6, half is 3, so we write (x + 3)² − 9 − 2 = 0, which simplifies to (x + 3)² = 11. Then x + 3 = ±√11, so x = −3 ± √11.

以 x² + 6x − 2 = 0 为例。x 的系数为 6，一半是 3，因此写作 (x + 3)² − 9 − 2 = 0，化简得 (x + 3)² = 11。于是 x + 3 = ±√11，所以 x = −3 ± √11。

When a ≠ 1, first factor out a from the first two terms. For 2x² − 4x + 1, write 2(x² − 2x) + 1, then complete the square inside: 2[(x − 1)² − 1] + 1 = 2(x − 1)² − 2 + 1 = 2(x − 1)² − 1.

当 a ≠ 1 时，先从前两项中提出 a。对于 2x² − 4x + 1，写成 2(x² − 2x) + 1，然后在括号内配方：2[(x − 1)² − 1] + 1 = 2(x − 1)² − 2 + 1 = 2(x − 1)² − 1。

4. The Quadratic Formula | 二次公式

The quadratic formula can solve any quadratic equation, including those that do not factorise. Given ax² + bx + c = 0, the solutions are given by:

二次公式可以解任意二次方程，包括那些不能因式分解的方程。对于 ax² + bx + c = 0，解由下式给出：

x = [−b ± √(b² − 4ac)] / (2a)

For example, solve 3x² + 2x − 5 = 0. Here a = 3, b = 2, c = −5. Substitute into the formula: x = [−2 ± √(4 + 60)] / 6 = [−2 ± √64] / 6 = [−2 ± 8] / 6, giving x = 1 or x = −5/3.

例如，解 3x² + 2x − 5 = 0。此处 a = 3，b = 2，c = −5。代入公式：x = [−2 ± √(4 + 60)] / 6 = [−2 ± √64] / 6 = [−2 ± 8] / 6，得到 x = 1 或 x = −5/3。

Always simplify surds where possible. If b² − 4ac is negative, the equation has no real roots, which is discussed in the next section.

始终尽可能化简根式。如果 b² − 4ac 为负，则方程没有实数根，这将在下一节讨论。

5. The Discriminant | 判别式

The discriminant is the value Δ = b² − 4ac. It determines the nature and number of roots of a quadratic equation without solving it fully.

判别式是 Δ = b² − 4ac 的值。它可以在不完全求解的情况下确定二次方程的根的性质与数量。

If Δ > 0, there are two distinct real roots. If Δ = 0, there is exactly one repeated real root (a double root). If Δ < 0, there are no real roots (the curve does not cross the x-axis).

若 Δ > 0，则有两个不同的实数根。若 Δ = 0，则恰好有一个实数重根（二重根）。若 Δ < 0，则没有实数根（曲线不与 x 轴相交）。

For a quadratic equation with constant coefficients, the discriminant is also used in problems about tangents and intersections. For example, x² + kx + 9 = 0 has equal roots when k² − 36 = 0, so k = ±6.

对于常系数二次方程，判别式也用于切线与交点问题。例如，x² + kx + 9 = 0 有等根时 k² − 36 = 0，因此 k = ±6。

6. Graphs of Quadratic Functions | 二次函数图像

The graph of y = ax² + bx + c is a parabola. If a > 0, the parabola opens upward (U-shaped). If a < 0, it opens downward (n-shaped). The roots of the equation are the x-intercepts of the graph.

y = ax² + bx + c 的图像是一条抛物线。若 a > 0，抛物线开口向上（U 形）。若 a < 0，则开口向下（n 形）。方程的根就是图像与 x 轴的交点横坐标。

The y-intercept is simply the constant term c, because when x = 0, y = c. This gives a quick check when sketching.

y 轴截距就是常数项 c，因为当 x = 0 时，y = c。这在画草图时提供了快速检验。

The axis of symmetry is the vertical line through the vertex. Its equation is x = −b/(2a). This can be derived from completing the square or from the quadratic formula.

对称轴是过顶点的竖直线。其方程为 x = −b/(2a)。这可以通过配方或二次公式推导得出。

7. Vertex and Turning Point | 顶点与转向点

The vertex is the maximum or minimum point of the parabola. In the completed square form y = a(x + p)² + q, the vertex is (−p, q). This is the turning point of the graph.

顶点是抛物线的最大值或最小值点。在配方形式 y = a(x + p)² + q 中，顶点为 (−p, q)。这就是图像的转向点。

For example, y = 2(x − 1)² + 5 has vertex (1, 5). Since a = 2 > 0, this is a minimum point. For y = −3(x + 2)² − 4, the vertex is (−2, −4) and it is a maximum because a = −3 < 0.

例如，y = 2(x − 1)² + 5 的顶点为 (1, 5)。由于 a = 2 > 0，这是一个最小值点。对于 y = −3(x + 2)² − 4，顶点为 (−2, −4)，且因为 a = −3 < 0，所以是最大值点。

To find the vertex from standard form, you can use the axis of symmetry x = −b/(2a), then substitute back to get y. For y = x² − 4x + 3, the axis is x = 2, and y = 4 − 8 + 3 = −1, so the vertex is (2, −1).

要从标准形式求顶点，可以通过对称轴 x = −b/(2a)，然后代回求出 y。对于 y = x² − 4x + 3，对称轴为 x = 2，y = 4 − 8 + 3 = −1，因此顶点为 (2, −1)。

8. Sketching a Quadratic Graph | 绘制二次函数图像

A clear sketch should show the y-intercept, the roots (if real), the vertex, and the axis of symmetry. Follow these steps:

清晰的草图应显示 y 轴截距、根（如果存在实数根）、顶点和对称轴。请遵循以下步骤：

Identify the shape from a (positive = U, negative = n). 根据 a 的正负确定开口方向（正为 U，负为 n）。
Write down the y-intercept (0, c). 写出 y 轴截距 (0, c)。
Solve ax² + bx + c = 0 to find roots; plot them. 解 ax² + bx + c = 0 求根并标出。
Find the vertex using x = −b/(2a) or completed square form. 用 x = −b/(2a) 或配方形式求顶点。
Draw the parabola through these points, labelled and smooth. 用平滑曲线穿过这些点，并标注。

If there are no real roots, the curve does not cross the x-axis. Still plot the vertex and y-intercept, and note whether the parabola lies entirely above or below the x-axis.

若没有实数根，曲线不与 x 轴相交。仍要标出顶点和 y 轴截距，并注意抛物线是完全在 x 轴上方还是下方。

9. Quadratic Inequalities | 二次不等式

Quadratic inequalities such as x² − 3x + 2 < 0 can be solved by first finding the roots of the corresponding equation. Here x² − 3x + 2 = 0 factorises as (x − 1)(x − 2) = 0, so the roots are 1 and 2.

二次不等式如 x² − 3x + 2 < 0 可通过先求对应方程的根来求解。这里 x² − 3x + 2 = 0 分解为 (x − 1)(x − 2) = 0，因此根为 1 和 2。

Sketch the parabola y = x² − 3x + 2. It is U-shaped and crosses the x-axis at 1 and 2. The inequality is negative where the graph is below the x-axis, so the solution is 1 < x < 2.

画出抛物线 y = x² − 3x + 2。它是 U 形，并在 1 和 2 处穿过 x 轴。不等式为负表示图像位于 x 轴下方，因此解为 1 < x < 2。

For ≥ or ≤, include the roots using square brackets in interval notation, e.g. [1, 2]. Remember that the inequality sign flips if you multiply or divide by a negative number, but this is rarely needed when solving by graphs.

对于 ≥ 或 ≤，在区间表示中用方括号包含端点，例如 [1, 2]。记住，若乘以或除以负数，不等号方向要改变，但用图像法求解时很少需要这样做。

10. Applications and Problem Solving | 应用与问题求解

Many real-world problems, such as projectile motion, area optimisation, and business profit, lead to quadratic equations.

许多实际问题，如抛体运动、面积优化和商业利润，都会引出二次方程。

For example, a rectangle has length (x + 3) cm and width (x − 2) cm, and its area is 50 cm². Then (x + 3)(x − 2) = 50, so x² + x − 6 = 50, giving x² + x − 56 = 0. Factorising yields (x + 8)(x − 7) = 0, so x = 7 (since length must be positive).

例如，一个矩形的长为 (x + 3) cm，宽为 (x − 2) cm，面积为 50 cm²。则 (x + 3)(x − 2) = 50，即 x² + x − 6 = 50，得到 x² + x − 56 = 0。因式分解得 (x + 8)(x − 7) = 0，所以 x = 7（因为长度必须为正）。

When setting up such problems, define a variable, form the quadratic equation, solve it, and always check whether each root is feasible in the original context.

在建立这类问题时，要先定义变量，建立二次方程，求解，并始终检查每个根在原问题情境中是否可行。

11. Common Mistakes and Tips | 常见错误与技巧

Students often forget to set the equation to zero before factorising or applying the quadratic formula. Always rearrange to the form ax² + bx + c = 0 first.

学生常常忘记在因式分解或使用二次公式前将方程化为零。始终先将方程整理为 ax² + bx + c = 0 的形式。

When using the quadratic formula, be careful with negative signs. For example, if b = −5, then −b = 5, and b² = 25, not −25. Write each substitution fully to avoid sign errors.

使用二次公式时要注意负号。例如，若 b = −5，则 −b = 5，且 b² = 25，而不是 −25。把每一步代入写完整，避免符号错误。

If the question asks for exact roots, leave surds simplified, e.g. x = 2 ± √3, unless a calculator answer is required. If asked for coordinates of the vertex, state both x and y, not just one.

如果题目要求精确根，应化简根式，例如 x = 2 ± √3，除非要求使用计算器答案。如果要求顶点坐标，要同时写出 x 和 y，而不仅仅是一个。

Practice using the discriminant to check the number of roots before solving. This saves time and helps you decide which method to use.

在求解前练习使用判别式来判断根的个数。这能节省时间，并帮助你决定使用哪种方法。

12. Summary | 总结

A quadratic equation can be solved by factorisation, completing the square, or the quadratic formula. The discriminant tells us about the nature of the roots. The graph of a quadratic function is a parabola whose vertex is found from completed square form or the axis of symmetry.

二次方程可以通过因式分解、配方或二次公式求解。判别式告诉我们根的性质。二次函数的图像是抛物线，其顶点可由配方形式或对称轴求得。

This topic is rich in marks at IGCSE, so ensure you can switch between forms and methods fluently. With regular practice, quadratics become one of the most reliable areas for full marks.

这一知识点在 IGCSE 中分值很高，因此请确保你能熟练地在不同形式和不同方法之间切换。通过定期练习，二次函数与方程会成为你最有把握获得满分的领域之一。

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Quadratic Functions and Equations | 二次函数与方程