Quadratics | 二次函数与二次方程

📚 Quadratics | 二次函数与二次方程

A quadratic is any expression of the form ax² + bx + c, where a, b and c are constants and a ≠ 0. In Edexcel A-Level Pure Mathematics, the quadratics topic underpins many later areas, including functions, differentiation and modelling. You are expected to solve quadratic equations by factorising, completing the square and using the quadratic formula, and to interpret the discriminant and sketch graphs confidently.

二次式是形如 ax² + bx + c 的表达式，其中 a、b、c 为常数且 a ≠ 0。在 Edexcel A-Level 纯数学中，二次函数是后续函数、微分和建模等内容的基础。你需要熟练掌握因式分解、配方法和求根公式三种解二次方程的方法，并能解释判别式的意义、准确画出二次函数图像。

1. Standard Form and Coefficients | 标准形式与系数

A quadratic expression in Edexcel Pure Mathematics is usually written as ax² + bx + c, where a, b and c are constants and a ≠ 0. The condition a ≠ 0 is essential because the expression would otherwise become linear. Always begin by rearranging an equation into this form so that the coefficients can be identified correctly.

在 Edexcel 纯数学中，二次式通常写作 ax² + bx + c，其中 a、b、c 为常数且 a ≠ 0。a ≠ 0 这一条件至关重要，否则表达式就变成一次式。解题时首先要将方程整理成这种标准形式，以便正确识别各项系数。

ax² + bx + c = 0, a ≠ 0

2. Solving by Factorising | 因式分解法

Factorising is often the fastest method when the quadratic has rational roots. You look for two numbers whose product equals ac and whose sum equals b, then split the middle term and factor by grouping. After factorising, set each bracket equal to zero and solve.

当二次方程有有理根时，因式分解通常是最快的方法。你寻找两个数，使它们的乘积等于 ac，和等于 b，然后拆分中项并分组分解。分解后令每个括号等于零并求解。

For example, x² − 5x + 6 = 0 factorises as (x − 2)(x − 3) = 0, giving x = 2 or x = 3.

例如，x² − 5x + 6 = 0 分解为 (x − 2)(x − 3) = 0，得 x = 2 或 x = 3。

Always check that the product of the constants gives the original constant term.
若常数项符号不对，说明因数选择有误，应重新检查。

3. Completing the Square | 配方法

Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. This form reveals the vertex of the graph and is essential for solving equations that do not factorise neatly. For x² + 6x + 1, halve the coefficient of x to get 3, square it to get 9, and write (x + 3)² − 9 + 1 = (x + 3)² − 8.

配方法将 ax² + bx + c 改写为 a(x + p)² + q 的形式。这种形式能直接显示图像的顶点，对于不能简单因式分解的方程也很重要。例如 x² + 6x + 1，取 x 的系数的一半 3，平方得 9，写作 (x + 3)² − 9 + 1 = (x + 3)² − 8。

x² + bx = (x + b/2)² − (b/2)²

When the leading coefficient is not 1, factor it out first, then complete the square inside the brackets.

当首项系数不为 1 时，先将其提出括号，再在括号内完成配方。

4. The Quadratic Formula | 求根公式

The quadratic formula gives the solutions of ax² + bx + c = 0 as x = (−b ± √(b² − 4ac)) / 2a. It always works, provided you substitute carefully with signs. You should use it when factorising is difficult or when the roots are irrational.

求根公式给出 ax² + bx + c = 0 的解为 x = (−b ± √(b² − 4ac)) / 2a。只要小心代入符号，它总能求解。当因式分解困难或根为无理数时，应使用该公式。

x = (−b ± √(b² − 4ac)) / 2a

The formula is given in the Edexcel formula booklet, but you must be able to apply it quickly and accurately. Watch for negative coefficients when substituting b and c.

Edexcel 公式表会给出此公式，但你必须能快速准确应用。代入 b 和 c 为负数时要特别小心符号。

5. The Discriminant and Nature of Roots | 判别式与根的性质

The discriminant D = b² − 4ac tells you how many real roots a quadratic equation has. Edexcel often asks questions such as ‘find the range of k for which the equation has two distinct real roots’, so you must set up and solve an inequality involving D.

判别式 D = b² − 4ac 可以判断二次方程有多少个实根。Edexcel 常考“求 k 的取值范围，使方程有两个不相等的实根”，因此需要建立并求解关于 D 的不等式。

D = b² − 4ac

Condition	Nature of roots	中文
D > 0	Two distinct real roots	两个不相等的实根
D = 0	One repeated real root	一个重根
D < 0	No real roots	没有实根

For example, the equation x² + kx + 4 = 0 has equal roots when D = k² − 16 = 0, so k = ±4.

例如，方程 x² + kx + 4 = 0 有等根时，D = k² − 16 = 0，因此 k = ±4。

6. Sketching Quadratic Graphs | 二次函数图像

A quadratic graph y = ax² + 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. To sketch it, find the y-intercept c, the x-intercepts if they exist, and the vertex by completing the square or using x = −b/2a. Label all key points clearly.

二次函数图像 y = ax² + bx + c 是抛物线。若 a > 0，抛物线开口向上，有最小值点；若 a < 0，开口向下，有最大值点。画图时要找 y 轴截距 c、x 轴截距（若存在），以及通过配方或 x = −b/2a 求出的顶点。清晰标注所有关键点。

Start by marking the y-intercept at (0, c).
先标出 y 轴截距 (0, c)。
Find roots by solving ax² + bx + c = 0, if the discriminant is non-negative.
若判别式非负，解 ax² + bx + c = 0 求根。
Plot the vertex and draw a smooth symmetric curve.
标出顶点，画出对称的光滑曲线。

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

From the completed square form a(x + p)² + q, the vertex is (−p, q) and the axis of symmetry is x = −p. From standard form, the axis is x = −b/2a, and substituting this into the equation gives the y-coordinate of the vertex. This is often the quickest way to find the maximum or minimum value of a quadratic expression.

由完全平方形式 a(x + p)² + q 可知，顶点为 (−p, q)，对称轴为 x = −p。由标准形式，对称轴为 x = −b/2a，将其代入方程可得顶点纵坐标。这通常是求二次式最大值或最小值的最快方法。

x = −b / 2a

For y = 2x² − 12x + 7, completing the square gives y = 2(x − 3)² − 11, so the minimum point is (3, −11).

对于 y = 2x² − 12x + 7，配方得 y = 2(x − 3)² − 11，因此最小值点为 (3, −11)。

8. Quadratic Inequalities | 二次不等式

To solve a quadratic inequality such as x² − 4x − 5 < 0, first solve the corresponding equation to find the critical values. Then sketch the parabola or use a sign table to determine where the graph is below or above the x-axis.

解二次不等式如 x² − 4x − 5 < 0，先解对应方程求出临界值。然后画抛物线或用符号表判断图像在 x 轴下方还是上方。

For x² − 4x − 5 < 0, the roots are x = −1 and x = 5, and the solution is −1 < x < 5 because the parabola opens upwards.

对于 x² − 4x − 5 < 0，根为 x = −1 和 x = 5，解集为 −1 < x < 5，因为抛物线开口向上。

Be careful with the direction of the inequality and whether the critical values should be included or excluded when the inequality is strict or non-strict.

要注意不等号的方向，以及当不等式为严格或非严格时，临界值是否应包含在解集中。

9. Hidden Quadratics | 隐藏二次方程

Hidden quadratics are equations that can be transformed into quadratic form by substitution. For example, x⁴ − 5x² + 4 = 0 can be written as u² − 5u + 4 = 0 by letting u = x². Solve for u, then substitute back to find x.

隐藏二次方程是指通过代换可化为二次形式的方程。例如 x⁴ − 5x² + 4 = 0 可令 u = x²，化为 u² − 5u + 4 = 0。解出 u 后，再代回求 x。

For u = 1 and u = 4, we obtain x² = 1 and x² = 4, so x = ±1 or x = ±2. Always check for extraneous or invalid solutions, especially when u cannot be negative.

由 u = 1 和 u = 4，得 x² = 1 和 x² = 4，所以 x = ±1 或 x = ±2。必须检查增根或无效解，尤其是当 u 不能为负数时。

10. Modelling with Quadratics | 二次函数建模

Quadratics can model many real-world situations, such as projectile motion, areas and profit. In modelling questions, identify the variables, form a quadratic equation or expression, then use the techniques above to maximise or solve.

二次函数可以建模许多实际问题，如抛体运动、面积和利润。在建模题中，先确定变量，建立二次方程或表达式，再运用上述技巧求最大值或求解。

For example, the height of a projectile might be modelled by h = −5t² + 20t + 1, where the maximum height is found from the vertex. After obtaining mathematical solutions, always interpret your answer in context and reject negative or non-real values where appropriate.

例如，抛体高度可建模为 h = −5t² + 20t + 1，最大高度可由顶点求出。得到数学解后，一定要结合实际问题解释答案，并适当地舍去负数或非实数解。

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Quadratics | 二次函数与二次方程