📚 Quadratic Equations: Solving, Graphs and the Discriminant | 二次方程：求解、图像与判别式
Quadratic equations are a central topic in AQA A-level Mathematics. They appear in pure algebra, coordinate geometry, modelling and calculus. This guide covers standard form, solving methods, the discriminant, graphs, inequalities and hidden quadratics, with clear links to AQA exam questions.
二次方程是 AQA A-level 数学的核心主题，出现在纯代数、坐标几何、建模和微积分中。本文涵盖标准形式、求解方法、判别式、图像、不等式和隐藏二次方程，并紧扣 AQA 考试题型。
1. Standard Form and Key Terms | 标准形式与核心术语
A quadratic equation in one variable is any equation that can be rearranged into the standard form ax² + bx + c = 0, where a, b and c are real constants and a ≠ 0.
一元二次方程是任何可以整理为标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为实常数且 a ≠ 0。
The condition a ≠ 0 is essential because it guarantees the x² term does not vanish. The coefficient a determines the shape and direction of the parabola. The constant c gives the y-intercept.
条件 a ≠ 0 至关重要，它保证 x² 项不会消失。系数 a 决定抛物线的形状和开口方向，常数 c 是 y 轴截距。
If b = 0, the equation is ax² + c = 0; if c = 0, it is ax² + bx = 0. Both are still quadratic and can be solved by isolation or factorisation.
若 b = 0，方程为 ax² + c = 0；若 c = 0，方程为 ax² + bx = 0。两者仍然是二次方程，可通过移项或因式分解求解。
2. Solving by Factorising | 因式分解法求解
If a quadratic can be written as (px + q)(rx + s) = 0, then the zero product property gives px + q = 0 or rx + s = 0. Solving each linear equation produces the roots.
如果二次式可以写成 (px + q)(rx + s) = 0，那么由零乘积性质可得 px + q = 0 或 rx + s = 0，解每个一次方程即可得到根。
Factorising is usually the fastest method when a = 1 or the coefficients are small integers. Always check by expanding the brackets.
当 a = 1 或系数为较小整数时，因式分解通常是最快的方法。求出后务必将括号展开检验。
Example: Solve x² − 5x + 6 = 0.
例：解 x² − 5x + 6 = 0。
Factorise: (x − 2)(x − 3) = 0, so x = 2 or x = 3.
因式分解：(x − 2)(x − 3) = 0，因此 x = 2 或 x = 3。
3. Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form a(x + p)² + q. This reveals the vertex of the parabola and is the key step in deriving the quadratic formula.
配方法把二次式写成 a(x + p)² + q 的形式。这种形式能直接显示抛物线的顶点，也是推导求根公式的关键步骤。
For a monic quadratic, use the identity:
对于首项系数为 1 的二次式，可以使用恒等式：
x² + bx + c = (x + b/2)² − (b/2)² + c
For the general case ax² + bx + c, first factor out a from the x² and x terms, then complete the square inside the bracket.
对于一般情况 ax² + bx + c，先从 x² 和 x 项中提出 a，再在括号内配方。
Example: Solve x² + 6x + 1 = 0. Write (x + 3)² − 9 + 1 = 0, so (x + 3)² = 8, giving x = −3 ± 2√2.
例：解 x² + 6x + 1 = 0。写成 (x + 3)² − 9 + 1 = 0，得 (x + 3)² = 8，因此 x = −3 ± 2√2。
4. The Quadratic Formula | 求根公式
For any quadratic equation ax² + bx + c = 0 with a ≠ 0, the roots are given by the quadratic formula:
对于任何 a ≠ 0 的二次方程 ax² + bx + c = 0，其根由求根公式给出：
x = (−b ± √(b² − 4ac)) / 2a
This formula works for every quadratic, including those that do not factorise over the integers and those with irrational roots.
该公式适用于所有二次方程，包括不能在整数范围内因式分解以及具有无理根的方程。
Example: Solve 2x² − 4x − 3 = 0. Here a = 2, b = −4, c = −3, so b² − 4ac = 16 + 24 = 40. Thus x = (4 ± √40) / 4 = 1 ± (√10)/2.
例：解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3，所以 b² − 4ac = 16 + 24 = 40。因此 x = (4 ± √40) / 4 = 1 ± (√10)/2。
5. The Discriminant and Nature of Roots | 判别式与根的性质
The discriminant D = b² − 4ac determines the nature of the roots without solving the equation. It comes directly from the square root part of the quadratic formula.
判别式 D = b² − 4ac 可以在不求解方程的情况下判断根的性质。它直接来自求根公式中的平方根部分。
| Discriminant D = b² − 4ac | Nature of roots |
|---|---|
| D > 0 | Two distinct real roots |
| D = 0 | One repeated real root (equal roots) |
| D < 0 | No real roots (two complex conjugate roots) |
AQA questions often ask for the number of real roots or for values of a parameter that give equal or real roots.
AQA 试题常要求判断实根个数，或求使方程有等根或实根的参数值。
Example: Find k such that x² + kx + 9 = 0 has equal roots. Set D = 0: k² − 36 = 0, so k = ±6.
例：求 k 使 x² + kx + 9 = 0 有等根。令 D = 0：k² − 36 = 0，所以 k = ±6。
6. Sketching Quadratic Graphs | 二次函数图像草图
The graph of y = ax² + bx + c is a parabola. If a > 0, the parabola opens upwards in a U-shape; if a < 0, it opens downwards in an inverted U-shape.
y = ax² + bx + c 的图像是抛物线。若 a > 0，抛物线开口向上，呈 U 形；若 a < 0，则开口向下，呈倒 U 形。
The y-intercept is (0, c). The x-intercepts are the real roots of ax² + bx + c = 0, provided D ≥ 0.
y 轴截距为 (0, c)。若 D ≥ 0，x 轴截距就是方程 ax² + bx + c = 0 的实根。
The vertex has x-coordinate −b / 2a and the line of symmetry is x = −b / 2a. Completing the square gives the vertex form a(x + p)² + q, where the vertex is (−p, q).
顶点的 x 坐标为 −b / 2a，对称轴为 x = −b / 2a。配方法给出顶点式 a(x + p)² + q，其中顶点为 (−p, q)。
7. Quadratic Inequalities | 二次不等式
To solve ax² + bx + c > 0 or ax² + bx + c < 0, first solve ax² + bx + c = 0 to find the critical values. Then sketch the parabola and identify the intervals where it lies above or below the x-axis.
求解 ax² + bx + c > 0 或 ax² + bx + c < 0 时，先解 ax² + bx + c = 0 得到临界值。然后画出抛物线草图，确定其位于 x 轴上方或下方的区间。
Use strict inequalities < or > to exclude the endpoints, and use ≤ or ≥ to include them. Always write the solution using interval notation or inequality notation as required.
用严格不等号 < 或 > 时排除端点，用 ≤ 或 ≥ 时包含端点。答案通常要求写成区间表示法或不等式表示法。
Example: Solve x² − x − 6 > 0. Factorise: (x − 3)(x + 2) = 0, so critical values are x = 3 and x = −2. The parabola opens upwards, so the solution is x < −2 or x > 3.
例：解 x² − x − 6 > 0。因式分解：(x − 3)(x + 2) = 0，临界值为 x = 3 和 x = −2。抛物线开口向上，因此解为 x < −2 或 x > 3。
8. Hidden Quadratics and Substitution | 隐藏二次方程与换元法
Some equations are not immediately quadratic but can be transformed into a quadratic by a suitable substitution. Typical hidden quadratics involve powers such as x⁴, 2^x or expressions with √x.
有些方程表面不是二次方程，但可通过适当换元转化为二次方程。常见的隐藏二次方程涉及 x⁴、2^x 或 √x 等幂或根式。
Example: Solve x⁴ − 5x² + 4 = 0. Let u = x², then u² − 5u + 4 = 0, so u = 1 or u = 4. Substituting back gives x² = 1 or x² = 4, hence x = ±1, ±2.
例：解 x⁴ − 5x² + 4 = 0。令 u = x²，得到 u² − 5u + 4 = 0，所以 u = 1 或 u = 4。代回得 x² = 1 或 x² = 4，因此 x = ±1, ±2。
For equations such as 2^{2x} − 5·2^x + 4 = 0, let y = 2^x. The equation becomes y² − 5y + 4 = 0, giving y = 1 or y = 4. Then 2^x = 1 gives x = 0, and 2^x = 4 gives x = 2.
对于 2^{2x} − 5·2^x + 4 = 0 这类方程，令 y = 2^x。方程变为 y² − 5y + 4 = 0，解得 y = 1 或 y = 4。再由 2^x = 1 得 x = 0，由 2^x = 4 得 x = 2。
9. Sum and Product of Roots | 根与系数的关系
If α and β are the roots of ax² + bx + c = 0, then the sum and product of the roots are given by:
若 α 和 β 是 ax² + bx + c = 0 的根，则根的和与积为：
α + β = −b / a and αβ = c / a
These relations are extremely useful for forming quadratic equations from given roots and for evaluating symmetric functions such as α² + β².
这些关系在已知根构造二次方程以及计算 α² + β² 等对称函数时非常有用。
Example: If the roots are 3 and −2, then α + β = 1 and αβ = −6, so a monic quadratic is x² − x − 6 = 0.
例：若根为 3 和 −2，则 α + β = 1，αβ = −6，因此首项系数为 1 的二次方程为 x² − x − 6 = 0。
10. Applications and Exam Tips | 应用与考试技巧
Quadratic equations model many real-world situations, including projectile motion, areas, optimisation and break-even problems. In applied questions, define variables clearly and translate the constraints into a quadratic equation.
二次方程可以用于许多实际情境的建模，包括抛体运动、面积、优化和盈亏平衡问题。在应用题中，要清晰地定义变量，并将约束条件转化为二次方程。
- If the quadratic factorises quickly, use factorisation; otherwise use the quadratic formula.
- Quote the quadratic formula accurately and
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