📚 Solving Quadratic Equations | 解二次方程
Quadratic equations appear throughout A-Level Mathematics, and mastering the methods used to solve them is essential for tackling a wide range of problems, from pure algebra to mechanics and modelling. This article guides you through factorising, the quadratic formula, completing the square and the discriminant, then moves on to more advanced topics such as hidden quadratics, quadratic inequalities, complex roots and Vieta's formulas, all aligned with the Edexcel specification.
二次方程贯穿于A-Level数学的各个部分，掌握求解它们的方法对于解决各类问题至关重要，无论是纯数代数，还是力学与建模题目都离不开它。本文将带你系统学习因式分解法、二次公式法、配方法、判别式分析，并进一步延伸到隐藏二次方程、二次不等式、复数根与韦达定理等进阶主题，内容紧扣爱德思（Edexcel）考试大纲。
1. What is a Quadratic Equation? | 什么是二次方程？
l constants and a ≠ 0. The condition a ≠ 0 matters: if a were zero the x² term would disappear and the equation would become linear, with at most one solution. The highest power of the unknown is therefore exactly two, and this is what gives quadratics their distinctive behaviour and their distinctive curved graphs.
二次方程是指可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为实常数且 a ≠ 0。a ≠ 0 这一条件非常关键：如果 a 为零，x² 项就会消失，方程退化为一次方程，最多只有一个解。因此未知数的最高次数恰好为二次，这正是二次方程具有独特性质与独特曲线图像的原因。
rearranges to 3x² − 11x − 1 = 0. Equally, 4/x + x = 5 can be multiplied by x to give x² − 5x + 4 = 0, provided x ≠ 0. Always expand brackets and collect every term on one side, usually the left, before you choose a method.
很多看起来不像二次方程的式子，经过整理后都可以化为标准形式。例如 3x(x − 2) = 5x + 1 展开得 3x² − 6x = 5x + 1，移项后为 3x² − 11x − 1 = 0。同样，4/x + x = 5 在 x ≠ 0 的前提下两边乘以 x，可得 x² − 5x + 4 = 0。因此在选择解法之前，务必先展开括号并把所有项移到同一侧（通常移到左边）。
he x-values where the parabola intersects the x-axis. These values are called the roots, solutions or zeros of the equation, and they are exactly the x-intercepts of the curve. A quadratic can cross the x-axis twice, touch it once, or miss it entirely, and that geometric fact is the key to understanding the discriminant later in this article.
y = ax² + bx + c 的图像是一条抛物线。解方程 ax² + bx + c = 0，就是寻找抛物线与 x 轴交点的横坐标。这些值称为方程的根、解或零点，它们恰好是曲线与 x 轴交点的 x 坐标。抛物线与 x 轴可能有两个交点、一个交点，也可能完全不相交，这一几何事实正是后文理解判别式的关键。
2. Roots, Factors and the Factor Theorem | 根、因式与因式定理
two roots x = α and x = β. This is why factorising is the fastest solution method whenever the factors are obvious integers, and it also explains the strong link between roots and factors that runs through the whole of algebra.
二次式常常可以写成两个一次因式的乘积：ax² + bx + c = a(x − α)(x − β)。令每个括号分别等于零，即可得到两个根 x = α 与 x = β。这就是为什么当因式为明显的整数时，因式分解法是最快的解法，同时也解释了贯穿整个代数体系的"根与因式一一对应"关系。
ckly by substitution: if p(3) = 0 then x = 3 is a root and (x − 3) is a factor. The remainder theorem is the general form of the same idea, saying that p(α) is the remainder when p(x) is divided by (x − α); when the remainder is zero, the division is exact.
因式定理指出：多项式 p(x) 含有因式 (x − α) 当且仅当 p(α) = 0。对于二次式而言，这意味着可以通过代入快速检验某个疑似根：若 p(3) = 0，则 x = 3 是根，且 (x − 3) 是因式。余数定理是同一思想的一般形式，即 p(x) 除以 (x − α) 的余数为 p(α)；当余数为零时，说明整除成立。
vertex lies halfway between them, at x = 3.5. Recognising this symmetry early makes sketching quick and gives you a built-in check on any algebraic answer you obtain.
根还能告诉你抛物线与 x 轴的交点位置。若分解式为 y = (x − 3)(x − 4)，曲线在 x = 3 与 x = 4 处与 x 轴相交；由对称性可知，顶点位于两者中点，即 x = 3.5。及早识别这种对称性可以让你快速作图，并为你得到的代数结果提供一个自检手段。
3. Solving by Factorising | 因式分解法求解
mbers p and q whose sum is b and whose product is c, then write x² + bx + c = (x + p)(x + q). For a non-monic quadratic ax² + bx + c, look for two numbers whose sum is b and whose product is ac, then split the middle term and factorise in pairs.
当根为有理数时，因式分解法最为高效，通常应首先尝试。对于首项系数为 1 的二次式 x² + bx + c，寻找两个数 p 与 q，使其和等于 b、积等于 c，则可写成 x² + bx + c = (x + p)(x + q)。对于首项系数不为 1 的二次式 ax² + bx + c，则寻找和等于 b、积等于 ac 的两个数，再拆分中间项并分组分解。
Here is a reliable step-by-step routine you can apply in the exam:
以下是可在考试中直接套用的可靠步骤：
- out any common factor, including a negative sign if it makes the quadratic easier to read.
- Calculate the product ac, then list factor pairs of ac until you find the pair whose sum equals b.
- Split the middle term using that pair, factorise the first two and last two terms, and take out the common bracket.
- Set each factor equal to zero and solve the resulting linear equations, then check by substitution.
- 把方程整理为标准形式 ax² + bx + c = 0，并提取公因式（必要时先提出负号，使系数更易处理）。
- 计算乘积 ac，列出 ac 的因数对，找到和为 b 的那一对。
- 用该因数对拆分中间项，分别对前两项与后两项提取公因式，再提取公共括号。
- 令每个因式等于零，解出相应的一次方程，最后代入原式检验。
= 0 and x = −½ or x = −3. Special patterns worth memorising are the difference of two squares, x² − k² = (x − k)(x + k), and the perfect square, x² ± 2kx + k² = (x ± k)².
例题：解 2x² + 7x + 3 = 0。此处 ac = 6，因数和为 7 的一对是 6 与 1。拆分得 2x² + 6x + x + 3 = 0，即 2x(x + 3) + 1(x + 3) = 0，从而 (2x + 1)(x + 3) = 0，故 x = −½ 或 x = −3。值得记住的特殊模式有平方差 x² − k² = (x − k)(x + k) 与完全平方 x² ± 2kx + k² = (x ± k)²。
4. The Quadratic Formula | 二次公式（求根公式）
, the quadratic formula gives the exact roots every time. For ax² + bx + c = 0 with a ≠ 0:
当二次式无法用有理数分解时，二次公式总能给出精确的根。对于 ax² + bx + c = 0 且 a ≠ 0：
x = (−b ± √b² − 4ac) / 2a
ra in Section 5. Because of the ± sign, the formula produces both roots at once, and it also works when the roots are irrational or complex, making it the most general method available to you.
该公式可由对一般式配方推导得到，这说明它并非凭空出现的"魔法"，而是第 5 节代数操作的直接结果。由于 ± 号的存在，公式一次性给出两个根；当根为无理数或复数时它依然适用，因此是适用范围最广的方法。
x² − 4x − 1 = 0. Here a = 1, b = −4, c = −1, so b² − 4ac = 16 + 4 = 20. Therefore x = (4 ± √20) / 2 = (4 ± 2√5) / 2 = 2 ± √5. Leaving the answer in surd form is expected in Edexcel pure papers unless the question explicitly asks for a decimal, in which case give three significant figures unless told otherwise.
例题：解 x² − 4x − 1 = 0。此处 a = 1，b = −4，c = −1，故 b² − 4ac = 16 + 4 = 20。因此 x = (4 ± √20) / 2 = (4 ± 2√5) / 2 = 2 ± √5。在爱德思纯数试卷中，除非题目明确要求小数，否则应保留根式形式；若需要小数，通常保留三位有效数字。
Common errors to avoid: forgetting the minus sign in front of b, dividing only part of the numerator by 2a, and mis-squaring a negative b. Write the formula down first, substitute inside brackets, and only then evaluate. A quick mental check that the two roots are roughly where the parabola crosses the axis will catch most arithmetic slips.
常见错误包括：漏掉 b 前面的负号、只把分子的一部分除以 2a、以及负数 b 平方出错。建议先把公式写出，代入时加括号，最后再计算。用"两根大致位于抛物线与 x 轴交点处"做一次心算检验，可以抓住大多数运算失误。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. It is the standard way to find the vertex of a parabola, the minimum or maximum value of a quadratic function, and it is also required for solving equations that do not factorise neatly. For a monic quadratic:
配方法把二次式改写为 a(x + p)² + q 的形式。它是求抛物线顶点、求二次函数最值（最大值或最小值）的标准方法，也是解那些不易分解的方程所必需的技巧。对于首项系数为 1 的二次式：
x² + bx + c = (x + b/2)² + c − b²/4
The rule is to halve the coefficient of x, put it inside the bracket, then subtract the square of that half-coefficient to keep the expression unchanged. If a ≠ 1, first factor a out of the first two terms, complete the square inside the bracket, then multiply back out and tidy up the constant term.
操作要点是：取 x 系数的一半放入括号内，再减去这个半系数的平方以保持表达式不变。若 a ≠ 1，先把前两项中的 a 提出，在括号内配方，再乘回去并整理常数项。
− 9] + 7 = 2(x − 3)² − 11. The vertex is therefore at (3, −11), and since a = 2 > 0 the curve opens upwards, so the minimum value of the expression is −11, occurring at x = 3.
例题：把 2x² − 12x + 7 化为配方形式。提出 2 得 2(x² − 6x) + 7。−6 的一半是 −3，故 x² − 6x = (x − 3)² − 9，于是 2[(x − 3)² − 9] + 7 = 2(x − 3)² − 11。因此顶点为 (3, −11)；由于 a = 2 > 0，抛物线开口向上，故该式的最小值为 −11，在 x = 3 处取得。
Completed-square form is also a solving method. From (x − 3)² = 11/2 you get x − 3 = ± √11/2, hence x = 3 ± √11/2. Edexcel frequently links this topic to 'show that' questions, to finding least and greatest values, and to proving that an expression is always positive, which follows immediately when the completed square has a positive leading coefficient and a positive constant term.
配方形式也可用于求解。由 (x − 3)² = 11/2 可得 x − 3 = ± √11/2，故 x = 3 ± √11/2。爱德思考试常把这一知识点与"证明题"、求最大值最小值，以及证明某表达式恒为正联系起来；当配方后首项系数为正且常数项为正时，恒为正的结论便一目了然。
6. The Discriminant and the Nature of Roots | 判别式与根的性质
is called the discriminant. It determines how many real roots the equation has, without your having to solve it. This makes the discriminant the single most powerful tool in quadratic exam questions about intersections, tangency and the number of solutions.
二次公式中根号下的表达式 Δ = b² − 4ac 称为判别式。它无需解方程就能决定方程有多少个实根，因此是处理交点、相切、解的个数这类二次方程考题时最有力的工具。
| Value of Δ = b² − 4ac | Nature of roots | Graph of y = ax² + bx + c |
|---|---|---|
| Δ > 0 | Two distinct real roots | Crosses the x-axis twice |
| Δ = 0 | One repeated real root, x = −b/2a | Touches the x-axis at the vertex |
| Δ < 0 | No real roots (two complex roots) | Does not meet the x-axis |
| Δ = b² − 4ac 的取值 | 根的性质 | y = ax² + bx + c 的图像 |
|---|---|---|
| Δ > 0 | 两个不相等的实根 | 与 x 轴有两个交点 |
| Δ = 0 | 一个重根，x = −b/2a | 在顶点处与 x 轴相切 |
| Δ < 0 | 无实根（两个复根） | 与 x 轴不相交 |
) = 0, and tangency means exactly one repeated root, so Δ = 36 − 4(7 − k) = 0, giving k = −2. Whenever a question mentions 'tangent', 'touches', 'equal roots' or 'exactly one intersection', write Δ = 0 immediately.
典型考题：求 k 的取值，使直线 y = 2x + k 与曲线 y = x² − 4x + 7 相切。令两式相等得 x² − 6x + (7 − k) = 0；相切意味着恰有一个重根，故 Δ = 36 − 4(7 − k) = 0，得 k = −2。凡题目出现"相切""只有一个交点""两个相等实根"等表述，应立即写出 Δ = 0。
n compute Δ = 16 − 28 = −12 < 0 and note that the leading coefficient is positive, so the parabola always lies above the x-axis. The completed-square route, (x + 2)² + 3, proves the same statement and is often considered the cleaner argument.
即使没有实根，判别式仍然提供信息。若要证明对一切实数 x 都有 x² + 4x + 7 > 0，可算出 Δ = 16 − 28 = −12 < 0，且首项系数为正，说明抛物线始终位于 x 轴上方。用配方形式 (x + 2)² + 3 也能证明同一结论，而且往往被视为更简洁的论证。
7. Hidden Quadratics and Substitution | 隐藏二次方程与换元法
relationship: if the highest power is double the middle power, the substitution u = (middle expression) turns the equation into a familiar quadratic. Recognising this pattern is a standard Edexcel skill.
隐藏二次方程是指本身并非二次形式，但经过换元后可化为二次方程的式子。识别线索是"重复出现的表达式"或"幂次成倍关系"：若最高次幂是中间项幂次的两倍，则可令 u =（中间表达式），把方程化为熟悉的二次方程。识别这一模式是爱德思考试的标准技能。
x², so u² − 5u + 4 = 0, giving u = 1 or u = 4. Since u = x², we get x² = 1 or x² = 4, so x
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