📚 Mastering Quadratic Equations: Methods and Exam Techniques | 掌握二次方程：方法与应试技巧
Quadratic equations appear throughout the IGCSE and A-Level Mathematics syllabus, from straightforward factorisation to challenging word problems. Being able to recognise the standard form, choose an efficient solving method, and interpret the discriminant is essential for top marks. This guide revises every core technique, shows worked examples step by step, and finishes with the exam habits that separate a grade 6 from a grade 9.
二次方程贯穿 IGCSE 与 A-Level 数学考试大纲，从直接的因式分解到有难度的应用题。能够识别标准形式、选择高效解法并解释判别式，是取得高分的关键。本文系统复习所有核心方法，逐步展示例题，并在最后总结能把 6 分提升到 9 分的应试习惯。
1. What is a Quadratic Equation? | 什么是二次方程？
e written in the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0.
二次方程是变量最高次数为 2 的多项式方程。它总能写成标准形式 ax² + bx + c = 0，其中 a、b、c 为常数且 a ≠ 0。
t has only one root instead of two. Examples include x² - 5x + 6 = 0 and 2x² + 3x - 2 = 0.
a ≠ 0 这个条件正是方程成为二次方程的关键。如果 a = 0，方程就退化为一次方程，只有一根而不是两根。例如 x² - 5x + 6 = 0 和 2x² + 3x - 2 = 0。
ro, and they correspond to the points where the curve y = ax² + bx + c crosses the x-axis.
由于次数为 2，二次方程最多有两个实根。这些根就是使表达式等于零的 x 值，也对应曲线 y = ax² + bx + c 与 x 轴的交点。
2. Standard Form and Coefficients | 标准形式与系数
a, b and c are called the quadratic coefficient, the linear coefficient and the constant term.
在使用任何方法之前，先把方程整理成 ax² + bx + c = 0。数字 a、b、c 分别称为二次项系数、一次项系数和常数项。
om both sides and reorder the terms to get -4x² + 3x - 7 = 0, so a = -4, b = 3 and c = -7.
考题常常给出未整理成标准形式的方程。例如 3x - 4x² = 7，两边同减 7 并整理项序，得到 -4x² + 3x - 7 = 0，所以 a = -4、b = 3、c = -7。
Coefficients must be read together with their signs; dropping a minus sign is the single most expensive mistake in this topic. Use these three steps every time:
系数必须连同符号一起看，漏掉负号是本章最"贵"的错误。每次都用下面三步：
- Move every term to the left-hand side so the right-hand side is 0. 把所有项移到左边，使右边等于 0。
- Write the terms in descending powers of x, from x² down to the constant. 按 x 的降幂排列，从 x² 排到常数项。
- Read off a, b and c with their signs, filling in 0 for any missing term. 带符号读出 a、b、c，缺项补 0。
a = 1, b = 0 and c = -9; it is still a quadratic equation because the x² term is present.
有些方程看起来缺项。在 x² - 9 = 0 中，a = 1、b = 0、c = -9；它仍然是二次方程，因为 x² 项存在。
3. Solving by Factorisation | 因式分解法
to try first. For x² + bx + c = 0, find two numbers whose sum is b and whose product is c.
当二次方程具有简单的根时，因式分解是最快的方法，也是考官期望你首先尝试的方法。对于 x² + bx + c = 0，要找到两个数，使它们的和为 b、积为 c。
cause -2 + (-3) = -5 and (-2) × (-3) = 6. Hence (x - 2)(x - 3) = 0, giving x = 2 or x = 3.
对于 x² - 5x + 6 = 0，这两个数是 -2 和 -3，因为 -2 + (-3) = -5，(-2) × (-3) = 6。因此 (x - 2)(x - 3) = 0，得到 x = 2 或 x = 3。
en include square roots written as x = ±r, cube-free integers and other convenient values.
一定要展开检验：(x - 2)(x - 3) = x² - 5x + 6，与原式一致。适合因式分解的根通常是整数或简单的分数，例如 x = ±r 这类平方根形式。
-4 and the sum b = 3, so the two numbers are 4 and -1. Rewrite the middle term and group:
当 a ≠ 1 时，可以拆中项。对于 2x² + 3x - 2 = 0，a × c = -4，而 b = 3，所以两个数是 4 和 -1。把中项拆开并分组：
2x² + 4x - x - 2 = 2x(x + 2) - 1(x + 2) = (2x - 1)(x + 2)
k), and taking out a common factor such as x² - 5x = x(x - 5) = 0, giving x = 0 or x = 5.
所以 x = 1/2 或 x = -2。有两个特殊结构值得记忆：平方差 x² - k² = (x - k)(x + k)，以及提公因式，例如 x² - 5x = x(x - 5) = 0，得到 x = 0 或 x = 5。
y x destroys the root x = 0, and you will lose a mark even if the other answer is correct.
警告：在 x² = 5x 这类方程中，绝不要两边同除以 x。除以 x 会丢掉根 x = 0，即使另一个根正确，也会失分。
4. Completing the Square | 配方法
Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. It works even when the roots are irrational, and it is the standard way to find the vertex of a parabola or to prove that an expression is always positive.
配方法把 ax² + bx + c 改写为 a(x + p)² + q 的形式。即使根是无理数它也能用，而且是求抛物线顶点或证明表达式恒正的标准方法。
= 1) the rule is: take half of b, square it, add it and subtract it. For x² + 6x + 1 = 0:
对于首项系数为 1 的二次式，法则是：取 b 的一半，平方后加上再减去。对于 x² + 6x + 1 = 0：
(x + 3)² - 9 + 1 = 0 → (x + 3)² = 8 → x = -3 ± √8 = -3 ± 2√2
3, -8), and since the squared term cannot be negative, the minimum value of y is -8. When a ≠ 1, first divide through by a, or factor a out of the first two terms only, then complete the square inside the bracket.
y = (x + 3)² - 8 的顶点是 (-3, -8)；由于平方项不可能为负，y 的最小值为 -8。当 a ≠ 1 时，先整体除以 a，或者只把前两项提出 a，再在括号内配方。
Exam questions using this method frequently ask you to show that x² + bx + c > 0 for all x. Write the expression as (x + p)² + q with q > 0, then state that the square is at least 0, so the whole expression is at least q, which is positive.
使用该方法的考题经常要求你证明 x² + bx + c > 0 对一切 x 成立。把表达式写成 (x + p)² + q，且 q > 0，然后说明平方项至少为 0，因此整个表达式至少为 q，而 q 为正数。
5. The Quadratic Formula | 求根公式
formula. It is derived by completing the square on the general equation ax² + bx + c = 0:
当因式分解行不通或数字不友好时，使用求根公式。它是通过对一般式 ax² + bx + c = 0 配方推导出来的：
x = (-b ± √b² - 4ac) / (2a)
+ 3x - 2 = 0. Here a = 2, b = 3 and c = -2, so b² - 4ac = 9 - 4(2)(-2) = 9 + 16 = 25, and √25 = 5. Therefore x = (-3 ± 5)/4, giving x = 2/4 = 1/2 or x = -8/4 = -2.
例题：解 2x² + 3x - 2 = 0。这里 a = 2、b = 3、c = -2，所以 b² - 4ac = 9 - 4(2)(-2) = 9 + 16 = 25，√25 = 5。因此 x = (-3 ± 5)/4，得到 x = 2/4 = 1/2 或 x = -8/4 = -2。
Mark-earning habits with the formula:
使用公式的得分习惯：
- Write the formula down before substituting; method marks are awarded for the correct structure. 代入前先写出公式，结构分就是靠这一步拿到的。
- Substitute with brackets around every value, especially negatives: (-3)² not -3². 代入时每个数值加括号，尤其是负数：写 (-3)² 而不是 -3²。
- Keep the whole calculation on one line so the examiner can follow it. 把整个过程写在同一行，方便考官跟随你的思路。
- Round only at the end, normally to 3 significant figures unless told otherwise. 只在最后一步取近似，除非另有说明，一般保留 3 位有效数字。
If the discriminant is negative you will be square-rooting a negative number, which has no real solution. In that case stop and state that there are no real roots.
如果判别式为负，就要对负数开平方，这在实数范围内无解。此时应停下并说明方程没有实根。
6. The Discriminant | 判别式
ed the discriminant. It tells you how many real roots a quadratic has before you solve it.
根号下的表达式 Δ = b² - 4ac 称为判别式。它能在求解之前告诉你方程有几个实根。
| Value of Δ | Number of real roots | Graph of y = ax² + bx + c |
| Δ > 0 | Two distinct real roots | Crosses the x-axis twice |
| Δ = 0 | One repeated real root | Touches the x-axis at the vertex |
| Δ < 0 | No real roots | Does not meet the x-axis |
. For x² + kx + 9 = 0 with a repeated root, set Δ = 0: k² - 36 = 0, so k² = 36 and k = ±6.
典型考题会给出含未知常数的方程，要求求出使方程有等根的值。对于有重根的 x² + kx + 9 = 0，令 Δ = 0：k² - 36 = 0，所以 k² = 36，k = ±6。
having two distinct real roots, remember two conditions: Δ > 0 and the coefficient of x² must be non-zero. Always state the discriminant first, then solve the resulting inequality.
对于 kx² + 2x + 3 = 0 有两个不同实根这类问题，要记住两个条件：Δ > 0，且 x² 的系数不能为零。先写出判别式，再解所得的不等式。
7. Graphs, Roots and the Parabola | 图像、根与抛物线
. If a > 0 it opens upwards and has a minimum point; if a < 0 it opens downwards and has a maximum point. The larger the value of |a|, the narrower the curve.
y = ax² + bx + c 的图像是一条抛物线。a > 0 时开口向上，有最小值点；a < 0 时开口向下，有最大值点。|a| 越大，曲线越窄。
it. Substituting that x-value back into the equation gives the y-coordinate of the turning point.
对称轴是竖直线 x = -b/(2a)，顶点就位于这条轴上。把该 x 值代回方程即得顶点纵坐标。
The roots found algebraically are exactly the x-intercepts of the graph, which links the three representations together. A useful checking habit is to sketch the parabola roughly: if your algebraic roots suggest two crossings but your sketch shows the vertex above the axis and the curve opening upwards, something is wrong.
代数求得的根正是图像与 x 轴交点的横坐标，这就把三种表示联系起来了。一个有用的检查习惯是粗略画出抛物线：如果代数结果给出两个交点，但草图显示开口向上且顶点在 x 轴上方，那么一定有地方出错了。
, rearrange to x² - 3x - 2 = 0, and solve; the number of intersections is governed by the discriminant of the combined equation.
图像题也可能要求解联立方程，例如求 y = x² - 2x + 1 与 y = x + 3 的交点。令两式相等，整理得 x² - 3x - 2 = 0 再求解；交点个数由合并后方程的判别式决定。
8. Sum and Product of Roots | 两根之和与两根之积
then their sum and product are given directly by the coefficients, with no need to solve the equation:
若 ax² + bx + c = 0 的两根为 α 和 β，则它们的和与积可以直接由系数给出，无需解方程：
α + β = -b/a and αβ = c/a
iven one root, you can find the other by substitution: if α = 1/2, then β = 3/2 ÷ 1/2 = 3.
对于 2x² - 7x + 3 = 0，α + β = 7/2，αβ = 3/2。已知一根时可用代入法求另一根：若 α = 1/2，则 β = 3/2 ÷ 1/2 = 3。
written as x² - (α + β)x + αβ = 0, or equivalently k(x - α)(x - β) = 0 for any non-zero k.
这些结论还能让你由根反推方程。以 α、β 为根的二次方程可写成 x² - (α + β)x + αβ = 0，或等价地写成 k(x - α)(x - β) = 0，其中 k 为非零常数。
² - 2αβ and substitute the coefficient values directly, which avoids solving for α and β separately.
高阶层题目常要求求 α² + β²。使用恒等式 α² + β² = (α + β)² - 2αβ，直接代入系数值即可，避免分别求出 α 和 β。
9. Forming Equations and Word Problems | 构造方程与应用题
Word problems are solved in five fixed steps: define the variable clearly, translate the words into an equation, rearrange into standard form, solve, then interpret the answer in context.
应用题按五个固定步骤求解：清楚定义变量、把文字翻译成方程、整理成标准形式、求解，最后结合情境解释答案。
th be w cm, so the length is w + 3. Then w(w + 3) = 40, which expands to w² + 3w - 40 = 0.
例：一个长方形的长比宽多 3 cm，面积为 40 cm²。设宽为 w cm，则长为 w + 3。于是 w(w + 3) = 40，展开得 w² + 3w - 40 = 0。
(w + 8)(w - 5) = 0 → w = -8 or w = 5
tive, w = 5 cm and the length is 8 cm. Checking: 5 × 8 = 40, which matches the given area.
由于宽度不能为负，取 w = 5 cm，长为 8 cm。检验：5 × 8 = 40，与题目给出的面积相符。
. In every case, write down the units and reject any root that contradicts the physical situation.
常见情境包括连续整数（n 与 n + 1）、用速度、路程、时间关系的问题（时间 = 路程 ÷ 速度）以及年龄问题。无论哪种，都要写上单位，并舍去与实际情况矛盾的根。
10. Hidden Quadratics and Substitution | 隐性二次方程与换元法
Some equations are not written as quadratics but can be turned into one by substitution. Recognising the pattern x², x⁴, or a repeated bracket is the key skill being tested.
有些方程并不以二次形式出现，但通过换元可以化为二次方程。识别 x²、x⁴ 或重复出现的括号结构，就是本题型考查的关键技能。
For x⁴ - 5x² + 4 = 0, let u = x². Then u²
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