📚 Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | IGCSE 数学：二次方程——因式分解、配方法与求根公式
Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in algebra, coordinate geometry, measurement problems and even in optimisation questions. Mastering the three core algebraic methods - factorising, completing the square and the quadratic formula - will give you a reliable toolkit for both Core and Extended papers.
二次方程是 IGCSE 数学中最重要的主题之一。它们出现在代数、坐标几何、测量问题甚至优化题中。掌握三种核心代数方法——因式分解、配方法和求根公式——将为你应对核心卷和扩展卷提供可靠的工具箱。
1. Recognising a Quadratic Equation | 识别二次方程
re constants and a ≠ 0. The condition a ≠ 0 is essential because if a were zero, the x² term would disappear and the equation would become linear. The constants a, b and c may be positive, negative, whole numbers, fractions or surds - the only rule is that the highest power of x must be exactly 2.
一个一元二次方程是任何可以整理成标准式 ax² + bx + c = 0 的方程，其中 a、b、c 是常数且 a ≠ 0。a ≠ 0 这一条件至关重要，因为如果 a 为零，x² 项就会消失，方程退化为一次方程。常数 a、b、c 可以是正数、负数、整数、分数或根式——唯一的规则是 x 的最高次幂必须正好是 2。
Before choosing a method, always rewrite the equation so that one side is zero. Many exam questions deliberately hide the quadratic by placing terms on both sides.
在选择方法之前，务必先把方程整理成一边为零的形式。很多考题会故意把项分散在等号两边，从而"隐藏"二次方程的身份。
- x² − 5x + 6 = 0 → already in standard form / 已是标准式，a = 1, b = −5, c = 6
- 2x² = 5x − 3 → rearrange to 2x² − 5x + 3 = 0 / 移项得 2x² − 5x + 3 = 0
- 3x(x − 2) = 7 → expand to 3x² − 6x − 7 = 0 / 展开得 3x² − 6x − 7 = 0
- (x + 4)² = 9 → expand to x² + 8x + 7 = 0 / 展开得 x² + 8x + 7 = 0
x = 4. Dropping the negative root is one of the most frequent lost marks in IGCSE papers.
常见的陷阱是把 x² = 16 误当成一次方程处理。它其实是二次方程，答案是 x = 4 或 x = −4，而不只是 x = 4。漏掉负根是 IGCSE 考卷中失分最常见的原因之一。
2. Roots, Solutions and Graphical Meaning | 根、解与图像含义
n you solve ax² + bx + c = 0, you are finding where the curve y = ax² + bx + c crosses the x-axis.
有几个词描述的是同一件事。二次方程的根、解、零点或 x 轴截距，都是使方程成立的 x 值。当你解 ax² + bx + c = 0 时，你实际上是在求曲线 y = ax² + bx + c 与 x 轴的交点。
The graph of any quadratic 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. A parabola can meet the x-axis twice, once (touching it) or not at all, which corresponds to two, one or zero real roots.
任何二次函数的图像都是抛物线。若 a > 0，抛物线开口向上，有最低点；若 a < 0，开口向下，有最高点。抛物线与 x 轴可能相交两次、相切一次或完全不相交，这分别对应两个、一个或零个实根。
y = ax² + bx + c, a ≠ 0 → parabola shape
re x = −7 and x = 12, but your sketch shows the curve crossing the axis near 0 and 3, something has gone wrong.
这一图像联系在扩展卷中威力很大：画个草图就能检验代数答案是否合理。如果因式分解给出的根是 x = −7 和 x = 12，而草图显示曲线在 0 和 3 附近穿过 x 轴，那一定出错了。
3. Method 1: Factorising - The Basics | 方法一：因式分解基础
t multiply to give c and add to give b. The factorised form is (x + p)(x + q) where pq = c and p + q = b.
因式分解通常是最快的方法，所以只要数字看起来"友好"，就应优先尝试。对于 x² + bx + c 的简单情形，你需要找到两个数：它们相乘等于 c，相加等于 b。分解后的形式是 (x + p)(x + q)，其中 pq = c 且 p + q = b。
- x² + 7x + 12 = (x + 3)(x + 4), since 3 × 4 = 12 and 3 + 4 = 7 / 因为 3 × 4 = 12，3 + 4 = 7
- x² − 5x + 6 = (x − 2)(x − 3), since (−2) × (−3) = 6 and (−2) + (−3) = −5
- x² − x − 12 = (x − 4)(x + 3), since (−4) × 3 = −12 and (−4) + 3 = −1
- x² + 3x − 10 = (x + 5)(x − 2), since 5 × (−2) = −10 and 5 + (−2) = 3
The sign pattern is a quick guide. If c is positive, the two numbers share the same sign, and that sign matches b. If c is negative, the two numbers have opposite signs, and the larger one in absolute value takes the sign of b.
符号规律可以快速指路。若 c 为正，两个数同号，且符号与 b 相同；若 c 为负，两个数异号，而绝对值较大的那个数取 b 的符号。
lly: (x + 3)(x + 4) = x² + 7x + 12. Ten seconds of checking prevents most careless errors.
务必在脑中展开检验：(x + 3)(x + 4) = x² + 7x + 12。花十秒检验，可以避免绝大多数粗心错误。
4. Factorising Harder Quadratics (a ≠ 1) | 系数不为 1 的二次式分解
When the coefficient of x² is not 1, the sum-and-product trick needs an adjustment. A dependable method is to multiply a by c, find two numbers that multiply to ac and add to b, split the middle term, then factorise by grouping.
当 x² 的系数不是 1 时，单纯的"和与积"技巧需要调整。一个可靠的方法是：先算 a × c，找两个数相乘等于 ac、相加等于 b，然后拆开中间项，再用分组法分解。
−210, and we need two numbers with product −210 and sum 11. Those numbers are 21 and −10.
例如：分解 6x² + 11x − 35。此时 ac = 6 × (−35) = −210，需要两个数乘积为 −210、和为 11。这两个数是 21 和 −10。
6x² + 21x − 10x − 35 = 3x(2x + 7) − 5(2x + 7) = (2x + 7)(3x − 5)
+ 11x − 35, which matches. Also memorise the special patterns, since examiners love them.
展开检验：(2x + 7)(3x − 5) = 6x² − 10x + 21x − 35 = 6x² + 11x − 35，完全吻合。另外要牢记特殊模式，考官非常喜欢考。
- Difference of two squares: a² − b² = (a + b)(a − b), e.g. 9x² − 25 = (3x + 5)(3x − 5)
- Perfect square: a² + 2ab + b² = (a + b)², e.g. 4x² + 12x + 9 = (2x + 3)²
- Common factor first: 2x² − 8 = 2(x² − 4) = 2(x + 2)(x − 2)
Always take out a common numerical factor before anything else. It shrinks the numbers and often reveals a difference of two squares you would otherwise miss.
永远先提取公因数。这样数字会变小，而且常常能暴露出你原本会忽略的平方差结构。
5. The Null Factor Law | 零因子定律
at least one of the expressions must be zero. In symbols: if AB = 0, then A = 0 or B = 0.
一旦完成因式分解，零因子定律就能收尾。它指出：如果两个表达式的乘积为零，那么至少有一个表达式为零。用符号表示：若 AB = 0，则 A = 0 或 B = 0。
0, giving x = 4 or x = −3. Both values should be stated clearly, usually as a final line.
解 x² − x − 12 = 0：分解得 (x − 4)(x + 3) = 0，于是 x − 4 = 0 或 x + 3 = 0，即 x = 4 或 x = −3。两个值都要清楚写出，通常作为最后一行结论。
approach rearranges to x² − 5x = 0, factorises to x(x − 5) = 0, and gives x = 0 or x = 5.
经典错误是两边同时除以 x。例如 x² = 5x，若除以 x 便得到 x = 5，丢掉了根 x = 0。正确做法是移项得 x² − 5x = 0，分解为 x(x − 5) = 0，得到 x = 0 或 x = 5。
Never divide by a variable unless you are certain it cannot be zero. In quadratic work it almost always can be, so always factorise instead.
除非能确定变量不为零，否则绝不要除以变量。在二次方程问题中它几乎总可以为零，所以永远选择因式分解。
6. Method 2: Completing the Square | 方法二：配方法
Completing the square rewrites a quadratic in the form (x + p)² + q. It works on every quadratic, including those that will not factorise with integers, and it exposes the vertex of the parabola directly.
配方法把二次式改写为 (x + p)² + q 的形式。它对所有二次式都适用，包括无法用整数分解的那些，而且能直接揭示抛物线的顶点。
p², then add c back. Written as a formula, this is x² + bx + c = (x + b/2)² − (b/2)² + c.
对于 x² + bx + c，做法是：把 b 除以 2 得到 p，减去 p²，再把 c 加回来。写成公式即 x² + bx + c = (x + b/2)² − (b/2)² + c。
- x² + 8x + 3 = (x + 4)² − 16 + 3 = (x + 4)² − 13 / 半数为 4，4² = 16
- x² − 6x + 1 = (x − 3)² − 9 + 1 = (x − 3)² − 8 / half of −6 is −3
- x² + 10x = (x + 5)² − 25 / c = 0 here, so nothing is added back
² − 6x) + 7, complete inside to get (x − 3)² − 9, so the result is 2(x − 3)² − 18 + 7 = 2(x − 3)² − 11.
当 a ≠ 1 时，先把 a 从 x² 项和 x 项中提出（常数项不参与），在括号内配方，再乘回去。对 2x² − 12x + 7：提出 2 得 2(x² − 6x) + 7，括号内配方得 (x − 3)² − 9，因此结果为 2(x − 3)² − 18 + 7 = 2(x − 3)² − 11。
A frequent slip is forgetting to multiply the subtracted square by the factor in front. If you take 2 out of the bracket, everything inside must be multiplied by 2 when you finish.
常见失误是忘记把减去的平方数乘回括号前的系数。既然提出了 2，最后就必须把括号内的一切乘以 2。
7. Solving and Finding the Vertex by Completing the Square | 用配方法解方程与求顶点
sides, and remember the ± sign. This is precisely where the quadratic formula comes from.
用配方法解方程时，先把常数移到等号右边，两边开平方，并记住 ± 号。求根公式正是由此推导出来的。
e x² + 8x + 3 = 0. Complete the square: (x + 4)² − 13 = 0, so (x + 4)² = 13, and x + 4 = ±√13. Therefore x = −4 + √13 or x = −4 − √13, approximately 0.39 or −8.39.
解 x² + 8x + 3 = 0。配方得 (x + 4)² − 13 = 0，即 (x + 4)² = 13，所以 x + 4 = ±√13。因此 x = −4 + √13 或 x = −4 − √13，约等于 0.39 或 −8.39。
− 13, the minimum point is (−4, −13) and the axis of symmetry is the vertical line x = −4.
配方式还能直接给出顶点坐标，无需额外计算。对 y = (x + 4)² − 13，最低点为 (−4, −13)，对称轴为竖直线 x = −4。
y = a(x + p)² + q → vertex at (−p, q)
. If the coefficient were negative, the same point would be a maximum. Exam questions often ask for the minimum value of a quantity, and this is the fastest route to the answer.
对 2(x − 3)² − 11，顶点为 (3, −11)，由于 a = 2 > 0 是最低点。若系数为负，同一个点就是最高点。考题常问某个量的最小值，这是最快的求解途径。
8. Method 3: The Quadratic Formula | 方法三：求根公式
When factorising fails or the numbers are unpleasant, use the quadratic formula. It solves every quadratic equation with real roots, provided the equation is first written in standard form.
当因式分解行不通或数字很难看时，就用求根公式。只要先把方程写成标准式，它能解出所有有实根的二次方程。
x = (−b ± √b² − 4ac) / (2a)
Example: solve 2x² + 5x − 3 = 0. Here a = 2, b = 5, c = −3, so b² − 4ac = 25 + 24 = 49 and √49 = 7. Then x = (−5 ± 7) / 4, giving x = 2/4 = 1/2 or x = −12/4 = −3.
例：解 2x² + 5x − 3 = 0。这里 a = 2，b = 5，c = −3，所以 b² − 4ac = 25 + 24 = 49，√49 = 7。于是 x = (−5 ± 7) / 4，得 x = 2 ÷ 4 = 1/2 或 x = −12 ÷ 4 = −3。
Example with surds: solve x² + 4x + 1 = 0. Then b² − 4ac = 16 − 4 = 12, so x = (−4 ± √12) / 2 = (−4 ± 2√3) / 2 = −2 ± √3. Leaving the answer in surd form is exact; decimals such as −0.27 and −3.73 are rounded and may lose marks if the question says 'exact'.
含根式的例子：解 x² + 4x + 1 = 0。则 b² − 4ac = 16 − 4 = 12，所以 x = (−4 ± √12) / 2 = (−4 ± 2√3) / 2 = −2 ± √3。保留根式是精确答案；像 −0.27 和 −3.73 这样的十进制是四舍五入结果，若题目要求"精确值"就可能失分。
ers. Writing −b with b = −7 gives 7, not −7, and calculator mistakes here cost easy marks.
代入时，负数一定要加括号。当 b = −7 时，−b 等于 7，而不是 −7。这种计算器输入失误会白白丢分。
9. The Discriminant | 判别式
The expression b² − 4ac inside the square root is called the discriminant, often written Δ. Its sign tells you how many real roots the equation has, before you do any solving.
根号内的表达式 b² − 4ac 称为判别式，常记作 Δ。它的符号可以在求解之前就告诉你方程有多少个实根。
| Discriminant / 判别式 | Number of real roots / 实根个数 | Graph / 图像 |
| b² − 4ac > 0 | Two distinct real roots / 两个不同实根 | Crosses the x-axis twice / 与 x 轴交于两点 |
| b² − 4ac = 0 | One repeated root / 一个重根 | Touches the axis / 与 x 轴相切 |
| b² − 4ac < 0 | No real roots / 无实根 | Does not meet the axis / 与 x 轴不相交 |
Typical exam wording: 'Find the values of k for which x² + kx + 9 = 0 has equal roots.' − 36 = 0 and k = ±6. The discriminant has turned an abstract question into simple algebra.
典型考题："求使 x² + kx + 9 = 0 有两个相等实根的 k 值。"相等实根要求 b² − 4ac = 0，即 k² − 36 = 0，所以 k = ±6。判别式把一个抽象问题变成了简单代数。
10. Choosing the Right Method | 如何选择合适的方法
Strong candidates do not use the same method every time. They scan the equation, decide in a few seconds, and then commit. The decision rules below cover almost every IGCSE question you will meet.
优秀考生不会每次都套用同一个方法。他们会先审视方程，几秒钟内做出判断，然后果断执行。下面的判断规则几乎覆盖你会在 IGCSE 中遇到的所有题目。
- Small integer coefficients and an obvious pair? Factorise - it is fastest. / 整数系数小、能一眼看出配对？用因式分解，最快。
- Asked for the vertex, minimum value or a proof? Complete the square. / 题目问顶点、最小值或要求证明？用配方法。
- Ugly numbers, decimals or a "give your answer to 3 significant figures" instruction? Use the formula. / 数字丑陋、含小数或要求"保留 3 位有效数字"？用求根公式。
- Asked how many roots, or for a value of k? Use the discriminant. / 问有几个根，或求参数 k？用判别式。
- Cannot factorise and the question says "exact form"? Complete the square or use the formula and keep the surd. / 无法分解且要求"精确形式"？用配方法或公式，保留根式。
If you have time, use two methods and check that the answers agree. Solving by factorising and then confirming with the formula takes barely a minute and converts a risky answer into a certain one.
如果时间允许，用两种方法交叉验证答案是否一致。先用因式分解求解，再用公式确认，只需不到一分钟，却能把一个冒险的答案变成确定的
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