📚 Solving Quadratic Equations | 二次方程求解方法精讲
Quadratic equations are central to IGCSE Mathematics and appear in many topics, including algebra, graphs in coordinate geometry, and real-life modelling problems. A quadratic equation has degree 2, and the syllabus expects you to solve them confidently by factorisation, completing the square, and the quadratic formula.
二次方程是 IGCSE 数学的核心内容，出现在许多主题中，包括代数、坐标几何中的函数图像以及实际建模问题。二次方程的次数为 2，课程大纲要求你能够熟练运用因式分解法、配方法和求根公式来求解。
In this article we explain each method step by step, compare the discriminant and graphs, and highlight common errors to help you revise efficiently.
在本文中，我们将逐步讲解每一种方法，比较判别式与图像，并强调常见错误，帮助你高效复习。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is a polynomial equation of degree 2. The highest power of the variable x is 2, and the equation can contain an x² term, an x term, and a constant term.
二次方程是一个二次多项式方程。变量 x 的最高次数是 2，方程可以包含 x² 项、x 项和常数项。
ic equations. The second example can be rearranged into standard form as 3x² − 2x − 8 = 0.
例如，x² + 5x + 6 = 0 和 3x² − 2x = 8 都是二次方程。第二个例子可以整理为标准形式 3x² − 2x − 8 = 0。
A quadratic equation can have zero, one, or two real solutions depending on its coefficients and discriminant. The word 'quadratic' comes from the Latin word 'quadratus', meaning square, because the variable is squared.
二次方程根据其系数和判别式，可以有零个、一个或两个实数解。'quadratic'（二次）一词来自拉丁语 'quadratus'，意思是"平方"，因为变量被平方。
Note the difference between a quadratic equation and a quadratic expression. An expression such as x² + 5x + 6 can be factorised but has no solutions; only an equation, which contains an equals sign, can be solved.
请注意二次方程与二次表达式的区别。像 x² + 5x + 6 这样的表达式可以因式分解，但没有"解"；只有包含等号的方程才能求解。
2. Standard Form and Coefficients | 标准形式与系数
and a ≠ 0. The condition a ≠ 0 is essential because if a = 0 the equation becomes linear.
标准形式为 ax² + bx + c = 0，其中 a、b 和 c 是常数且 a ≠ 0。a ≠ 0 这个条件非常重要，因为如果 a = 0，方程就变成了一次方程。
quadratic equations, such as (x + 2)(x − 3) = 0, which can be expanded first if necessary.
正确识别 a、b 和 c 是使用求根公式或求判别式的第一步。你还应注意隐藏的二次方程，例如 (x + 2)(x − 3) = 0，必要时可以先展开。
| Equation | a | b | c |
|---|---|---|---|
| x² + 5x + 6 = 0 | 1 | 5 | 6 |
| 2x² − 3x − 5 = 0 | 2 | −3 | −5 |
| 3x² − 2x − 8 = 0 | 3 | −2 | −8 |
stance, 3x² − 2x = 8 must be rewritten as 3x² − 2x − 8 = 0, so that c = −8 rather than +8.
在读系数之前，务必先把方程整理成标准形式。例如 3x² − 2x = 8 必须改写成 3x² − 2x − 8 = 0，这样 c = −8，而不是 +8。
3. Solving by Factorisation | 因式分解法
Factorisation is the fastest method when the roots are integers or simple fractions. The key idea is the null factor law: if the product of two factors is zero, then at least one factor must be zero.
当根为整数或简单分数时，因式分解是最快的方法。核心思想是零因子法则：如果两个因式的乘积为零，那么至少有一个因式为零。
If (x − p)(x − q) = 0, then x = p or x = q
+ 6 = 0, the numbers 2 and 3 multiply to 6 and add to 5, so x² + 5x + 6 = (x + 2)(x + 3).
对于 a = 1 的 x² + bx + c = 0，要找两个数，使它们的乘积等于 c，和等于 b。在 x² + 5x + 6 = 0 中，2 和 3 相乘得 6，相加得 5，所以 x² + 5x + 6 = (x + 2)(x + 3)。
ts should be written down, and a quick substitution back into the original equation confirms they work.
令每个括号等于零，得到 x + 2 = 0 或 x + 3 = 0，所以 x = −2 或 x = −3。两个根都要写出来，把答案代回原方程快速验证即可确认无误。
, so split −3x into −5x + 2x, giving 2x² − 5x + 2x − 5 = x(2x − 5) + 1(2x − 5) = (2x − 5)(x + 1). Hence x = 5/2 or x = −1.
当 a ≠ 1 时，可以用带首项系数的"乘积与和"法，或拆分中间项。对 2x² − 3x − 5 = 0，乘积 a × c = −10，和为 −3，于是把 −3x 拆成 −5x + 2x，得到 2x² − 5x + 2x − 5 = x(2x − 5) + 1(2x − 5) = (2x − 5)(x + 1)。因此 x = 5/2 或 x = −1。
4. Completing the Square | 配方法
Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. It is the method required for finding the vertex of a parabola and for proving that a quadratic is always positive or always negative.
配方法把 ax² + bx + c 改写成 a(x + p)² + q 的形式。它是求抛物线顶点、以及证明某个二次式恒正或恒负时必须使用的方法。
alf of b, square it, and add and subtract it. The identity is x² + bx = (x + b/2)² − b²/4.
对于首项系数为 1 的二次式 x² + bx + c，取 b 的一半，平方后加上再减去。恒等式为 x² + bx = (x + b/2)² − b²/4。
= 9, so x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7. Therefore (x + 3)² = 7 and x = −3 ± √7.
例：解 x² + 6x + 2 = 0。6 的一半是 3，3² = 9，所以 x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7。因此 (x + 3)² = 7，x = −3 ± √7。
+ 2)² − 4] + 1 = 2(x + 2)² − 7. The vertex of the curve y = 2(x + 2)² − 7 is at (−2, −7).
当 a ≠ 1 时，先只把 a 从含 x 的项中提出：2x² + 8x + 1 = 2(x² + 4x) + 1 = 2[(x + 2)² − 4] + 1 = 2(x + 2)² − 7。曲线 y = 2(x + 2)² − 7 的顶点在 (−2, −7)。
Completing the square also gives the turning-point form used to answer questions such as 'show that x² − 4x + 7 > 0 for all x', because (x − 2)² + 3 is always at least 3.
配方法还能给出"顶点式"，用来回答如"证明对任意 x 都有 x² − 4x + 7 > 0"这类问题，因为 (x − 2)² + 3 恒不小于 3。
5. The Quadratic Formula | 求根公式
ot factorise. It is derived by completing the square on the general form ax² + bx + c = 0.
求根公式适用于所有二次方程，包括无法因式分解的方程。它正是对一般形式 ax² + bx + c = 0 配方推导出来的。
x = (−b ± √b² − 4ac) / 2a, a ≠ 0
To use it, write the equation in standard form, identify a, b and c with their signs, then substitute carefully. A common slip is to forget that −b changes the sign of b when b itself is negative.
使用时，先把方程化为标准形式，带符号写出 a、b、c，再小心代入。常见的失误是忘记 −b 会改变 b 的符号，尤其是 b 本身为负时。
Example: solve 2x² − 3x − 5 = 0. Here a = 2, b = −3, c = −5, so b² − 4ac = 9 + 40 = 49 and √49 = 7. Thus x = (3 ± 7) / 4, giving x = 2.5 or x = −1, matching the factorised answer.
例：解 2x² − 3x − 5 = 0。这里 a = 2，b = −3，c = −5，所以 b² − 4ac = 9 + 40 = 49，√49 = 7。因此 x = (3 ± 7) / 4，得到 x = 2.5 或 x = −1，与因式分解的结果一致。
Leave surds in exact form unless the question asks for a decimal, and round only at the final step. Keep at least three significant figures, or follow the accuracy stated in the paper.
除非题目要求小数，否则应保留根式的精确形式，并且只在最后一步取近似值。至少保留三位有效数字，或遵循试卷中给出的精度要求。
6. The Discriminant | 判别式
discriminant. Its value tells you how many real roots the equation has without solving it.
根号下的表达式 Δ = b² − 4ac 称为判别式。它的取值可以在不解方程的情况下告诉你方程有几个实数根。
| Discriminant | Number of real roots | Graph of y = ax² + bx + c |
|---|---|---|
| b² − 4ac > 0 | Two distinct real roots | Crosses the x-axis twice |
| b² − 4ac = 0 | One repeated root | Touches the x-axis once |
| b² − 4ac < 0 | No real roots | Does not meet the x-axis |
x² + 2x + 1 = 0, Δ = 4 − 4 = 0 and the only root is x = −1, a repeated root at the vertex.
对 x² + 5x + 6 = 0，Δ = 25 − 24 = 1 > 0，所以有两个根 −2 和 −3。对 x² + 2x + 1 = 0，Δ = 4 − 4 = 0，唯一的根是 x = −1，这是位于顶点处的重根。
Discriminant questions often involve an unknown constant, for example 'find k such that x² + kx + 9 = 0 has equal roots'. Setting k² − 36 = 0 gives k = 6 or k = −6.
判别式题目常含有未知常数，例如"求使 x² + kx + 9 = 0 有等根的 k"。令 k² − 36 = 0，得到 k = 6 或 k = −6。
7. Quadratic Graphs and Roots | 二次函数图像与根
x + c. Solving an equation and finding intercepts are therefore the same task seen from two angles.
方程 ax² + bx + c = 0 的根，正是曲线 y = ax² + bx + c 与 x 轴交点的横坐标。因此"解方程"与"求截距"是从两个角度看同一件事。
wnwards and has a maximum point. The curve is symmetric about the vertical line x = −b/2a.
所有二次函数图像都是抛物线。若 a > 0，开口向上，有最低点；若 a < 0，开口向下，有最高点。曲线关于直线 x = −b/2a 对称。
y = a(x + p)² + q has vertex (−p, q). The y-intercept is always c, found by setting x = 0.
配方可以直接给出顶点坐标：y = a(x + p)² + q 的顶点为 (−p, q)。y 轴截距总是 c，令 x = 0 即可得到。
draw y = x² − 2x − 3 and y = 2x + 1 and read off the x-values of the intersection points.
图像法也能解原本不是二次方程的问题。要解 x² − 2x − 3 = 2x + 1，可以画 y = x² − 2x − 3 与 y = 2x + 1，然后读出交点横坐标。
8. Hidden Quadratics and Substitution | 隐藏的二次方程与换元法
Some equations are quadratic in disguise. Recognising the pattern allows you to apply the same three methods and gain easy marks.
有些方程只是"伪装"成其他形式，本质仍是二次方程。识别出这种结构，就能套用同样的三种方法，轻松得分。
- x² + 4 = 0: let u = x², solve u² − 5u + 4 = 0, then x² = 1 or x² = 4, so x = ±1 or x = ±2.
- x⁴ − 5x² + 4 = 0：令 u = x²，解 u² − 5u + 4 = 0，得 x² = 1 或 x² = 4，所以 x = ±1 或 x = ±2。
- (x + 2)(x − 3) = 0: already factorised, so the roots are x = −2 and x = 3.
- (x + 2)(x − 3) = 0：已经分解好，根为 x = −2 和 x = 3。
- x + 6/x = 5: multiply by x to get x² − 5x + 6 = 0, so x = 2 or x = 3 with x ≠ 0.
- x + 6/x = 5：两边乘以 x 得 x² − 5x + 6 = 0，所以 x = 2 或 x = 3，且 x ≠ 0。
When you substitute, always convert back to the original variable, and check that no solution makes a denominator zero or lies outside the domain given in the question.
换元之后一定要换回原变量，并检查是否有解使分母为零，或不在题目给定的定义域内。
9. Word Problems and Modelling | 应用题与建模
Many exam questions set quadratics in context: areas of rectangles, projectile heights, profit functions and speed-time problems. The mathematics is the same; the extra step is interpreting the answer.
许多考题把二次方程放在实际情境中：矩形面积、抛体高度、利润函数、速度—时间问题。数学过程相同，多出的一步是对答案作出解释。
x − 40 = 0, giving (x + 8)(x − 5) = 0 and x = 5 (since x = −8 is impossible for a length).
一个矩形的长为 (x + 3) 厘米，宽为 x 厘米，面积为 40 平方厘米。则 x(x + 3) = 40，即 x² + 3x − 40 = 0，分解得 (x + 8)(x − 5) = 0，所以 x = 5（x = −8 作为长度不可能）。
onds. It hits the ground when h = 0, so −5t² + 20t = 0, −5t(t − 4) = 0, and t = 4 seconds.
一个上抛的小球在 t 秒后的高度为 h = −5t² + 20t 米。落地时 h = 0，即 −5t² + 20t = 0，−5t(t − 4) = 0，所以 t = 4 秒。
Always reject solutions that are physically impossible, define your variable clearly in words, and finish with a sentence that states the answer in the context of the question.
凡是物理上不可能的解都要舍去，先用文字清楚定义变量，最后用一句话结合题目情境给出答案。
10. Common Errors and Exam Tips | 常见错误与考试技巧
Most lost marks come from sign slips rather than from not knowing the method. Reading your own working line by line is often worth more than extra practice questions.
失分大多来自符号错误，而不是不会方法。逐行检查自己的运算，往往比多做几道题更有价值。
- Forgetting to rearrange into ax² + bx + c = 0 before identifying a, b and c.
- 在读系数之前忘记整理成 ax² + bx + c = 0。
- Dividing both sides by x, which loses the root x = 0.
- 两边同时除以 x，从而丢掉 x = 0 这个根。
- Writing only one root when the equation has two.
- 方程有两个根时只写一个。
- Sign errors when substituting b = −3 into −b and into b².
- 把 b = −3 代入 −b 和 b² 时出现符号错误。
- Rounding too early, which makes the final answer inaccurate.
- 过早取近似值，导致最终答案不准确。
In a 'show that' question, do not cross-multiply backwards from the given answer to your own starting line; start from the given information and work forward with clear, justified steps.
在"证明"类题目中，不要从给出的结论倒推回起点；应从头信息出发，用清晰、有依据的步骤向前推导。
11. Practice Questions with Worked Answers | 练习题与详解
Try these before reading the solutions. They cover each method and the main exam question styles.
先自己动手做，再看解答。这些题覆盖了各种方法以及主要的考试题型。
- Q1. Solve x² + 5x + 6 = 0 by factorisation.
- Q1. 用因式分解解 x² + 5x + 6 = 0。
- Q2. Solve 3x² − 2x − 8 = 0 by the quadratic formula, giving exact answers.
- Q2. 用求根公式解 3x² − 2x − 8 = 0，给出精确答案。
- Q3. Write x² + 6x + 2 in the form (x + p)² + q and hence solve x² + 6x + 2 = 0.
- Q3. 把 x² + 6x + 2 写成 (x + p)² + q 的形式，并由此解 x² + 6x + 2 = 0。
- Q4. Find the values of k for which x² + kx + 9 = 0 has equal roots.
- Q4. 求使 x² + kx + 9 = 0 有等根的 k 值。
Answers: Q1 gives (x + 2)(x + 3) = 0, so x = −2 or x = −3. Q2 gives x = (2 ± √4 + 96) / 6 = (2 ± 10) / 6, so x = 2 or x = −4/3. Q3 gives (x + 3)² − 7 = 0, so x = −3 ± √7. Q4 requires k² − 36 = 0, so k = 6 or k = −6.
答案：Q1 得 (x + 2)(x + 3) = 0，所以 x = −2 或 x = −3。Q2 得 x = (2 ± √4 + 96) / 6 = (2 ± 10) / 6，所以 x = 2 或 x = −4/3。Q3 得 (x + 3)² − 7 = 0，所以 x = −3 ± √7。Q4 需要 k² − 36 = 0，所以 k = 6 或 k = −6。
12. Quick Revision Summary | 复习速查总结
Use this checklist the night before the exam. If you can complete every line from memory, you are ready for any quadratic question the paper can offer.
考前一夜可以用这份清单自查。如果每一行都能凭记忆写出来，那么试卷上任何二次方程题目你都能应对。
- Standard form: ax² + bx + c = 0 with a ≠ 0.
- 标准形式：ax² + bx + c = 0，且 a ≠ 0。
- Factorisation: find two numbers with the right product and sum, then use the null factor law.
- 因式分解：找两个数使乘积与和都正确，再用零因子法则。
- mpleting the square: x² + bx = (x + b/2)² − b²/4, giving vertex (−p, q) for a(x + p)² + q.
- 配方法：x² + bx = (x + b/2)² − b²/4，a(x + p)² + q 的顶点为 (−p, q)。
- Quadratic formula: x = (−b ± √b² − 4ac) / 2a.
- 求根公式：x = (−b ± √b² − 4ac) / 2a。
- Discriminant: Δ > 0 two roots, Δ = 0 one repeated root, Δ < 0 no real roots.
- 判别式：Δ > 0 两个根，Δ = 0 一个重根，Δ < 0 无实根。
- Always check the answer by substitution and reject impossible solutions in context.
- 务必代回检验，并在实际情境中舍去不可能的解。
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