📚 Mastering Quadratic Equations | 掌握二次方程
Quadratic equations appear in almost every IGCSE Mathematics examination, whether tested directly as an algebra question or hidden inside a graph, mensuration or problem-solving context. To achieve a top grade, you must master factorisation, the quadratic formula and completing the square, and know exactly when to use each method.
二次方程几乎出现在每一份IGCSE数学试卷中，无论是直接以代数题的形式考查，还是隐藏在图像、测量或应用题情境里。要拿到高分，你必须熟练掌握因式分解法、求根公式和配方法，并清楚知道在什么情况下使用哪一种方法。
1. What Is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is any equation that can be rearranged into the standard form shown below, where a, b and c are constants and a is not equal to zero. The highest power of the unknown x is always 2.
二次方程是任何可以整理成下列标准形式的方程，其中 a、b、c 为常数，且 a 不等于零。未知数 x 的最高次数始终是 2。
ax² + bx + c = 0, a ≠ 0
0 are not. A quadratic equation can have two real roots, one repeated root or no real roots, depending on the value of the discriminant.
例如，3x² − 5x + 2 = 0 是二次方程，而 4x + 1 = 0 和 x³ − 2x = 0 都不是。二次方程可以有两个不同的实根、一个重根或没有实根，这取决于判别式的值。
2. The Factorisation Method | 因式分解法
Factorisation is the fastest method when the expression factorises easily. The key principle is the zero product property: if the product of two factors is zero, then at least one of the factors must be zero.
当表达式容易分解时，因式分解法是最快捷的方法。关键原理是零积性质：如果两个因式的乘积为零，那么至少有一个因式为零。
can be written as (x + 2)(x + 3) = 0. Setting each factor to zero gives x = −2 or x = −3.
考虑 x² + 5x + 6 = 0。我们要找两个数，它们相乘得 6，相加得 5；这两个数是 2 和 3。因此方程可以写成 (x + 2)(x + 3) = 0。令每个因式为零，得到 x = −2 或 x = −3。
(x + 2)(x + 3) = 0 ⇒ x = −2 or x = −3
ve but c is positive, both numbers are negative, for example x² − 5x + 6 = (x − 2)(x − 3).
务必在脑中展开因式进行检验。展开 (x + 2)(x + 3) 得到 x² + 5x + 6，验证了答案。如果 b 为负数而 c 为正数，则两个数都是负数，例如 x² − 5x + 6 = (x − 2)(x − 3)。
3. The Quadratic Formula | 求根公式
The quadratic formula works for every quadratic equation, including those that cannot be factorised. You are expected to know and apply it without a calculator in most IGCSE papers.
求根公式适用于所有二次方程，包括那些无法因式分解的方程。在大多数IGCSE试卷中，你需要不借助计算器记住并运用该公式。
x = (−b ± √b² − 4ac) ÷ (2a)
= −4 and c = −3. Substituting these values gives x = (4 ± √(16 − 4 × 2 × (−3))) ÷ 4 = (4 ± √40) ÷ 4, which simplifies to approximately x = 2.58 or x = −0.58.
用求根公式解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。代入这些值得到 x = (4 ± √(16 − 4 × 2 × (−3))) ÷ 4 = (4 ± √40) ÷ 4，化简后约为 x = 2.58 或 x = −0.58。
Be extremely careful with negative signs inside the formula. Notice that −b becomes positive when b is negative, and that subtracting 4ac requires correct handling of a negative c. Write out every substitution step to avoid sign errors.
在公式中要极其小心负号。注意当 b 为负数时，−b 会变成正数；而在减去 4ac 时，要正确处理负的 c。写出每一个代入步骤，以避免符号错误。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form p(x + q)² + r. This form reveals the turning point and is also the starting point for solving equations when factorisation fails.
配方法把二次方程改写成 p(x + q)² + r 的形式。这种形式直接显示顶点坐标，也是因式分解失败时解方程的出发点。
nt gives x² + 6x + 1 = (x + 3)² − 8. Setting this to zero gives (x + 3)² = 8, so x + 3 = ±√8 and x = −3 ± 2√2.
对于 x² + 6x + 1 = 0，取 6 的一半得 3，然后把 x² + 6x 写成 (x + 3)² − 9。加上原来的常数项，得到 x² + 6x + 1 = (x + 3)² − 8。令其为零，得 (x + 3)² = 8，因此 x + 3 = ±√8，x = −3 ± 2√2。
x² + 6x + 1 = 0 ⇒ (x + 3)² − 8 = 0 ⇒ x = −3 ± 2√2
− 4] + 5 = 2(x + 2)² − 3. This process is essential for finding the vertex of a parabola.
当 x² 的系数不为 1 时，先把该系数提出。例如，2x² + 8x + 5 可以写成 2(x² + 4x) + 5，再写成 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3。这个过程对于求抛物线的顶点至关重要。
5. The Discriminant | 判别式
The discriminant is the value of b² − 4ac, the part of the quadratic formula under the square root. It tells you the number and type of roots without solving the full equation.
判别式就是 b² − 4ac 的值，即求根公式中根号下的部分。它可以在不解完整方程的情况下，告诉你根的个数和类型。
| Value of b² − 4ac | Number of real roots | Graph interpretation |
| Positive (> 0) | Two distinct real roots | Curve cuts the x-axis twice |
| Zero (= 0) | One repeated root | Curve touches the x-axis once |
| Negative (< 0) | No real roots | Curve never meets the x-axis |
ated root. Set the discriminant to zero: k² − 4 × 1 × 9 = k² − 36 = 0, so k = 6 or k = −6.
对于方程 x² + kx + 9 = 0，求使方程有重根时 k 的值。令判别式为零：k² − 4 × 1 × 9 = k² − 36 = 0，所以 k = 6 或 k = −6。
6. Harder Factorisation When a ≠ 1 | 首项系数不为1的因式分解
3 = 6. Find two numbers whose product is 6 and whose sum is 7; those numbers are 1 and 6.
当 x² 的系数不为 1 时，因式分解需要系统的方法。考虑 2x² + 7x + 3。将 a 和 c 相乘：2 × 3 = 6。找两个数，其乘积为 6，和为 7；这两个数是 1 和 6。
Finally factor out the common bracket to obtain (2x + 1)(x + 3) = 0, so x = −½ or x = −3.
用这两个数拆开中间项：2x² + 1x + 6x + 3。分组为 (2x² + x) + (6x + 3)，再对每组分别因式分解：x(2x + 1) + 3(2x + 1)。最后提出公因式得到 (2x + 1)(x + 3) = 0，所以 x = −½ 或 x = −3。
2x² + 7x + 3 = (2x + 1)(x + 3)
(x + 3) expands to 2x² + 6x + x + 3 = 2x² + 7x + 3, which matches the original expression.
这种方法通常称为 ac 法。务必展开两个外项进行验证：(2x + 1)(x + 3) 展开为 2x² + 6x + x + 3 = 2x² + 7x + 3，与原表达式一致。
7. Graphs of Quadratic Functions | 二次函数图像
and has a maximum point. The x-coordinate of the turning point is given by x = −b ÷ (2a).
二次函数的图像是抛物线。若 a > 0，抛物线开口向上，有最小值点；若 a < 0，抛物线开口向下，有最大值点。顶点（转向点）的 x 坐标为 x = −b ÷ (2a)。
oots at x = 2, and substituting x = 2 gives y = 4 − 8 + 3 = −1, the minimum point (2, −1).
要画出 y = x² − 4x + 3 的草图，先令 x = 0 求 y 截距，得 y = 3。然后通过因式分解求根：(x − 1)(x − 3) = 0，因此曲线与 x 轴交于 x = 1 和 x = 3 两点。对称轴位于两根中点 x = 2 处，代入 x = 2 得 y = 4 − 8 + 3 = −1，即最小值点 (2, −1)。
y = x² − 4x + 3 = (x − 1)(x − 3), vertex at (2, −1)
If the equation has no real roots, the graph does not cross the x-axis; to sketch it, compute the turning point by completing the square and find the y-intercept. Examiners award method marks for the axis of symmetry and correct vertex even when the overall shape is imperfect.
如果方程没有实根，图像与 x 轴不相交；要画草图，可以通过配方法计算顶点并求出 y 截距。即便整体形状不够完美，考官也会根据对称轴和顶点是否正确来给方法分。
8. Quadratic Word Problems | 二次方程应用题
and width (x − 2) cm, and its area is 30 cm². This gives the equation (x + 4)(x − 2) = 30.
应用题需要你把现实情境转化为二次方程。一个矩形的长为 (x + 4) cm，宽为 (x − 2) cm，面积为 30 cm²。由此得到方程 (x + 4)(x − 2) = 30。
² + 2x − 38 = 0. Using the quadratic formula with a = 1, b = 2 and c = −38 gives x = (−2 ± √4 + 152 ÷ 2, so x ≈ 5.25 or x ≈ −7.25. Since a length cannot be negative, only x ≈ 5.25 is valid.
展开得 x² + 2x − 8 = 30，即 x² + 2x − 38 = 0。用求根公式，a = 1，b = 2，c = −38，得到 x = (−2 ± √4 + 152) ÷ 2 = (−2 ± √156) ÷ 2，因此 x ≈ 5.25 或 x ≈ −7.25。由于长度不能为负，只有 x ≈ 5.25 有效。
Always check whether a negative root makes sense in the context. In projectile motion problems, the time t is usually positive; in geometry problems, lengths and areas are positive. Reject invalid roots clearly in your final answer.
始终检查负根在具体情境中是否有意义。在抛体运动问题中，时间 t 通常为正；在几何问题中，长度和面积为正。在最终答案中明确舍弃无效根。
9. Common Mistakes | 常见错误
+ bx + c = 0 before factorising or applying the formula. Move all terms to one side first.
忘记先把方程整理成 ax² + bx + c = 0 的形式，就直接进行因式分解或代入公式。应先把所有项移到一边。
example, x² − 3x = 0 must be solved by factorising as x(x − 3) = 0, not by dividing by x.
当 x = 0 是其中一个根时，两边同时除以 x，导致丢根。例如，x² − 3x = 0 必须通过因式分解 x(x − 3) = 0 来求解，而不能两边除以 x。
Sign errors when substituting into the quadratic formula, especially with negative values of b and c. Write the substitution line fully before simplifying.
代入求根公式时出现符号错误，尤其是 b、c 为负数的情况。在化简前，先完整写出代入行。
hat (x + 3)² = x² + 9. The correct expansion is x² + 6x + 9. Remember the middle term 2ab.
误以为 (x + 3)² = x² + 9。正确的展开是 x² + 6x + 9。不要忘记中间项 2ab。
Stating only one root when the equation has two. Every quadratic equation has two roots, counting repeated roots, in the set of complex numbers.
只写一个根，而方程有两个根。在复数范围内，每个二次方程都有两个根（重根按两个计）。
10. Exam Tips | 考试技巧
Read the question carefully to decide which method is expected. If the question says "factorise", you must show factorisation and not use the formula. If it says "give your answers correct to 2 decimal places", you should use the quadratic formula or completing the square, as factorisation may not give the required accuracy.
仔细读题，判断题目期望使用哪种方法。如果题目要求"因式分解"，你必须展示因式分解过程，而不能用求根公式。如果题目说"答案精确到小数点后两位"，你应使用求根公式或配方法，因为因式分解可能无法给出所需精度。
When solving by the quadratic formula, always work out b² and 4ac separately before combining them. This reduces arithmetic errors. For graphical questions, label the roots, the y-intercept and the turning point clearly, and use a smooth curve rather than straight line segments.
使用求根公式时，先分别算出 b² 和 4ac，再合并。这可以减少计算错误。对于图像题，清晰标出根、y 截距和顶点，用平滑曲线而不是直线段连接各点。
Finally, check your answers by substituting them back into the original equation. A quick check takes ten seconds and can earn you full marks instead of method marks only.
最后，把答案代回原方程进行检验。快速检验只需十秒钟，却能让你拿到满分，而不只是方法分。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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