📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are among the most important topics in IGCSE Mathematics. They appear in algebra, graphing, coordinate geometry and even in the foundations of calculus. Mastering the different methods of solving quadratics will help you secure marks in both Paper 2 and Paper 4 of the IGCSE Mathematics examination.
二次方程是IGCSE数学中最重要的主题之一。它们出现在代数、绘图、坐标几何甚至微积分的基础内容中。掌握解二次方程的不同方法，能帮助你在IGCSE数学Paper 2和Paper 4中都稳定拿分。
1. Standard Form of a Quadratic Equation | 二次方程的标准形式
A quadratic equation is any equation that can be written in the standard form:
ax² + bx + c = 0, where a ≠ 0
is the unknown variable. The highest power of x is 2, which is why the equation is called 'quadratic', from the Latin word quadratus meaning 'square' The coefficient a must not be zero; if a = 0, the equation becomes linear, not quadratic.
二次方程是可以写成标准形式的任何方程：
ax² + bx + c = 0，其中 a ≠ 0
这里 a、b、c 是常数，x 是未知数。x 的最高次数是2，因此方程被称为“二次”（quadratic，源自拉丁语 quadratus，意为“平方”）。由于最高次数是2，二次方程最多有两个不同的解。系数 a 不能为零；若 a = 0，方程就变成了一次方程，而不是二次方程。
attempting to solve, always check that the equation is exactly in the standard form. Sometimes you will need to expand brackets, collect like terms or move terms from one side to the other first.
例如，在方程 3x² − 2x + 5 = 0 中，a = 3，b = −2，c = 5。在尝试求解之前，务必检查方程是否恰好是标准形式。有时你需要先展开括号、合并同类项或把项移到等号的另一侧。
2. Solving by Factorisation | 因式分解法求解
Factorisation is often the quickest method when the quadratic has simple integer factors. The idea is to write the quadratic expression as a product of two linear brackets, then set each bracket equal to zero.
因式分解法通常是最快捷的解法，前提是二次表达式能分解成简单的整数因式。其核心思想是把二次表达式写成两个一次括号的乘积，然后令每个括号等于零。
a = 1, look for two numbers that multiply to give c and add to give b. For example, solve:
x² − 5x + 6 = 0
We need two numbers whose product is 6 and whose sum is −5. Those numbers are −2 and −3. Therefore:
(x − 2)(x − 3) = 0
x = 2 or x = 3. Always verify by substituting each value back into the original equation.
当 a = 1 时，找两个数，它们相乘等于 c，相加等于 b。例如，解：
x² − 5x + 6 = 0
我们需要两个数，它们乘积为6，和为−5。这两个数是 −2 和 −3。因此：
(x − 2)(x − 3) = 0
令每个括号等于零，得 x − 2 = 0 或 x − 3 = 0，所以 x = 2 或 x = 3。务必把每个解代回原方程验证。
+ 3 = 0. The factor pairs must produce 2x² from the first terms and 3 from the constants:
(2x + 3)(x + 1) = 0
am, if you cannot spot factors quickly, switch to another method rather than wasting time.
当 a ≠ 1 时，因式分解稍难一些。考虑 2x² + 5x + 3 = 0。因式组合必须使首项得到 2x²，常数项得到 3：
(2x + 3)(x + 1) = 0
展开括号得 2x² + 2x + 3x + 3 = 2x² + 5x + 3，说明分解正确。因此 x = −3/2 或 x = −1。在考试中，如果你无法快速看出因式，就不要浪费时间，应立即改用其他方法。
Check the sign of c: if c is positive, both numbers have the same sign; if c is negative, one is positive and one is negative.
检查 c 的符号：若 c 为正，两个数同号；若 c 为负，则一正一负。
Always expand your factorised answer to confirm it matches the original expression.
始终展开你的分解结果，以确认与原表达式一致。
3. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q. This method is useful for finding the turning point of a parabola and for solving equations when factorisation fails.
配方法将二次式重写为 (x + p)² + q 的形式。这种方法适用于求抛物线的顶点，也适用于因式分解行不通时的求解。
For a quadratic of the form x² + bx + c, the rule is:
(x + b/2)² − (b/2)² + c
Solve x² + 6x + 5 = 0 by completing the square. Here b = 6, so b/2 = 3:
(x + 3)² − 9 + 5 = 0
(x + 3)² = 4
member that the square root of a positive number has both a positive and a negative value.
对于 x² + bx + c 形式的二次式，规则是：
(x + b/2)² − (b/2)² + c
用配方法解 x² + 6x + 5 = 0。这里 b = 6，所以 b/2 = 3：
(x + 3)² − 9 + 5 = 0
(x + 3)² = 4
两边开平方得 x + 3 = ±2，所以 x = −1 或 x = −5。记住正数开平方既有正根也有负根。
− 8x + 3 = 0 becomes 2(x² − 4x) + 3 = 0, and then complete the square inside the brackets.
如果 x² 的系数不是1，先把它提出来。例如，2x² − 8x + 3 = 0 可写成 2(x² − 4x) + 3 = 0，然后在括号内配方。
4. The Quadratic Formula | 求根公式
even when factorisation is not possible. For ax² + bx + c = 0, the solutions are given by:
x = (−b ± √b² − 4ac) / (2a)
This formula is on the IGCSE formula sheet, but you must be confident in substituting values correctly and simplifying the result, including surds.
求根公式对所有二次方程都适用，即使因式分解不可行。对于 ax² + bx + c = 0，解为：
x = (−b ± √b² − 4ac) / (2a)
这个公式印在IGCSE公式表上，但你必须熟练地正确代入数值并化简结果，包括根式（无理数）的处理。
ample: solve 2x² − 4x − 6 = 0. Here a = 2, b = −4 and c = −6. First find the discriminant:
b² − 4ac = (−4)² − 4 × 2 × (−6) = 16 + 48 = 64
Now substitute into the formula:
x = (4 ± √64) / 4 = (4 ± 8) / 4
case the same quadratic could also have been solved by factorisation: 2(x − 3)(x + 1) = 0.
例题：解 2x² − 4x − 6 = 0。这里 a = 2，b = −4，c = −6。先求判别式：
b² − 4ac = (−4)² − 4 × 2 × (−6) = 16 + 48 = 64
再代入公式：
x = (4 ± √64) / 4 = (4 ± 8) / 4
得 x = (4 + 8)/4 = 3 或 x = (4 − 8)/4 = −1。注意本例也可以用因式分解求解：2(x − 3)(x + 1) = 0。
he discriminant is not a perfect square, leave your answer as a surd, for example x = (3 ± √5)/2.
当判别式不是完全平方数时，答案保留根式形式，例如 x = (3 ± √5)/2。
Use a calculator to check your final answers as decimals, but show the exact surd form in your working.
可以用计算器把最终答案化为小数进行检验，但在解题过程中要展示精确的根式形式。
5. The Discriminant | 判别式
− 4ac. It tells us how many real roots a quadratic equation has without solving it fully.
判别式是求根公式中被开平方的部分，通常记作 Δ = b² − 4ac。它告诉我们二次方程有多少个实数根，而无需完整求解。
| Value of Δ 判别式 Δ 的值 |
Number of real roots 实数根的个数 |
Graph relationship 图像关系 |
|---|---|---|
| Δ > 0 | Two distinct real roots 两个不同的实数根 |
Parabola crosses the x-axis at two points 抛物线与x轴有两个交点 |
| Δ = 0 | One repeated root 一个重根 |
Parabola touches the x-axis at one point 抛物线与x轴相切于一点 |
| Δ < 0 | No real roots 没有实数根 |
Parabola never touches the x-axis 抛物线与x轴没有交点 |
If a question asks you to 'find the range of values of k for which the equation has two distinct real roots' example, x² + kx + 4 = 0 has two distinct real roots when k² − 16 > 0, so k > 4 or k < −4.
如果题目要求“求使方程有两个不同实数根的 k 的取值范围”，就令 Δ > 0 并解对应不等式。例如，x² + kx + 4 = 0 有两个不同实数根的条件是 k² − 16 > 0，即 k > 4 或 k < −4。
6. Graphs of Quadratic Functions | 二次函数图像
opens downward with a maximum point. The x-coordinate of the turning point is x = −b/(2a).
y = ax² + bx + c 的图像是抛物线。若 a > 0，抛物线开口向上，有最低点；若 a < 0，抛物线开口向下，有最高点。顶点的 x 坐标为 x = −b/(2a)。
the turning point is at (−p, q). For example, y = (x − 3)² − 4 has its minimum at (3, −4).
二次方程的解对应其图像与 x 轴的交点。y 轴截距总是常数项 c。配方后 y = (x + p)² + q，顶点坐标为 (−p, q)。例如，y = (x − 3)² − 4 的最低点为 (3, −4)。
To sketch a quadratic graph, it is helpful to follow these steps:
State the y-intercept: write down c.
写出 y 轴截距：即 c 的值。
Solve the equation ax² + bx + c = 0 to find the x-intercepts, if they exist.
解方程 ax² + bx + c = 0 求 x 轴截距（如果存在）。
Use completing the square or x = −b/(2a) to find the turning point.
用配方法或 x = −b/(2a) 求顶点坐标。
Join these points with a smooth curve, paying attention to whether the parabola opens up or down.
用平滑曲线连接这些点，注意抛物线开口向上还是向下。
7. Word Problems with Quadratic Equations | 二次方程应用题
Many IGCSE problems require you to form a quadratic equation from a real-life situation and then solve it. A common type involves area and dimensions.
许多IGCSE题目要求你从实际情境中列出二次方程，再求解。常见类型涉及面积和尺寸。
Example: A rectangle has a length 2 cm longer than its width, and its area is 15 cm². Let the width be x cm. Then the length is x + 2 cm, so:
x(x + 2) = 15
x² + 2x − 15 = 0
(x + 5)(x − 3) = 0
ince a length cannot be negative, reject x = −5. The width is 3 cm and the length is 5 cm.
例题：一个长方形的长比宽长2 cm，面积为 15 cm²。设宽为 x cm，则长为 x + 2 cm，于是：
x(x + 2) = 15
x² + 2x − 15 = 0
(x + 5)(x − 3) = 0
所以 x = −5 或 x = 3。由于长度不能为负数，舍去 x = −5。宽为 3 cm，长为 5 cm。
the ground is found by setting h = 0, giving t(12 − 5t) = 0, so t = 0 or t = 2.4 seconds.
抛体问题也很常见。若一个球被抛出后，t 秒时的高度 h 米满足 h = 12t − 5t²，求球落地的时间可令 h = 0，得 t(12 − 5t) = 0，所以 t = 0 或 t = 2.4 秒。
When solving word problems, define your variable clearly, write down the equation, solve it, and then decide which solution is valid in the context. Always include units in your final answer.
解应用题时，要清晰定义未知数，写出方程，求解，然后判断哪个解在情境中成立。最终答案必须带单位。
8. Quadratic Inequalities | 二次不等式
Quadratic inequalities are an extension topic that appears in the extended IGCSE syllabus. To solve them, first solve the corresponding quadratic equation to find the critical values, then test intervals on a number line.
二次不等式是IGCSE扩展考纲中的进阶内容。求解时，先解对应的二次方程求出临界值，然后在数轴上测试各个区间。
bola y = x² − 5x + 6 opens upward, so it is below the x-axis between the roots. Therefore:
2 < x < 3
For x² − 5x + 6 > 0, the parabola is above the x-axis outside the roots, so the solution is:
x < 2 or x > 3
解 x² − 5x + 6 < 0。方程 x² − 5x + 6 = 0 的根为 x = 2 和 x = 3。抛物线 y = x² − 5x + 6 开口向上，因此在两根之间位于 x 轴下方。因此：
2 < x < 3
对于 x² − 5x + 6 > 0，抛物线在两根之外位于 x 轴上方，所以
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