📚 Solving Quadratic Equations | 解二次方程
Quadratic equations are a fundamental part of the IGCSE Edexcel Mathematics syllabus. This guide covers the essential methods for solving them, including factorisation, the quadratic formula, completing the square, and using graphs. Each method is explained step by step with worked examples to help you master this important topic.
二次方程是 IGCSE Edexcel 数学教学大纲中的基础内容。本指南涵盖解二次方程的核心方法：因式分解法、二次公式法、配方法以及图像解法。每种方法均配有逐步讲解和例题，帮助你掌握这一重要考点。
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 form ax² + bx + c = 0 f the variable x is 2. Examples include x² - 5x + 6 = 0, 2x² + 3x - 1 = 0, and x² - 4 = 0.
二次方程是指可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数，且 a ≠ 0。变量 x 的最高次数是 2。例如：x² - 5x + 6 = 0、2x² + 3x - 1 = 0、x² - 4 = 0。
ax² + bx + c = 0
It is crucial to rearrange the equation into this form before attempting to solve it. If the equation is given differently, expand and simplify first.
在解方程之前，务必将其整理成标准形式。如果方程以其他形式给出，先展开并化简。
2. Method 1: Factorisation | 方法一：因式分解法
o binomials, then set each factor equal to zero because if a × b = 0, then a = 0 or b = 0.
当二次式具有简单的整数因子时，因式分解是最快的方法。其原理是将二次式写成两个二项式的乘积，然后令每个因式等于零，因为如果 a × b = 0，则 a = 0 或 b = 0。
Example 1: Solve x² - 5x + 6 = 0
Find two numbers that multiply to give 6 and add to give -5. These are -2 and -3.
找到两个数，相乘得 6，相加得 -5。这两个数是 -2 和 -3。
Write the factorised form: (x - 2)(x - 3) = 0
写出因式分解形式：(x - 2)(x - 3) = 0
Set each factor to zero: x - 2 = 0 or x - 3 = 0
令每个因式等于零：x - 2 = 0 或 x - 3 = 0
Solve: x = 2 or x = 3
解得：x = 2 或 x = 3
4 that add to -5 (they are -1 and -4), then split the middle term and factor by grouping.
对于 a ≠ 1 的二次式，如 2x² - 5x + 2 = 0，需谨慎使用乘积-求和法。将 a 与 c 相乘（2 × 2 = 4），找到 4 的因子且相加为 -5（它们是 -1 和 -4），然后拆分中间项并分组因式分解。
3. Method 2: The Quadratic Formula | 方法二：二次公式法
mpleting the square and gives solutions directly. For ax² + bx + c = 0, the solutions are:
二次公式适用于任何二次方程，包括无法因式分解的方程。该公式由配方法推导而来，可直接给出解。对于 ax² + bx + c = 0，解为：
x = (-b ± √b² - 4ac) / 2a
t compute b² - 4ac, then take the square root, then apply the ±, and finally divide by 2a.
注意运算顺序。先计算 b² - 4ac，然后开平方，再处理 ±，最后除以 2a。
Example 2: Solve 2x² + 3x - 1 = 0
Here a = 2, b = 3, c = -1. Substitute into the formula:
这里 a = 2，b = 3，c = -1。代入公式：
x = (-3 ± √(3² - 4×2×(-1))) / (2×2) = (-3 ± √17) / 4
Thus the two solutions are x ≈ 0.28 and x ≈ -1.78 (to 2 decimal places).
因此两个解为 x ≈ 0.28 和 x ≈ -1.78（精确到两位小数）。
4. Method 3: Completing the Square | 方法三：配方法
Completing the square rewrites a quadratic in the form (x + p)² + q. This method is particularly useful for solving equations and for finding the vertex of a parabola. It also helps derive the quadratic formula.
配方法将二次式改写为 (x + p)² + q 的形式。此方法特别适用于解方程以及求抛物线的顶点，也有助于推导二次公式。
Example 3: Solve x² + 6x + 1 = 0 by completing the square.
Move the constant term to the other side: x² + 6x = -1
将常数项移到另一边：x² + 6x = -1
he coefficient of x (6/2 = 3) and square it (3² = 9). Add 9 to both sides: x² + 6x + 9 = 8
取 x 系数的一半（6/2 = 3）并平方（3² = 9）。两边同时加 9：x² + 6x + 9 = 8
Write the left side as a perfect square: (x + 3)² = 8
将左边写成完全平方：(x + 3)² = 8
Take the square root: x + 3 = ±√8
两边开平方：x + 3 = ±√8
Solve: x = -3 ± 2√2
解得：x = -3 ± 2√2
2x² - 4x + 1 = 0 becomes 2(x² - 2x) + 1 = 0, then complete the square inside the bracket.
当 a 不是 1 时，先提取 a。例如，2x² - 4x + 1 = 0 变为 2(x² - 2x) + 1 = 0，然后在括号内配方。
5. Method 4: Using Graphs | 方法四：图像法
e x-axis. If the graph touches the x-axis at one point, there is one repeated root. If it does not cross, there are no real solutions.
二次方程 ax² + bx + c = 0 的解，是图像 y = ax² + bx + c 与 x 轴交点的 x 坐标。若图像与 x 轴相切于一点，则有一个重根；若不相交，则无实数解。
You may be asked to estimate roots from a given graph, or to draw the graph yourself by plotting points. This method is often used when exact roots are not required.
题目可能要求你从给定图像估算根，或自己描点作图。当不需要精确根时，常用此法。
6. The Discriminant | 判别式
The expression under the square root in the quadratic formula, b² - 4ac, is called the discriminant. It tells us the nature of the roots without solving the equation.
二次公式中根号内的表达式 b² - 4ac 称为判别式。它无需解方程即可判断根的性质。
| Discriminant | Nature of Roots | 根的性质 |
| b² - 4ac > 0 | Two distinct real roots | 两个不相等的实数根 |
| b² - 4ac = 0 | One repeated real root | 一个重根（两个相等实数根） |
| b² - 4ac < 0 | No real roots | 无实数根 |
If the discriminant is a perfect square, the quadratic can be factorised. Every quadratic curve has a vertical line of symmetry, and the discriminant affects how the curve sits relative to the x-axis.
若判别式是一个完全平方数，则二次式可以因式分解。每条二次曲线都有竖直对称轴，判别式决定曲线与 x 轴的相对位置。
7. Solving Equations by Substitution | 换元法解方程
0, which factorises to (y - 2)(y - 3) = 0, so y = 2 or y = 3. Since y = x², we have x = ±√2 or x = ±√3.
有时二次形式出现在更复杂的方程中。例如，方程 x⁴ - 5x² + 6 = 0 是关于 x² 的二次方程。令 y = x²，则方程变为 y² - 5y + 6 = 0，因式分解得 (y - 2)(y - 3) = 0，所以 y = 2 或 y = 3。因为 y = x²，所以 x = ±√2 或 ±√3。
ations like (x + 1)² - 3(x + 1) - 4 = 0. Let u = x + 1, solve for u, then substitute back.
换元法同样适用于 (x + 1)² - 3(x + 1) - 4 = 0 这类方程。令 u = x + 1，解出 u，再代回求 x。
8. Forming Quadratic Equations | 列二次方程
ving gives x = 7 or x = -10. Since width cannot be negative, x = 7 cm, and length = 10 cm.
许多应用题需要你根据已知信息列出二次方程。例如：一个长方形的长比宽多 3 cm，面积为 70 cm²。设宽为 x，则长为 x + 3。面积 = x(x + 3) = 70，所以 x² + 3x - 70 = 0。解得 x = 7 或 x = -10。因为宽不可能为负，所以 x = 7 cm，长为 10 cm。
Always check whether solutions make sense in the context of the problem. Reject any negative or impossible values.
务必检验解是否符合实际情境，舍去负数或不可能的值。
9. Common Mistakes and Tips | 常见错误与提示
Forgetting to set the equation to zero before factorising.
因式分解前忘记将方程化为零。
Sign errors when substituting negative numbers into the quadratic formula. Always write the formula carefully and use brackets.
将负数代入二次公式时出现符号错误。务必仔细书写公式并使用括号。
Losing solutions when taking square roots: remember that √x² = ±x if you are solving an equation, not just the positive root.
开平方时丢失解：解方程时 √x² = ±x，不能只取正根。
Not checking roots by substituting back into the original equation.
不将解代回原方程检验。
In the exam, if a question says "give your answers correct to 2 decimal places", use the quadratic formula. If it says "solve by factorisation", show the factorised form explicitly.
考试中，若题目要求“答案精确到两位小数”，应使用二次公式。若要求“用因式分解法解”，需明确写出因式分解形式。
10. Practice Questions | 练习题目
Try these questions on your own, then check your answers below.
请独立尝试以下题目，然后核对答案。
Solve x² - 7x + 12 = 0 (by factorisation)
用因式分解法解 x² - 7x + 12 = 0
Solve 2x² - 3x - 5 = 0 (using the quadratic formula)
用二次公式解 2x² - 3x - 5 = 0
Solve x² + 8x + 9 = 0 by completing the square.
用配方法解 x² + 8x + 9 = 0。
Find the value of k for which x² + kx + 16 = 0 has exactly one repeated root.
求 k 的值，使 x² + kx + 16 = 0 有唯一重根。
Answers:
1. x = 3 or x = 4
1. x = 3 或 x = 4
2. x = 2.5 or x = -1 (using formula: x = (3 ± 7)/4)
2. x = 2.5 或 x = -1（公式给出 x = (3 ± 7)/4）
3. (x + 4)² - 7 = 0, so x = -4 ± √7
3. (x + 4)² - 7 = 0，所以 x = -4 ± √7
4. Discriminant = k² - 4×1×16 = 0, so k = ±8
4. 判别式 = k² - 4×1×16 = 0，所以 k = ±8
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