📚 Solving Quadratic Equations | 解二次方程
Quadratic equations appear frequently in the Edexcel IGCSE Mathematics syllabus. You need to be able to solve them by factorisation, by using the quadratic formula, and by completing the square. This guide explains each method step by step, including common pitfalls and graph-based questions.
二次方程在 Edexcel IGCSE 数学考纲中频繁出现。你需要掌握因式分解法、二次公式法和配方法三种解法。本指南将逐步讲解每种方法，包括常见错误和与图像相关的题型。
1. Standard Form of a Quadratic Equation | 二次方程的标准形式
A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of x is 2. In the IGCSE exam, you must first rearrange the equation into this standard form before solving.
二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。x 的最高次数为 2。在 IGCSE 考试中，你必须先将方程整理成这种标准形式，然后才能求解。
ax² + bx + c = 0
For example, 2x² + 5x - 3 = 0 is already in standard form. The equation 3x² + x = 2 must be rewritten as 3x² + x - 2 = 0.
例如，2x² + 5x - 3 = 0 已经是标准形式。而方程 3x² + x = 2 必须改写为 3x² + x - 2 = 0。
2. Solving by Factorisation | 因式分解法求解
Factorisation is often the quickest method when the quadratic has simple integer roots. You look for two numbers that multiply to give a × c and add to give b. Once the quadratic is written as a product of two brackets, set each bracket equal to zero.
当二次方程具有简单的整数根时，因式分解通常是最快的方法。你需要找到两个数，它们相乘等于 a × c，相加等于 b。将二次式写成两个括号的乘积后，令每个括号等于零即可。
Example: Solve x² - 5x + 6 = 0. Two numbers that multiply to 6 and add to -5 are -2 and -3. So (x - 2)(x - 3) = 0, giving x = 2 or x = 3.
例：解 x² - 5x + 6 = 0。相乘等于 6 且相加等于 -5 的两个数是 -2 和 -3。因此 (x - 2)(x - 3) = 0，得到 x = 2 或 x = 3。
For a quadratic where a ≠ 1, such as 2x² + 7x + 3 = 0, multiply a and c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7: they are 1 and 6. Split the middle term: 2x² + x + 6x + 3 = 0, then factor by grouping: x(2x + 1) + 3(2x + 1) = 0, so (x + 3)(2x + 1) = 0. Therefore x = -3 or x = -1/2.
对于 a ≠ 1 的二次方程，例如 2x² + 7x + 3 = 0，先将 a 与 c 相乘：2 × 3 = 6。找到两个数相乘等于 6 且相加等于 7：它们是 1 和 6。拆开中间项：2x² + x + 6x + 3 = 0，再分组因式分解：x(2x + 1) + 3(2x + 1) = 0，所以 (x + 3)(2x + 1) = 0。因此 x = -3 或 x = -1/2。
3. Special Cases: Difference of Two Squares | 特殊情形：平方差
When a quadratic has no x term and both a and c are perfect squares, you can use the difference of two squares. The formula is a² - b² = (a - b)(a + b).
当二次方程没有 x 项，且 a 与 c 都是完全平方数时，可以使用平方差公式。公式为 a² - b² = (a - b)(a + b)。
Example: Solve x² - 16 = 0. This becomes (x - 4)(x + 4) = 0, so x = 4 or x = -4.
例：解 x² - 16 = 0。可以写成 (x - 4)(x + 4) = 0，所以 x = 4 或 x = -4。
This method also works for expressions like 9x² - 25 = 0. Write it as (3x - 5)(3x + 5) = 0, giving x = 5/3 or x = -5/3.
这个方法也适用于像 9x² - 25 = 0 这样的表达式。写成 (3x - 5)(3x + 5) = 0，得到 x = 5/3 或 x = -5/3。
4. Completing the Square | 配方法
Completing the square rewrites x² + bx + c in the form (x + p)² + q. The key step is to take half of the coefficient of x, square it, and then adjust the constant term.
配方法将 x² + bx + c 改写为 (x + p)² + q 的形式。关键步骤是取 x 系数的一半，平方，然后调整常数项。
Example: Solve x² + 6x - 7 = 0. Half of 6 is 3, so (x + 3)² - 9 - 7 = 0, i.e. (x + 3)² - 16 = 0. Then (x + 3)² = 16, so x + 3 = ± 4, giving x = 1 or x = -7.
例：解 x² + 6x - 7 = 0。6 的一半是 3，所以 (x + 3)² - 9 - 7 = 0，即 (x + 3)² - 16 = 0。于是 (x + 3)² = 16，所以 x + 3 = ± 4，得到 x = 1 或 x = -7。
When the coefficient of x² is not 1, first factor out a. For 2x² - 4x - 3 = 0, write 2(x² - 2x) - 3 = 0. Complete the square inside the bracket: (x - 1)² - 1, so 2[(x - 1)² - 1] - 3 = 0, which simplifies to 2(x - 1)² - 5 = 0.
当 x² 的系数不是 1 时，先提取 a。对于 2x² - 4x - 3 = 0，写成 2(x² - 2x) - 3 = 0。在括号内配方：(x - 1)² - 1，所以 2[(x - 1)² - 1] - 3 = 0，化简为 2(x - 1)² - 5 = 0。
5. The Quadratic Formula | 二次公式
The quadratic formula solves any quadratic equation. It is particularly useful when factorisation is difficult. The formula is:
二次公式可以求解任何二次方程，在因式分解困难时尤其有用。公式如下：
x = (−b ± √b² − 4ac) / (2a)
You must memorise this formula for the Edexcel IGCSE exam. Substitute the values of a, b and c from ax² + bx + c = 0. Simplify carefully, especially the square root.
你必须牢记这个公式以应对 Edexcel IGCSE 考试。将 ax² + bx + c = 0 中的 a、b、c 代入，小心化简，尤其是平方根部分。
Example: Solve 2x² + 5x - 3 = 0. Here a = 2, b = 5, c = -3. Then x = (−5 ± √(25 − 4×2×(−3))) / (2×2) = (−5 ± √49)/4 = (−5 ± 7)/4. So x = 2/4 = 1/2 or x = −12/4 = −3.
例：解 2x² + 5x - 3 = 0。这里 a = 2，b = 5，c = -3。那么 x = (−5 ± √(25 − 4×2×(−3))) / (2×2) = (−5 ± √49)/4 = (−5 ± 7)/4。因此 x = 2/4 = 1/2 或 x = −12/4 = −3。
6. The Discriminant | 判别式
The part of the quadratic formula under the square root, b² - 4ac, is called the discriminant. It tells you how many real roots the quadratic equation has.
二次公式中平方根下的部分 b² - 4ac 称为判别式。它告诉我们二次方程有几个实数根。
- If b² - 4ac > 0, there are two distinct real roots. 若 b² - 4ac > 0，方程有两个不相等的实数根。
- If b² - 4ac = 0, there is exactly one repeated real root. 若 b² - 4ac = 0，方程有一个二重实数根。
- If b² - 4ac < 0, there are no real roots. 若 b² - 4ac < 0，方程没有实数根。
Example: For x² + 2x + 5 = 0, the discriminant is 2² - 4×1×5 = 4 - 20 = -16, which is negative. So there are no real roots.
例：对于 x² + 2x + 5 = 0，判别式为 2² - 4×1×5 = 4 - 20 = -16，是负数，因此没有实数根。
7. Solving Quadratic Equations by Factorising When a ≠ 1 | 当 a ≠ 1 时的因式分解
In the Edexcel IGCSE papers, you may be asked to solve quadratics where the coefficient of x² is not 1. One reliable method is the “ac” method.
在 Edexcel IGCSE 试卷中，你可能会被要求解 x² 系数不为 1 的二次方程。一种可靠的方法是“ac”法。
Example: Solve 3x² + 11x + 6 = 0. Multiply a and c: 3×6 = 18. Find two numbers that multiply to 18 and add to 11: 2 and 9. Rewrite: 3x² + 2x + 9x + 6 = 0. Group: x(3x + 2) + 3(3x + 2) = 0. Thus (x + 3)(3x + 2) = 0, so x = -3 or x = -2/3.
例：解 3x² + 11x + 6 = 0。将 a 与 c 相乘：3×6 = 18。找到两个数相乘等于 18 且相加等于 11：2 和 9。重写：3x² + 2x + 9x + 6 = 0。分组：x(3x + 2) + 3(3x + 2) = 0。因此 (x + 3)(3x + 2) = 0，所以 x = -3 或 x = -2/3。
8. Quadratic Graphs and Roots | 二次图像与根
The solutions of a quadratic equation ax² + bx + c = 0 are the x-coordinates where the graph of y = ax² + bx + c crosses the x-axis. These points are called the roots or zeros of the function.
二次方程 ax² + bx + c = 0 的解就是二次函数 y = ax² + bx + c 的图像与 x 轴交点的横坐标，这些点称为函数的根或零点。
If the discriminant is positive, the graph crosses the x-axis twice. If it is zero, the graph touches the x-axis at the vertex. If it is negative, the graph never touches the x-axis.
如果判别式为正，图像与 x 轴相交两次。如果判别式为零，图像在顶点处与 x 轴相切。如果判别式为负，图像与 x 轴没有交点。
Example: The graph of y = x² - 4x + 3 intersects the x-axis at x = 1 and x = 3, because x² - 4x + 3 = (x - 1)(x - 3).
例：函数 y = x² - 4x + 3 的图像与 x 轴相交于 x = 1 和 x = 3，因为 x² - 4x + 3 = (x - 1)(x - 3)。
9. Deriving the Quadratic Formula by Completing the Square | 通过配方推导二次公式
You are not required to derive the quadratic formula in the exam, but understanding the derivation helps you see why it works. Start with ax² + bx + c = 0. Divide by a: x² + (b/a)x + c/a = 0. Complete the square:
考试不要求你推导二次公式，但理解推导过程有助于明白其原理。从 ax² + bx + c = 0 开始。除以 a：x² + (b/a)x + c/a = 0。配方：
(x + b/(2a))² − b²/(4a²) + c/a = 0
Rearrange and simplify to obtain the quadratic formula. This derivation also shows that the discriminant b² - 4ac determines the nature of the roots.
整理并化简即可得到二次公式。这个推导也说明判别式 b² - 4ac 决定了根的性质。
10. Solving Word Problems with Quadratics | 用二次方程解应用题
Many IGCSE questions present a practical problem that leads to a quadratic equation. You should define the unknown, form the equation from the given information, solve it, and then check whether each solution makes sense in the context.
许多 IGCSE 题目会给出一个实际问题，最终导出二次方程。你需要定义未知数，根据已知条件列出方程，求解，并检查每个解在具体情境中是否合理。
Example: The area of a rectangle is 28 cm² and its length is 3 cm more than its width. Let the width be x cm. Then the length is x + 3 cm. The area gives x(x + 3) = 28, so x² + 3x - 28 = 0. Factorise: (x + 7)(x - 4) = 0, so x = -7 or x = 4. Since width cannot be negative, the width is 4 cm and the length is 7 cm.
例：一个矩形的面积是 28 cm²，它的长比宽多 3 cm。设宽为 x cm，则长为 x + 3 cm。面积关系得 x(x + 3) = 28，即 x² + 3x - 28 = 0。因式分解：(x + 7)(x - 4) = 0，所以 x = -7 或 x = 4。由于宽不能为负数，所以宽为 4 cm，长为 7 cm。
11. Common Mistakes to Avoid | 常见错误避免
Students often make the same mistakes when solving quadratics. Here are the most important ones to avoid:
学生在解二次方程时经常犯同样的错误。以下是需要避免的最重要的几点：
- Forgetting to set the equation to zero before factorising. 忘记在因式分解前将方程化为零。
- Mixing up signs when using the quadratic formula, especially with negative b or c. 使用二次公式时弄错符号，尤其是 b 或 c 为负数时。
- Dropping the ± sign when taking the square root. 求平方根时遗漏 ± 号。
- Misinterpreting "no real roots" as "no solutions at all" – always state the discriminant is negative. 将“无实数根”误认为“完全没有解”——要说明判别式为负。
- Forgetting to check whether a negative answer is valid in a word problem. 在应用题中忘记检查负解是否合理。
Always test your solutions by substituting them back into the original equation.
解题后务必将解代回原方程进行检验。
12. Practice Questions | 练习题目
Try these questions to test your understanding. They are similar to Edexcel IGCSE questions.
尝试以下题目来测试你的理解。这些题目与 Edexcel IGCSE 题型相似。
1. Solve x² - 7x + 12 = 0.
2. Solve 2x² + 5x - 12 = 0 using factorisation.
3. Solve 3x² - 4x - 5 = 0 using the quadratic formula. Give your answers to 2 decimal places.
4. Find the value of the discriminant for 4x² + 4x + 1 = 0. How many real roots does the equation have?
5. The product of two consecutive positive integers is 240. Find the integers.
Answers: 1. x = 3 or x = 4. 2. (2x - 3)(x + 4) = 0, so x = 1.5 or x = -4. 3. a=3, b=-4, c=-5, discriminant = 16 + 60 = 76, x = (4 ± √76)/6 ≈ 2.12 or -0.79. 4. Discriminant = 16 - 16 = 0, so one repeated real root. 5. 15 and 16, because x(x+1) = 240 gives x² + x - 240 = 0, factorising to (x - 15)(x + 16) = 0, so x = 15 (positive).
答案：1. x = 3 或 x = 4。2. (2x - 3)(x + 4) = 0，所以 x = 1.5 或 x = -4。3. a=3, b=-4, c=-5，判别式 = 16 + 60 = 76，x = (4 ± √76)/6 ≈ 2.12 或 -0.79。4. 判别式 = 16 - 16 = 0，所以有一个二重实数根。5. 15 和 16，因为 x(x+1) = 240 得 x² + x - 240 = 0，因式分解为 (x - 15)(x + 16) = 0，所以 x = 15（取正数）。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图