📚 Solving Quadratic Equations | 解二次方程
Quadratic equations appear frequently in the IGCSE mathematics syllabus. Mastering them is essential not only for exam success but also for understanding curves, motion, and many real-world problems. This revision guide walks you through every important method step by step.
二次方程在 IGCSE 数学考纲中频繁出现。掌握它们不仅对考试成功至关重要，也有助于理解曲线、运动以及许多现实问题。本复习指南将一步一步带你掌握每一种重要方法。
1. What is a Quadratic Equation? | 什么是二次方程？
e a, b and c are numbers, and a ≠ 0. If a = 0, the equation becomes linear, not quadratic.
二次方程是变量最高次数为 2 的方程。它的一般形式是 ax² + bx + c = 0，其中 a、b、c 是数，且 a ≠ 0。如果 a = 0，方程就变成一次方程，而不是二次方程。
For example, 3x² − 5x + 2 = 0 is quadratic, but 2x + 4 = 0 is linear.
例如，3x² − 5x + 2 = 0 是二次方程，而 2x + 4 = 0 是一次方程。
2. Standard Form and Key Terms | 标准形式与关键术语
oefficients: a is the coefficient of x², b is the coefficient of x, and c is the constant term.
在求解之前，总是先将方程整理成标准形式 ax² + bx + c = 0。a、b、c 称为系数：a 是 x² 的系数，b 是 x 的系数，c 是常数项。
If a is positive, the parabola opens upward.
如果 a 为正，抛物线开口向上。
If a is negative, the parabola opens downward.
如果 a 为负，抛物线开口向下。
The solutions of the equation are the x-coordinates where the graph crosses the x-axis.
方程的解就是图像与 x 轴交点的 x 坐标。
3. Solving by Factorisation | 因式分解法
uct of two brackets. For example, x² + 5x + 6 = 0 can be factorised as (x + 2)(x + 3) = 0.
当二次式能写成两个括号相乘时，因式分解法通常是最快的方法。例如，x² + 5x + 6 = 0 可以分解为 (x + 2)(x + 3) = 0。
s is 0, at least one factor must be 0. So x + 2 = 0 or x + 3 = 0, giving x = −2 or x = −3.
然后运用零积性质：如果两个因子的乘积为 0，那么至少有一个因子为 0。所以 x + 2 = 0 或 x + 3 = 0，得到 x = −2 或 x = −3。
(x + p)(x + q) = x² + (p + q)x + pq
When the coefficient of x² is not 1, use the “multiply ac and factor” method or solving by grouping.
当 x² 的系数不是 1 时，可以使用“相乘 ac 再分解”的方法，或者用分组分解法。
4. Solving by Completing the Square | 配方法
Completing the square rewrites ax² + bx + c in the form a(x + h)² + k. This is especially useful for finding turning points and solving equations without factorisation.
配方法将 ax² + bx + c 改写为 a(x + h)² + k 的形式。这在求顶点坐标以及解无法因式分解的方程时尤其有用。
e x² + 6x = (x + 3)² − 9. Thus the equation becomes (x + 3)² − 9 + 2 = 0, or (x + 3)² = 7.
对于简单的二次方程 x² + 6x + 2 = 0，取 6 的一半得 3，再平方得 9。把 x² + 6x 写成 (x + 3)² − 9。于是方程变成 (x + 3)² − 9 + 2 = 0，即 (x + 3)² = 7。
Then take the square root: x + 3 = ±√7, so x = −3 ± √7.
然后开平方：x + 3 = ±√7，所以 x = −3 ± √7。
5. The Quadratic Formula | 二次求根公式
luding those that cannot be factorised. When ax² + bx + c = 0, the solutions are given by:
二次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。当 ax² + bx + c = 0 时，解为：
x = (−b ± √b² − 4ac) / (2a)
You must memorise this formula for IGCSE. Substitute the values of a, b and c carefully, then simplify the square root.
你必须牢记这个公式以应对 IGCSE 考试。小心代入 a、b、c 的值，然后化简根号内的部分。
b = −4, c = −6. Then b² − 4ac = 16 + 48 = 64, so x = (4 ± 8) / 4, giving x = 3 or x = −1.
例如，解 2x² − 4x − 6 = 0。这里 a = 2，b = −4，c = −6。那么 b² − 4ac = 16 + 48 = 64，所以 x = (4 ± 8) / 4，得到 x = 3 或 x = −1。
6. The Discriminant | 判别式
The expression b² − 4ac under the square root is called the discriminant. It tells us how many real roots the equation has.
根号内 b² − 4ac 的表达式称为判别式。它告诉我们方程有多少个实数根。
| Discriminant value | Number of real roots |
| b² − 4ac > 0 | Two distinct real roots |
| b² − 4ac = 0 | One repeated real root |
| b² − 4ac < 0 | No real roots |
In IGCSE exams, you may be asked to determine the nature of the roots without solving the equation.
在 IGCSE 考试中，你可能会被要求不解方程而判断根的情况。
7. Graphs of Quadratic Functions | 二次函数图像
rve called a parabola. The sign of a decides whether the curve has a minimum point or a maximum point.
y = ax² + bx + c 的图像是一条光滑曲线，叫做抛物线。a 的正负决定了曲线有最低点还是最高点。
When a > 0, the parabola has a minimum turning point.
当 a > 0 时，抛物线有最低转折点。
When a < 0, the parabola has a maximum turning point.
当 a < 0 时，抛物线有最高转折点。
The curve is symmetrical about a vertical line through the turning point.
曲线关于经过转折点的竖直线对称。
The solutions of the quadratic are the x-intercepts of the graph. If the discriminant is negative, the graph does not cross the x-axis at all.
二次方程的解就是图像与 x 轴的交点。如果判别式为负，图像完全不与 x 轴相交。
8. Roots and Turning Points | 根与顶点
For a quadratic in standard form, the sum of the roots is −b/a and the product of the roots is c/a. These relationships can be useful for quick checks.
对于标准形式的二次方程，两根之和为 −b/a，两根之积为 c/a。这些关系可用于快速检查。
ways x = −b / (2a). Substitute this value back into the equation to find the y-coordinate.
转折点的 x 坐标始终是 x = −b / (2a)。将这个值代回方程即可求出 y 坐标。
k, then the turning point is simply (h, k). Remember the sign inside the bracket changes.
如果二次式写成配方法形式 y = a(x − h)² + k，那么转折点就是 (h, k)。注意括号内符号的变化。
9. Applying Quadratics to Word Problems | 二次方程应用题
ven as 30. Then (x + 3)(x − 2) = 30, which expands to x² + x − 6 = 30, or x² + x − 36 = 0.
许多现实问题会引出二次方程。例如，长为 x + 3、宽为 x − 2 的矩形面积可能给出为 30。那么 (x + 3)(x − 2) = 30，展开得 x² + x − 6 = 30，即 x² + x − 36 = 0。
When solving word problems, always check that the answer makes sense in the context. A length cannot be negative, so discard any negative solution that does not fit the problem.
解应用题时，一定要检查答案是否符合实际背景。长度不能为负，所以要舍去不合适的负数解。
10. Common Mistakes and Tips | 常见错误与提示
Here is a list of common mistakes students make in IGCSE exams:
以下是 IGCSE 考试中学生常见的错误列表：
Forgetting to rearrange the equation into standard form before using the formula.
在使用公式前忘记将方程整理成标准形式。
Dropping the negative signs when substituting b into the quadratic formula.
代入求根公式时漏掉 b 前面的负号。
Forgetting the ± sign when taking square roots.
开平方时忘记 ± 符号。
Mistaking (x + 3)² = x² + 9 instead of x² + 6x + 9.
误以为 (x + 3)² = x² + 9，而实际上等于 x² + 6x + 9。
Always check your answers by substituting them back into the original equation.
始终通过把答案代回原方程来检验。
11. Practice Questions | 练习题
Try these questions yourself before looking at the solutions.
先自己尝试以下题目，再看答案。
Solve x² − 7x + 12 = 0 by factorisation.
用因式分解法解 x² − 7x + 12 = 0。
Solve 3x² + 5x − 2 = 0 using the quadratic formula.
用求根公式解 3x² + 5x − 2 = 0。
Find the turning point of y = x² − 4x + 5.
求 y = x² − 4x + 5 的转折点。
Determine the number of real roots of 2x² + 4x + 7 = 0.
判断 2x² + 4x + 7 = 0 的实数根个数。
A rectangle has length x + 4 and width x − 1. Its area is 28. Find x.
一个矩形长为 x + 4，宽为 x − 1，面积为 28。求 x。
Answers:
答案：
x = 3 or x = 4.
x = 3 或 x = 4。
x = −2 or x = 1/3.
x = −2 或 x = 1/3。
(2, 1).
(2, 1)。
Discriminant = 16 − 56 = −40 < 0, so no real roots.
判别式 = 16 − 56 = −40 < 0，所以没有实数根。
x = 3 (since x = −8 is impossible).
x = 3（因为 x = −8 不符合实际）。
12. Summary | 总结
To succeed with quadratic equations, remember these key steps: write the equation in standard form, choose the most efficient method, check the discriminant to predict the nature of roots, and always verify your final answers.
要成功解决二次方程，请记住这些关键步骤：先把方程写成标准形式，选择最有效的方法，用判别式判断根的性质，并始终验证最终答案。
With regular practice, solving quadratics will become quick and confident. Use this guide alongside your own exercises to strengthen your IGCSE revision.
通过规律练习，解二次方程会变得快速而自信。将本指南与你的练习册结合使用，以加强你的 IGCSE 复习。
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