📚 Solving Quadratic Equations | 解一元二次方程
Welcome to this TutorHao revision guide. In this article you will learn how to solve quadratic equations by factorisation, completing the square and the quadratic formula. You will also understand the discriminant and how quadratic equations relate to graphs and real-life problems.
欢迎阅读 TutorHao 复习指南。本文将讲解如何通过因式分解、配方法和求根公式来解一元二次方程，并帮助你理解判别式、二次函数图像以及实际生活中的应用。
1. What Is a Quadratic Equation? | 什么是一元二次方程
c = 0, where a, b and c are constants and a ≠ 0. The highest power of the variable x is 2.
一元二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 是常数，且 a ≠ 0。变量 x 的最高次数为 2。
dratic, while 2x² + 7 = 0 is also quadratic because it can be written as 2x² + 0x + 7 = 0.
例如，3x² − 5x + 2 = 0 是二次方程，而 2x² + 7 = 0 也是二次方程，因为它可以写成 2x² + 0x + 7 = 0。
equation such as x + 5 = 0 is linear, not quadratic, because the highest power of x is 1.
像 x + 5 = 0 这样的方程是一次方程，而不是二次方程，因为 x 的最高次数是 1。
2. Expanding and Factorising Quadratics | 展开与因式分解
sion. The general expansion of two linear brackets is (x + p)(x + q) = x² + (p + q)x + pq.
要用因式分解法求解二次方程，首先需要熟练掌握展开。两个一次括号的展开公式为：(x + p)(x + q) = x² + (p + q)x + pq。
For example, (x + 2)(x + 3) = x² + 5x + 6, because 2 + 3 = 5 and 2 × 3 = 6.
例如，(x + 2)(x + 3) = x² + 5x + 6，因为 2 + 3 = 5，并且 2 × 3 = 6。
at multiply to 6 and add to 5; these numbers are 2 and 3, so x² + 5x + 6 = (x + 2)(x + 3).
因式分解是展开的逆过程。要对 x² + 5x + 6 进行因式分解，需要找到两个数：它们相乘得 6，相加得 5。这两个数是 2 和 3，所以 x² + 5x + 6 = (x + 2)(x + 3)。
for two brackets (2x + 1)(x + 3). Check: 2x × x = 2x², 2x × 3 + 1 × x = 7x, and 1 × 3 = 3.
当 x² 的系数不是 1 时，例如 2x² + 7x + 3，我们需要找到两个括号 (2x + 1)(x + 3)。检验：2x × x = 2x²，2x × 3 + 1 × x = 7x，且 1 × 3 = 3。
3. Solving by Factorisation | 用因式分解法求解
The key idea is the null factor law: if the product AB = 0, then A = 0 or B = 0.
关键概念是零积性质：如果 AB = 0，那么必有 A = 0 或 B = 0。
)(x + 2). Then use the null factor law: either x − 5 = 0 or x + 2 = 0, so x = 5 or x = −2.
解方程 x² − 3x − 10 = 0。首先对左边进行因式分解：x² − 3x − 10 = (x − 5)(x + 2)。然后利用零积性质：x − 5 = 0 或 x + 2 = 0，所以 x = 5 或 x = −2。
ck. When x = 5, 25 − 15 − 10 = 0; when x = −2, 4 + 6 − 10 = 0. Both solutions are correct.
你应该把解代回原方程进行检验。当 x = 5 时，25 − 15 − 10 = 0；当 x = −2 时，4 + 6 − 10 = 0。两个解都正确。
4. Solving by Completing the Square | 用配方法求解
Completing the square rewrites a quadratic in the form (x + m)² + n. This is useful when factorisation is difficult.
配方法将二次式改写为 (x + m)² + n 的形式。当因式分解较困难时，这种方法非常有用。
The basic identity is x² + bx = (x + b/2)² − (b/2)².
基本恒等式是 x² + bx = (x + b/2)² − (b/2)²。
o (x + 3)² = 16. Taking square roots gives x + 3 = 4 or x + 3 = −4, hence x = 1 or x = −7.
用配方法解 x² + 6x − 7 = 0。首先两边加 7：x² + 6x = 7。然后完成配方：(x + 3)² − 9 = 7。因此 (x + 3)² = 16。两边开平方得 x + 3 = 4 或 x + 3 = −4，所以 x = 1 或 x = −7。
Notice that we must include both the positive and negative square roots.
请注意，开平方时必须同时取正根和负根。
5. The Quadratic Formula | 求根公式
any quadratic equation ax² + bx + c = 0, the solutions are given by the quadratic formula:
对于任意一元二次方程 ax² + bx + c = 0，其解可以用求根公式表示为：
x = (−b ± √b² − 4ac) / (2a)
is not straightforward. For example, solve 2x² + 3x − 2 = 0. Here a = 2, b = 3 and c = −2.
当因式分解不方便时，就使用这个公式。例如，解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。
First calculate the discriminant part: b² − 4ac = 9 − 4 × 2 × (−2) = 9 + 16 = 25. Then √25 = 5. Substituting into the formula gives x = (−3 ± 5) / 4.
先计算判别式部分：b² − 4ac = 9 − 4 × 2 × (−2) = 9 + 16 = 25。则 √25 = 5。代入公式得 x = (−3 ± 5) / 4。
Therefore x = (2)/4 = 0.5 or x = (−8)/4 = −2.
因此 x = 2/4 = 0.5 或 x = −8/4 = −2。
6. The Discriminant | 判别式
e discriminant. It tells us how many real roots a quadratic equation has without solving it.
表达式 Δ = b² − 4ac 称为判别式。它可以在不解方程的情况下判断一元二次方程有多少个实数根。
If Δ > 0, there are two distinct real roots.
如果 Δ > 0，方程有两个不相等的实数根。
If Δ = 0, there is exactly one repeated real root.
如果 Δ = 0，方程有两个相等的实数根，也可以说是一个重根。
If Δ < 0, there are no real roots.
如果 Δ < 0，方程没有实数根。
| Discriminant / 判别式 | Nature of roots / 根的情况 | Graph / 图像 |
|---|---|---|
| Δ > 0 | Two distinct real roots / 两个不等实数根 | Crosses x-axis twice / 与 x 轴有两个交点 |
| Δ = 0 | One repeated real root / 一个重根 | Touches x-axis once / 与 x 轴相切于一点 |
| Δ < 0 | No real roots / 无实数根 | Does not meet x-axis / 与 x 轴不相交 |
r example, x² + 2x + 5 = 0 has discriminant Δ = 4 − 20 = −16 < 0, so it has no real roots.
例如，x² + 2x + 5 = 0 的判别式为 Δ = 4 − 20 = −16 < 0，所以方程没有实数根。
7. Graphical Interpretation | 图像表示
x² + bx + c = 0 are the x-coordinates of the points where the parabola crosses the x-axis.
函数 y = ax² + bx + c 的图像是一条抛物线。方程 ax² + bx + c = 0 的解，就是抛物线与 x 轴交点的 x 坐标。
it opens downward. The axis of symmetry is x = −b/(2a), and the vertex lies on this line.
当 a > 0 时，抛物线开口向上；当 a < 0 时，抛物线开口向下。对称轴为 x = −b/(2a)，顶点就在这条直线上。
xis at x = 1 and x = 3, which are the roots of x² − 4x + 3 = 0 because (x − 1)(x − 3) = 0.
例如，y = x² − 4x + 3 的图像在 x = 1 和 x = 3 处与 x 轴相交，这正好是方程 x² − 4x + 3 = 0 的根，因为 (x − 1)(x − 3) = 0。
8. Forming Quadratic Equations from Word Problems | 从应用题构造一元二次方程
Many real-world problems can be solved by turning the words into a quadratic equation. A rectangle has a length that is 3 cm more than its width, and its area is 40 cm². Find the width.
许多实际问题都可以通过转化为一元二次方程来求解。例如，一个长方形的长比宽多 3 cm，面积为 40 cm²。求宽。
be x cm. Then the length is x + 3 cm. Since area = length × width, we write x(x + 3) = 40.
设宽为 x cm，则长为 x + 3 cm。因为面积 = 长 × 宽，所以我们有 x(x + 3) = 40。
5) = 0. Therefore x = −8 or x = 5. A negative width is not possible, so the width is 5 cm.
整理得 x² + 3x − 40 = 0。因式分解得 (x + 8)(x − 5) = 0。因此 x = −8 或 x = 5。宽度不能为负数，所以宽为 5 cm。
Always check whether your solutions make sense in the context of the problem.
在应用题中，一定要检查求出的解是否符合实际意义。
9. Common Mistakes and Exam Tips | 常见错误与考试技巧
Forgetting that the coefficient of x² cannot be zero; if it is zero, the equation is not quadratic.
忘记 x² 的系数不能为零；如果为零，方程就不是二次方程。
Dividing both sides of an equation by x, which can lose a root such as x = 0.
两边同时除以 x，这样可能会丢失一个根，例如 x = 0。
the ± sign when taking square roots in the quadratic formula or in completing the square.
在使用求根公式或配方法开平方时漏掉了正负号 ±。
Making sign errors when substituting negative values into b² − 4ac.
把负数代入 b² − 4ac 时出现符号错误。
In word problems, giving both mathematical solutions without rejecting those that are impossible.
在应用题中，没有舍去不符合实际意义的数学解。
In the exam, always write down the quadratic formula first if you use it, then substitute the values clearly. This shows the examiner your method and helps you avoid careless mistakes.
考试时，如果使用求根公式，请先写出公式，再清楚地代入数值。这样既能让阅卷老师看到你的思路，也能帮助你避免粗心错误。
10. Practice Questions | 练习
Try these questions on your own, then check the answers below.
请先独立完成下面这些题目，再对照答案。
| Question / 题目 | Answer / 答案 |
|---|---|
| Solve x² − 7x + 12 = 0 | x = 3 or x = 4 |
| Use the quadratic formula to solve x² + 4x − 21 = 0 | x = 3 or x = −7 |
| Find the discriminant of 2x² − 4x + 3 = 0 and state the number of real roots | Δ = 16 − 24 = −8, so no real roots |
on question 2, remember to set a = 1, b = 4, c = −21, so Δ = 16 + 84 = 100 and √100 = 10.
如果你在做第 2 题时出错，请记住你要取 a = 1，b = 4，c = −21，所以 Δ = 16 + 84 = 100，且 √100 = 10。
11. Summary and Key Formula | 总结与核心公式
e, or the quadratic formula. The discriminant Δ = b² − 4ac tells you how many roots exist.
解一元二次方程时，可以使用因式分解、配方或求根公式。判别式 Δ = b² − 4ac 可以判断根的个数。
The most important formula to remember is:
最重要的公式是：
x = (−b ± √b² − 4ac) / (2a)
Always factorise carefully, avoid sign errors, and check your answers in context. With regular practice, solving quadratic equations becomes a reliable skill in your IGCSE Mathematics exam.
请仔细因式分解，避免符号错误，并在原方程或应用题中检验答案。通过经常练习，解一元二次方程将成为你在 IGCSE 数学考试中一项可靠的能力。
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