📚 Solving Quadratic Equations | 解一元二次方程
Quadratic equations are one of the most heavily tested topics in IGCSE Mathematics. From factorisation to the quadratic formula, mastering this topic is essential for Paper 2 and Paper 4 success.
一元二次方程是 IGCSE 数学中考查频率最高的考点之一。从因式分解到求根公式，掌握这一主题对 Paper 2 和 Paper 4 取得高分至关重要。
1. Standard Form of a Quadratic Equation | 一元二次方程的标准形式
A quadratic equation in one unknown is an equation that can be written in the standard form:
一元二次方程是含有一个未知数、且可以写成以下标准形式的方程：
ax² + bx + c = 0
where a, b and c are constants, and a ≠ 0. The highest power of x is 2, which is what makes the equation "quadratic".
其中 a、b、c 为常数，且 a ≠ 0。未知数 x 的最高次数为 2，这正是“二次”名称的由来。
2. Expanding Double Brackets | 展开双括号
Before you can factorise, you must be fluent in expanding. To expand (x + p)(x + q), multiply each term in the first bracket by each term in the second bracket.
在掌握因式分解之前，你必须熟练展开括号。展开 (x + p)(x + q) 时，需要将第一个括号中的每一项分别乘以第二个括号中的每一项。
(x + p)(x + q) = x² + (p + q)x + pq
Notice that the coefficient of x is the sum of 3 and 5, and the constant term is their product.
例如，(x + 3)(x + 5) = x² + 8x + 15。注意 x 的系数是 3 和 5 的和，常数项则是它们的乘积。
3. Factorising Quadratics Where a = 1 | 当 a = 1 时的因式分解
When the coefficient of x² is 1, we look for two numbers that multiply to give c and add to give b.
当 x² 的系数为 1 时，我们需要找到两个数，使它们的乘积等于 c，并且它们的和等于 b。
need two numbers whose product is 12 and whose sum is 7. These numbers are 3 and 4. Hence:
对于方程 x² + 7x + 12 = 0，我们需要找到两个数，其乘积为 12、和为 7。这两个数分别是 3 和 4。因此：
x² + 7x + 12 = (x + 3)(x + 4)
ative. For x² − x − 6 = 0, the numbers are −3 and 2, because (−3)(2) = −6 and −3 + 2 = −1.
如果常数项为负数，则两个数中一个为正、一个为负。对于 x² − x − 6 = 0，两个数是 −3 和 2，因为 (−3)(2) = −6 且 −3 + 2 = −1。
4. Factorising Quadratics Where a ≠ 1 | 当 a ≠ 1 时的因式分解
When the coefficient of x² is not 1, the process requires more care. One reliable method is the "product-and-sum" approach.
当 x² 的系数不等于 1 时，因式分解需要更加细心。一种可靠的方法是“乘积与和”法。
numbers whose product is 6 and whose sum is 7: these are 1 and 6. Rewrite the middle term:
对于 2x² + 7x + 3 = 0，先计算 a 与 c 的乘积：2 × 3 = 6。找到两个数，使其乘积为 6、和为 7：这两个数是 1 和 6。重写中间项：
2x² + 7x + 3 = 2x² + x + 6x + 3
(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3). Always check your expansion to confirm the result.
然后分组因式分解：x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)。务必通过展开来验证结果是否正确。
5. Solving by Factorisation | 用因式分解法求解
The key principle is the zero product property: if the product of two expressions is zero, then at least one of them must be zero.
关键原理是零积性质：如果两个表达式的乘积为零，那么其中至少有一个必须为零。
² + 7x + 12 = 0, first factorise: (x + 3)(x + 4) = 0. Then set each bracket equal to zero:
解 x² + 7x + 12 = 0 时，先因式分解：(x + 3)(x + 4) = 0。然后令每一个括号分别等于零：
x + 3 = 0 或 x + 4 = 0
Therefore x = −3 or x = −4. Remember to write both solutions clearly, separated by "or". Substitute each answer back into the original equation to verify.
因此 x = −3 或 x = −4。注意用“或”清晰写出两个解。将每个答案代回原方程进行验算。
6. The Quadratic Formula | 求根公式
atic formula, which works for any quadratic equation written in the form ax² + bx + c = 0.
并非每个二次方程都能轻松因式分解。此时应使用求根公式，它适用于任何写成 ax² + bx + c = 0 形式的二次方程。
x = (−b ± √b² − 4ac) / 2a
For example, solve 2x² + 3x − 2 = 0. Here a = 2, b = 3 and c = −2. Substituting gives:
例如，解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2。代入得：
x = (−3 ± √9 + 16) / 4 = (−3 ± 5) / 4
So x = (−3 + 5)/4 = 0.5, or x = (−3 − 5)/4 = −2. Write the answer as x = 0.5 or x = −2.
因此 x = (−3 + 5)/4 = 0.5，或 x = (−3 − 5)/4 = −2。答案应写为 x = 0.5 或 x = −2。
7. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + h)² + k. This method is especially useful for finding turning points.
配方是将二次式改写为 a(x + h)² + k 的形式。这一方法在求顶点坐标时特别有用。
For x² + 6x + 5, take half of 6 to get 3, then square it to get 9. Rewrite:
对于 x² + 6x + 5，取 6 的一半得到 3，再平方得到 9。改写为：
x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4
3 = ±2, giving x = −1 or x = −5. Note that the turning point of the graph is at (−3, −4).
解方程 (x + 3)² − 4 = 0 时，先移项得到 (x + 3)² = 4，再得 x + 3 = ±2，即 x = −1 或 x = −5。注意该抛物线的顶点坐标为 (−3, −4)。
8. The Discriminant and the Nature of Roots | 判别式与根的性质
The expression b² − 4ac inside the quadratic formula is called the discriminant, usually denoted Δ. It tells us the number and type of roots without fully solving the equation.
求根公式中的 b² − 4ac 被称为判别式，通常记作 Δ。它无需完整解方程即可判断根的个数和类型。
There are three cases to remember:
需要记住以下三种情况：
| Discriminant Δ | Nature of roots | 判别式 Δ | 根的性质 |
| Δ > 0 | Two distinct real roots | Δ > 0 | 两个不相等的实数根 |
| Δ = 0 | One repeated real root | Δ = 0 | 两个相等实数根（重根） |
| Δ < 0 | No real roots | Δ < 0 | 无实数根 |
If a question asks you to "show that the equation has no real roots", simply calculate the discriminant and state that it is negative.
如果题目要求“证明方程无实数根”，只需计算判别式并说明其值为负即可。
9. Graphs of Quadratic Functions | 二次函数的图像
x² + bx + c is a parabola. If a > 0, the parabola opens upwards; if a < 0, it opens downwards.
二次函数 y = ax² + bx + c 的图像是一条抛物线。当 a > 0 时，抛物线开口向上；当 a < 0 时，开口向下。
correspond to the x-intercepts of the graph. The turning point lies halfway between the roots and can be found by completing the square.
方程 ax² + bx + c = 0 的根对应于图像与 x 轴的交点。顶点位于两个根的中点处，也可以通过配方法求得。
-axis at x = 1 and x = 3. The axis of symmetry is x = 2, and the minimum point is (2, −1).
例如，y = x² − 4x + 3 的图像与 x 轴交于 x = 1 和 x = 3。对称轴为 x = 2，最低点为 (2, −1)。
10. Solving Word Problems | 解应用题
Quadratic equations often appear in geometry and number problems. Read the question carefully, define a variable, and translate the information into a quadratic equation.
二次方程常出现在几何和数字类应用题中。审题时要仔细，设定未知数，并将题中信息转化为二次方程。
Example: The length of a rectangle is 3 cm longer than its width, and its area is 40 cm². Let the width be x cm. Then the length is (x + 3) cm, so:
例如：一个长方形的长比宽长 3 cm，面积为 40 cm²。设宽为 x cm，则长为 (x + 3) cm，因此：
x(x + 3) = 40, so x² + 3x − 40 = 0
tive, the width is 5 cm and the length is 8 cm. Always reject solutions that do not make sense in the context.
因式分解得 (x + 8)(x − 5) = 0，即 x = −8 或 x = 5。由于长度不能为负，所以宽为 5 cm，长为 8 cm。务必舍去在题意中不合理的解。
11. Common Mistakes to Avoid | 常见错误与易错点
In the exam, many marks are lost through avoidable errors. Here are the most common ones:
在考试中，许多分数都因可避免的错误而丢失。以下是最常见的几类失误：
- etting to rearrange the equation to the form ax² + bx + c = 0 before applying the formula. 忘记先将方程化为 ax² + bx + c = 0 的形式，就直接套用求根公式。
- Confusing the signs when substituting negative values of b and c into the formula. 将 b、c 的负值代入公式时混淆正负号。
- Dividing both sides by x, which loses the solution x = 0. 两边同时除以 x，导致丢失 x = 0 这个解。
- Writing only one solution when there are two distinct roots. 当有两个不同根时，只写出了其中一个解。
- Forgetting to check answers by substitution. 忘记通过代回原方程来检验答案。
y x. Instead, rearrange to x² − 5x = 0, factorise to x(x − 5) = 0, and get x = 0 or x = 5.
例如对于 x² = 5x，绝不能两边同除以 x。应该移项为 x² − 5x = 0，因式分解为 x(x − 5) = 0，得到 x = 0 或 x = 5。
12. Practice Questions | 巩固练习
Try these exam-style questions to test your understanding:
请尝试以下模拟考试题型，以检验你的理解程度：
- Solve x² − 9x + 20 = 0 by factorisation. 用因式分解法解 x² − 9x + 20 = 0。
- Solve 3x² + 5x − 2 = 0 using the quadratic formula. 用求根公式解 3x² + 5x − 2 = 0。
- Express x² − 8x + 3 in the form (x − a)² + b, and hence state the coordinates of the turning point. 将 x² − 8x + 3 化为 (x − a)² + b 的形式，并由此写出顶点坐标。
- Find the value of k for which the equation x² + kx + 9 = 0 has exactly one real root. 求 k 的值，使方程 x² + kx + 9 = 0 恰有一个实数根。
Answers: 1) x = 4 or 5 2) x = 1/3 or −2 3) (x − 4)² − 13, turning point (4, −13) 4) k = ±6
ers: 1) x = 4 or 5; 2) x = 1/3 or −2; 3) (x − 4)² − 13, turning point (4, −13); 4) k = ±6.
答案：1) x = 4 或 5；2) x = 1/3 或 −2；3) (x − 4)² − 13，顶点为 (4, −13)；4) k = ±6。
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