📚 Mastering Quadratic Equations and Functions | 精通二次方程与函数
Quadratic equations and functions form one of the most essential topics in the IGCSE Mathematics syllabus. From factorisation to the quadratic formula, understanding this area unlocks success in algebra, graphs, and problem-solving.
二次方程与函数是IGCSE数学课程中最核心的专题之一。从因式分解到求根公式，掌握这一领域是学好代数、函数图像以及实际应用题的关键。
1. The Standard Form of a Quadratic Equation | 二次方程的标准形式
A quadratic equation in one variable can be written in the standard form:
一元二次方程可以写成如下标准形式：
ax² + bx + c = 0, a ≠ 0
Here, a, b, and c are real constants, and a must not be zero. If a = 0, the equation becomes linear, not quadratic.
其中 a、b、c 是实数常数，且 a ≠ 0。若 a = 0，方程就变成一次方程而非二次方程。
For example, x² − 5x + 6 = 0 is quadratic because the highest power of x is 2.
例如，x² − 5x + 6 = 0 是二次方程，因为 x 的最高次数为 2。
2. Solving by Factorisation | 用因式分解法求解
Factorisation is often the quickest method when the quadratic expression can be written as a product of two linear factors.
当二次表达式可以写成两个一次因式的乘积时，因式分解往往是最快捷的方法。
Consider x² − 5x + 6 = 0. We look for two numbers whose product is 6 and whose sum is −5.
以 x² − 5x + 6 = 0 为例，我们寻找两个数，它们乘积为 6，和为 −5。
(x − 2)(x − 3) = 0
Then, by the zero product property:
接着，利用零积性质：
If x − 2 = 0, then x = 2.
若 x − 2 = 0，则 x = 2。
If x − 3 = 0, then x = 3.
若 x − 3 = 0，则 x = 3。
Always expand your factors to check: (x − 2)(x − 3) = x² − 5x + 6. ✔
务必展开因式来检查：(x − 2)(x − 3) = x² − 5x + 6。✔
3. Factorising Quadratics with a ≠ 1 | 当 a ≠ 1 时的因式分解
coefficient of x² is not 1, factorisation requires extra care. Consider 2x² + 7x + 3 = 0.
当 x² 的系数不为 1 时，因式分解需要格外小心。例如 2x² + 7x + 3 = 0。
Method: Multiply a and c o get 2 × 3 = 6. Find two numbers whose product is 6 and whose sum is 7: they are 1 and 6.
方法：将 a 与 c 相乘得到 2 × 3 = 6。寻找乘积为 6 且和为 7 的两个数：它们就是 1 和 6。
Rewrite the middle term:
改写中间项：
2x² + x + 6x + 3 = 0
Group and factor:
分组并因式分解：
x(2x + 1) + 3(2x + 1) = 0
(2x + 1)(x + 3) = 0
Thus x = −1/2 or x = −3.
因此 x = −1/2 或 x = −3。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This is especially useful for finding turning points and solving equations.
配方法将二次式改写为 a(x + p)² + q 的形式。这在求顶点坐标和解方程时特别有用。
For x² + 6x + 2 = 0, take half of 6, which is 3, and square it to get 9.
对于 x² + 6x + 2 = 0，取 6 的一半得 3，再平方得到 9。
x² + 6x + 9 − 9 + 2 = 0
(x + 3)² − 7 = 0
Then solve: (x + 3)² = 7, so x + 3 = ±√7, giving x = −3 ± √7.
然后求解：(x + 3)² = 7，因此 x + 3 = ±√7，即 x = −3 ± √7。
If the coefficient of x² is not 1, factor it out first before completing the square.
如果 x² 的系数不是 1，请先提取该系数，再进行配方。
5. The Quadratic Formula | 求根公式
The quadratic formula works for every quadratic equation, even when factorisation is difficult or impossible.
求根公式适用于所有二次方程，即使因式分解困难甚至无法分解时也能使用。
x = [−b ± √b² − 4ac] / (2a)
For the equation 2x² − 4x − 3 = 0, identify a = 2, b = −4, c = −3.
对于方程 2x² − 4x − 3 = 0，先确定 a = 2，b = −4，c = −3。
x = [4 ± √16 + 24] / 4 = [4 ± √40] / 4
Simplify: x = (4 ± 2√10) / 4 = 1 ± √10 / 2.
化简得：x = (4 ± 2√10) / 4 = 1 ± √10 / 2。
Memorise the formula carefully — one sign error can cost you the whole mark.
请牢记该公式——一个符号错误就可能导致全题失分。
6. The Discriminant | 判别式
The expression b² − 4ac under the square root is called the discriminant, often denoted by Δ.
根号下的表达式 b² − 4ac 称为判别式，常用 Δ 表示。
The discriminant tells us how many real roots a quadratic equation has:
判别式告诉我们二次方程有多少个实数根：
| Discriminant Δ | Nature of Roots | 根的性质 |
| Δ > 0 | Two distinct real roots | 两个不相等的实数根 |
| Δ = 0 | One repeated real root | 一个重根（两个相等实数根） |
| Δ < 0 | No real roots | 没有实数根 |
For example, x² + 2x + 5 = 0 has Δ = 4 − 20 = −16 < 0, so it has no real solutions.
例如，x² + 2x + 5 = 0 中 Δ = 4 − 20 = −16 < 0，因此该方程没有实数解。
7. Graphs of Quadratic Functions | 二次函数图像
ax² + bx + c is a parabola. If a > 0, it opens upward; if a < 0, it opens downward.
二次函数 y = ax² + bx + c 的图像是一条抛物线。若 a > 0，抛物线开口向上；若 a < 0，抛物线开口向下。
Key features of the parabola include:
抛物线的主要特征包括：
The axis of symmetry is x = −b/(2a).
对称轴为 x = −b/(2a)。
axis of symmetry; its y-coordinate is found by substituting x = −b/(2a) into the function.
顶点（拐点）位于对称轴上；将 x = −b/(2a) 代入函数即可求出其 y 坐标。
The y-intercept is c.
y 截距为 c。
The x-intercepts are the real roots of ax² + bx + c = 0.
x 截距就是方程 ax² + bx + c = 0 的实数根。
For y = x² − 4x + 3, the axis is x = 2 and the vertex is (2, −1).
对于 y = x² − 4x + 3，对称轴为 x = 2，顶点为 (2, −1)。
8. Transformations of Parabolas | 抛物线的变换
ing the completed square form y = a(x − h)² + k, we can identify transformations directly.
利用配方式 y = a(x − h)² + k，我们可以直接判断图像变换。
The graph shifts horizontally by h units: right if h > 0, left if h < 0.
图像水平平移 h 个单位：h > 0 向右移，h < 0 向左移。
The graph shifts vertically by k units: up if k > 0, down if k < 0.
图像垂直平移 k 个单位：k > 0 向上移，k < 0 向下移。
The value of a controls the width and direction of the parabola. A larger |a| makes it narrower; a negative a flips it upside down.
a 的值控制抛物线的宽度与方向。|a| 越大抛物线越窄；a 为负值时抛物线上下翻转。
the parabola y = x² shifted 3 units right, 1 unit up, and made narrower by a factor of 2.
例如，y = 2(x − 3)² + 1 是由 y = x² 向右平移 3 个单位、向上平移 1 个单位，并按因子 2 收窄得到的抛物线。
9. Solving Quadratic Inequalities | 解二次不等式
Quadratic inequalities are solved by first finding the roots, then sketching the parabola to determine the sign of the expression.
解二次不等式时，先求出根，再画出抛物线草图以确定表达式的符号。
Consider x² − x − 6 < 0. Factorise to get (x − 3)(x + 2) < 0.
考虑 x² − x − 6 < 0。因式分解得 (x − 3)(x + 2) < 0。
and x = 3. Since the parabola opens upward, the expression is negative between the roots.
根为 x = −2 和 x = 3。由于抛物线开口向上，表达式在两根之间为负。
−2 < x < 3
For x² − x − 6 > 0, the solution is x < −2 or x > 3.
对于 x² − x − 6 > 0，解为 x < −2 或 x > 3。
rict (< or >) or inclusive (≤ or ≥), as this affects whether the roots are included.
始终注意不等式是严格不等号（< 或 >）还是包含等号（≤ 或 ≥），因为这决定了根是否包含在解集中。
10. Applications and Word Problems | 应用与文字题
Quadratic equations appear in geometry, physics, and business contexts. A classic problem: a rectangle has area 40 cm² and its length is 6 cm more than its width. Find the width.
二次方程在几何、物理和商业情境中经常出现。经典题目：一个矩形面积为 40 cm²，其长比宽多 6 cm。求宽。
Let the width be x. Then length = x + 6, so:
设宽为 x，则长 = x + 6，于是：
x(x + 6) = 40
x² + 6x − 40 = 0
e: (x + 10)(x − 4) = 0, giving x = −10 or x = 4. Since width cannot be negative, x = 4 cm.
因式分解得：(x + 10)(x − 4) = 0，所以 x = −10 或 x = 4。因为宽度不能为负，故 x = 4 cm。
ight h(t) = −5t² + 20t + 2 can be solved to find when the object reaches a certain height.
在抛体运动中，高度函数 h(t) = −5t² + 20t + 2 可用于求解物体到达某一高度的时刻。
11. Exam Tips and Common Mistakes | 考试技巧与常见错误
Many students lose marks in quadratic questions due to avoidable errors. Keep these tips in mind:
许多学生在二次方程题目中因可避免的错误而失分。请牢记以下建议：
Always rearrange the equation into standard form ax² + bx + c = 0 before solving.
求解前务必先将方程整理为标准形式 ax² + bx + c = 0。
When using the quadratic formula, double-check the signs of a, b, and c, especially when b or c is negative.
使用求根公式时，仔细检查 a、b、c 的符号，特别是当 b 或 c 为负数时。
Do not forget the '±' sign in the quadratic formula — it gives both roots.
不要遗漏求根公式中的“±”号——它给出两个根。
When solving a word problem, check that your answer makes sense in the context (for example, lengths cannot be negative).
解文字题时，检查答案在情境中是否合理（例如长度不能为负）。
In graphical questions, clearly label the vertex, intercepts, and axis of symmetry.
在图像题中，清楚地标出顶点、截距和对称轴。
If the question asks for answers to 2 decimal places, use the quadratic formula rather than factorisation to avoid rounding too early.
如果题目要求答案保留两位小数，请使用求根公式而非因式分解，以避免过早四舍五入。
12. Practice Problems | 练习题目
Try these problems to consolidate your understanding. Solutions are shown below each question.
尝试以下题目以巩固理解。每题下方附有解答。
Problem 1: Solve x² − 7x + 12 = 0.
题目1：求方程 x² − 7x + 12 = 0 的解。
Solution: (x − 3)(x − 4) = 0, so x = 3 or x = 4.
解答：(x − 3)(x − 4) = 0，所以 x = 3 或 x = 4。
Problem 2: Find the discriminant of 2x² + 3x − 5 = 0 and state the number of real roots.
题目2：求 2x² + 3x − 5 = 0 的判别式，并说明其实数根的个数。
Solution: Δ = 9 − 4(2)(−5) = 9 + 40 = 49 > 0, so there are two distinct real roots.
解答：Δ = 9 − 4(2)(−5) = 9 + 40 = 49 > 0，因此有两个不相等的实数根。
Problem 3: Write y = x² + 4x + 7 in completed square form and find the minimum point.
题目3：将 y = x² + 4x + 7 写成配方式，并求最小值点。
Solution: y = (x + 2)² + 3, so the minimum point is (−2, 3).
解答：y = (x + 2)² + 3，因此最小值为 (−2, 3)。
Problem 4: The sum of a number and its reciprocal is 17/4. Find the number.
题目4：一个数与它的倒数之和为 17/4。求这个数。
+ 4 = 17x, so 4x² − 17x + 4 = 0. Factorise: (4x − 1)(x − 4) = 0, giving x = 1/4 or x = 4.
解答：设这个数为 x。则 x + 1/x = 17/4。两边乘以 4x：4x² + 4 = 17x，即 4x² − 17x + 4 = 0。因式分解得：(4x − 1)(x − 4) = 0，因此 x = 1/4 或 x = 4。
Mastering quadratics takes practice, but every method you learn builds a stronger mathematical foundation. Review these techniques regularly, and they will become second nature.
精通二次方程需要不断练习，但你学到的每种方法都会为数学基础添砖加瓦。定期复习这些技巧，它们就会变得熟能生巧。
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