📚 Quadratic Equations and Graphs | 二次方程与图像
A quadratic equation is one of the most important topics in IGCSE Mathematics. It appears in algebra, graphs, problem solving and even in real-life modelling. A strong understanding of quadratics will help you solve equations, sketch curves and interpret turning points with confidence.
二次方程是 IGCSE 数学中最重要的主题之一。它出现在代数、函数图像、问题解决甚至现实生活建模中。扎实掌握二次方程能帮助你自信地解方程、画函数图像并理解顶点。
1. Recognising a Quadratic Expression | 识别二次表达式
A quadratic expression is an algebraic expression where the highest power of the variable is 2. Its general form is ax² + bx + c, where a, b and c are constants and a ≠ 0.
二次表达式是变量的最高次数为 2 的代数表达式。它的一般形式是 ax² + bx + c，其中 a、b、c 是常数，且 a ≠ 0。
- x² + 5x + 6 is quadratic because the highest power is 2.
- 3x² − 2x + 1 is quadratic.
- 4x + 7 is linear, not quadratic.
- x³ + 2x² − 1 is cubic, not quadratic.
- x² + 5x + 6 是二次式，因为最高次数是 2。
- 3x² − 2x + 1 是二次式。
- 4x + 7 是一次式，不是二次式。
- x³ + 2x² − 1 是三次式，不是二次式。
2. Expanding Double Brackets | 展开双括号
Expanding two binomials is often the first step towards factorising and solving quadratic equations. Use the FOIL method: multiply the First, Outer, Inner and Last terms, then collect like terms.
展开两个二项式通常是因式分解和解二次方程的第一步。使用 FOIL 方法：先乘首项、外项、内项和末项，然后合并同类项。
Expand (x + 3)(x + 2):
展开 (x + 3)(x + 2)：
(x + 3)(x + 2) = x² + 2x + 3x + 6 = x² + 5x + 6
careful with negative signs. For example, (x − 4)(x + 2) = x² + 2x − 4x − 8 = x² − 2x − 8.
注意负号。例如 (x − 4)(x + 2) = x² + 2x − 4x − 8 = x² − 2x − 8。
3. Factorising Quadratics | 二次式因式分解
Factorising is the reverse of expanding. To factorise x² + bx + c, find two numbers that multiply to give c and add to give b.
因式分解是展开的逆运算。要对 x² + bx + c 进行因式分解，找出两个数，它们相乘得 c，相加得 b。
Factorise x² + 7x + 10. We need two numbers with product 10 and sum 7. The numbers are 2 and 5, so:
对 x² + 7x + 10 进行因式分解。我们需要两个数，乘积为 10，和为 7。这两个数是 2 和 5，所以：
x² + 7x + 10 = (x + 2)(x + 5)
5x + 3, split the middle term: 2x² + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1).
当 a ≠ 1 时，因式分解更复杂。例如 2x² + 5x + 3，拆分中间项：2x² + 2x + 3x + 3 = 2x(x + 1) + 3(x + 1) = (2x + 3)(x + 1)。
4. Solving by Factorisation | 用因式分解解二次方程
factorise the other side and use the zero product rule: if a × b = 0, then a = 0 or b = 0.
要用因式分解法解二次方程，首先使一边等于 0。然后对另一边因式分解，并使用零乘积法则：若 a × b = 0，则 a = 0 或 b = 0。
Solve x² − 5x + 6 = 0:
解方程 x² − 5x + 6 = 0：
x² − 5x + 6 = (x − 2)(x − 3) = 0
x − 2 = 0 or x − 3 = 0
x = 2 or x = 3
Always check your answers by substituting them back into the original equation.
始终把答案代回原方程进行检验。
5. The Quadratic Formula | 二次公式
be factorised easily, use the quadratic formula. For ax² + bx + c = 0, the solutions are:
当二次方程不容易因式分解时，使用二次公式。对于 ax² + bx + c = 0，解为：
x = (−b ± √b² − 4ac) ÷ 2a
Note: In this article, √b² − 4ac means the square root of the entire expression b² − 4ac.
注意：在本文中，√b² − 4ac 表示整个表达式 b² − 4ac 的平方根。
Solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, c = −2:
解 2x² + 3x − 2 = 0。这里 a = 2，b = 3，c = −2：
x = (−3 ± √(3² − 4×2×(−2))) ÷ (2×2)
x = (−3 ± √9 + 16) ÷ 4
x = (−3 ± √25) ÷ 4
x = (−3 ± 5) ÷ 4
x = 1/2 or x = −2
This method always works for any quadratic equation, even when the roots are irrational or complex.
这个方法对任何二次方程都有效，即使根是无理数或复数。
6. The Discriminant | 判别式
The expression under the square root in the quadratic formula, b² − 4ac, is called the discriminant. It tells you how many real roots a quadratic equation has.
二次公式中平方根下的表达式 b² − 4ac 称为判别式。它告诉你二次方程有多少个实数根。
| Discriminant | Number of real roots | Graph behaviour |
|---|---|---|
| b² − 4ac > 0 | 2 distinct real roots | Curve crosses x-axis twice |
| b² − 4ac = 0 | 1 repeated real root | Curve touches x-axis once |
| b² − 4ac < 0 | No real roots | Curve never crosses x-axis |
| 判别式 | 实数根的个数 | 图像行为 |
|---|---|---|
| b² − 4ac > 0 | 2 个不同实根 | 曲线穿过 x 轴两次 |
| b² − 4ac = 0 | 1 个重根 | 曲线与 x 轴相切一次 |
| b² − 4ac < 0 | 无实根 | 曲线不与 x 轴相交 |
mple, x² + 2x + 5 = 0 has discriminant 2² − 4×1×5 = 4 − 20 = −16, so it has no real roots.
例如，x² + 2x + 5 = 0 的判别式为 2² − 4×1×5 = 4 − 20 = −16，因此它没有实根。
7. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This form makes it easy to find the turning point and solve the equation.
配方法把二次式写成 a(x + p)² + q 的形式。这种形式便于求顶点和解方程。
Rewrite x² + 6x + 1 in completed square form:
把 x² + 6x + 1 写成配方法形式：
x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8
Now solve x² + 6x + 1 = 0:
现在解 x² + 6x + 1 = 0：
(x + 3)² − 8 = 0
(x + 3)² = 8
x + 3 = ±√8
x = −3 ± 2√2
Completing the square is especially useful for finding maximum or minimum values of quadratic functions.
配方法对求二次函数的最大值或最小值特别有用。
8. Graphs of Quadratic Functions | 二次函数图像
. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, it opens downwards and has a maximum point.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上，有最小值点。如果 a < 0，抛物线开口向下，有最大值点。
substituting x = 0. The x-intercepts are the real roots of the equation ax² + bx + c = 0.
y 轴截距是常数项 c，通过代入 x = 0 得到。x 轴截距是方程 ax² + bx + c = 0 的实根。
- For y = x² + 5x + 6, the y-intercept is 6.
- The roots are x = −2 and x = −3, so the x-intercepts are (−2, 0) and (−3, 0).
- 对于 y = x² + 5x + 6，y 轴截距是 6。
- 它的根是 x = −2 和 x = −3，因此 x 轴截距是 (−2, 0) 和 (−3, 0)。
Sketch the curve by plotting the intercepts, the turning point and a symmetrical shape.
通过标出截距、顶点和对称形状来绘制曲线草图。
9. Turning Point and Axis of Symmetry | 顶点与对称轴
ten as y = a(x + p)² + q, the turning point is (−p, q) and the axis of symmetry is x = −p.
每条抛物线都有一个顶点和一条竖直对称轴。如果二次函数写成 y = a(x + p)² + q，顶点为 (−p, q)，对称轴为 x = −p。
For y = (x + 3)² − 8, the turning point is (−3, −8) and the axis of symmetry is x = −3.
对于 y = (x + 3)² − 8，顶点是 (−3, −8)，对称轴是 x = −3。
c is in standard form y = ax² + bx + c, the x-coordinate of the turning point is given by:
如果二次函数是一般形式 y = ax² + bx + c，顶点的 x 坐标由下式给出：
x = −b ÷ 2a
Substitute this x-value back into the equation to find the y-coordinate of the turning point.
把这个 x 值代回方程，即可求出顶点的 y 坐标。
10. Solving Graphically and Applications | 图像解法与应用
y = ax² + bx + c crosses the x-axis. These x-values are the solutions of ax² + bx + c = 0.
二次方程可以通过图像求解：找出曲线 y = ax² + bx + c 与 x 轴的交点。这些 x 值就是 ax² + bx + c = 0 的解。
Quadratic functions also model many real-life situations such as projectile motion, area problems and profit maximisation. In these problems, the maximum or minimum value often answers the practical question.
二次函数还能模拟许多现实生活情境，如抛体运动、面积问题和利润最大化。在这些问题中，最大值或最小值通常就是实际问题的答案。
a = −20 ÷ (2×−5) = 2 seconds. Substituting t = 2 gives h = −5(2)² + 20(2) + 1 = 21 metres.
例题：一个球被抛出，t 秒后的高度 h 米满足 h = −5t² + 20t + 1。最大高度出现在 t = −b ÷ 2a = −20 ÷ (2×−5) = 2 秒处。代入 t = 2 得 h = −5(2)² + 20(2) + 1 = 21 米。
11. Exam-Style Tips | 考试技巧
In IGCSE exams, always show your working clearly. If a question asks for answers correct to a given number of decimal places, use the quadratic formula rather than factorisation when the roots are not integers.
在 IGCSE 考试中，务必清晰地展示解题过程。如果题目要求答案精确到指定小数位，且根不是整数，应使用二次公式而不是因式分解。
Check the sign of a before sketching a parabola. A positive a gives a ∪ shape; a negative a gives an ∩ shape. Always label the intercepts and turning point when asked to sketch a graph.
画抛物线之前先检查 a 的符号。a 为正时图像是 ∪ 形；a 为负时图像是 ∩ 形。当题目要求画草图时，务必标出截距和顶点。
For word problems, define your variable first, form the quadratic equation, solve it, and then reject any solution that does not make sense in the context, such as a negative length or time.
对于文字题，先定义变量，建立二次方程，解方程，然后舍去在情境中没有意义的解，例如负的长度或时间。
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