📚 Quadratic Equations & Functions | 二次方程与函数
Quadratic equations and functions form one of the most examinable topics in IGCSE Mathematics. This guide covers everything from factorisation and the quadratic formula to graph transformations and word problems, with careful attention to the techniques that earn full marks in the examination.
二次方程与函数是 IGCSE 数学中考查频率最高的内容之一。本指南涵盖因式分解、二次公式、图像变换与应用题等全部考点，并特别关注能够在考试中获取满分的解题技巧。
1. The General Form | 一般形式
A quadratic expression is written in the form ax² + bx + c, where a ≠ 0. The coefficient a, b and c are real numbers, with a being the coefficient of x², b the coefficient of x, and c the constant term. A quadratic equation is formed when we set this expression equal to zero: ax² + bx + c = 0.
二次表达式的一般形式为 ax² + bx + c，其中 a ≠ 0。系数 a、b、c 为实数，a 为 x² 的系数，b 为 x 的系数，c 为常数项。当该表达式等于零时即构成二次方程：ax² + bx + c = 0。
Recognising the standard form is essential. For example, 2x² – 5x + 3 = 0 has a = 2, b = -5, c = 3. Some equations must first be rearranged into this form before solving, such as 3x² = 2x + 1, which becomes 3x² – 2x – 1 = 0.
识别标准形式至关重要。例如，2x² – 5x + 3 = 0 中 a = 2，b = -5，c = 3。有些方程需要先移项化为标准形式再求解，例如 3x² = 2x + 1 化为 3x² – 2x – 1 = 0。
2. Solving by Factorisation | 因式分解法
Factorisation is often the quickest method when the quadratic has rational roots. For a quadratic ax² + bx + c, look for two numbers whose product equals a × c and whose sum equals b. If the quadratic is monic (a = 1), simply find two numbers that multiply to give c and add to give b.
当二次方程具有有理数根时，因式分解通常是最快捷的方法。对于 ax² + bx + c，寻找两个数，使其乘积等于 a × c，和等于 b。若为首项系数为 1 的二次式，只需找到两个数相乘等于 c，相加等于 b。
Solve x² – 7x + 12 = 0. The two numbers are -3 and -4 because (-3) × (-4) = 12 and (-3) + (-4) = -7. Therefore (x – 3)(x – 4) = 0. By the zero product property, either x – 3 = 0 or x – 4 = 0, giving x = 3 or x = 4.
解 x² – 7x + 12 = 0。两个数为 -3 和 -4，因为 (-3) × (-4) = 12，(-3) + (-4) = -7。因此 (x – 3)(x – 4) = 0。根据零乘积性质，x – 3 = 0 或 x – 4 = 0，解得 x = 3 或 x = 4。
For a non-monic quadratic like 2x² + 5x – 3 = 0, we require two numbers whose product is 2 × (-3) = -6 and whose sum is 5. The numbers are 6 and -1. Rewrite the middle term: 2x² + 6x – x – 3 = 0. Then factorise by grouping: 2x(x + 3) – 1(x + 3) = 0, so (2x – 1)(x + 3) = 0. Hence x = ½ or x = -3.
对于非首一二次式如 2x² + 5x – 3 = 0，需找两个数使其乘积为 2 × (-3) = -6，和为 5。这两个数是 6 和 -1。改写中间项：2x² + 6x – x – 3 = 0。然后分组因式分解：2x(x + 3) – 1(x + 3) = 0，即 (2x – 1)(x + 3) = 0。因此 x = ½ 或 x = -3。
If a × b = 0, then a = 0 or b = 0 | 若 a × b = 0，则 a = 0 或 b = 0
3. The Quadratic Formula | 求根公式
When factorisation is difficult or impossible, the quadratic formula provides a universal solution. For ax² + bx + c = 0, the solutions are given by:
当因式分解困难或不可行时，求根公式提供了通用解法。对于 ax² + bx + c = 0，解为：
x = (-b ± √(b² – 4ac)) / 2a
This formula is derived by completing the square on the general form. It works for every quadratic equation, whether the roots are rational, irrational, or even non-real (though IGCSE typically focuses on real roots).
该公式通过对一般形式配方法推导得出。它适用于所有二次方程，无论根是有理数、无理数，甚至非实数（不过 IGCSE 通常侧重于实根）。
Solve 3x² – 4x – 2 = 0 using the formula. Here a = 3, b = -4, c = -2. Substitute into the formula:
使用求根公式解 3x² – 4x – 2 = 0。这里 a = 3，b = -4，c = -2。代入公式：
x = (4 ± √(16 + 24)) / 6 = (4 ± √40) / 6 = (4 ± 2√10) / 6 = (2 ± √10) / 3
The two solutions are x = (2 + √10)/3 and x = (2 – √10)/3. Always simplify surds where possible and check whether the question asks for exact values or decimal approximations.
两个解为 x = (2 + √10)/3 和 x = (2 – √10)/3。务必尽可能化简根式，并注意题目要求精确值还是小数近似值。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form a(x + p)² + q. This form is extremely useful for finding the vertex of a parabola and solving equations when factorisation fails and the quadratic formula is not preferred.
配方法将二次式改写为 a(x + p)² + q 的形式。这种形式在求抛物线顶点以及因式分解失败且不便使用求根公式时极为有用。
For x² – 6x + 5, take half of the coefficient of x: half of -6 is -3. Square it to get 9. Rewrite: x² – 6x + 9 – 9 + 5 = (x – 3)² – 4. Thus the vertex of y = x² – 6x + 5 is at (3, -4), and the equation x² – 6x + 5 = 0 becomes (x – 3)² = 4, giving x = 3 ± 2, so x = 5 or x = 1.
对于 x² – 6x + 5，取 x 系数的一半：-6 的一半为 -3。平方得 9。重写：x² – 6x + 9 – 9 + 5 = (x – 3)² – 4。因此 y = x² – 6x + 5 的顶点为 (3, -4)，方程 x² – 6x + 5 = 0 化为 (x – 3)² = 4，得 x = 3 ± 2，即 x = 5 或 x = 1。
When a ≠ 1, factor out a first. For 2x² + 8x + 3, write 2(x² + 4x) + 3. Complete the square inside: 2[(x + 2)² – 4] + 3 = 2(x + 2)² – 8 + 3 = 2(x + 2)² – 5. The vertex is (-2, -5).
当 a ≠ 1 时，先提取 a。对于 2x² + 8x + 3，写成 2(x² + 4x) + 3。在括号内配方：2[(x + 2)² – 4] + 3 = 2(x + 2)² – 8 + 3 = 2(x + 2)² – 5。顶点为 (-2, -5)。
5. The Discriminant | 判别式
The expression b² – 4ac inside the quadratic formula is called the discriminant, often denoted by Δ (Delta). Its value determines the nature of the roots without solving the equation.
求根公式中的表达式 b² – 4ac 称为判别式，常用 Δ（德尔塔）表示。不解方程即可由其值判定根的性质。
| Discriminant | 判别式 | Nature of Roots | 根的性质 |
| b² – 4ac > 0 | Two distinct real roots | 两个不等实根 |
| b² – 4ac = 0 | One repeated real root | 一个重根（两个相等实根） |
| b² – 4ac < 0 | No real roots | 无实根 |
For x² + 6x + 9 = 0, the discriminant is 36 – 36 = 0, indicating one repeated root. Indeed, (x + 3)² = 0 gives x = -3 as a double root. This also means the parabola touches the x-axis at exactly one point.
对于 x² + 6x + 9 = 0，判别式为 36 – 36 = 0，表明有一个重根。实际上，(x + 3)² = 0 给出 x = -3 作为二重根。这也意味着抛物线恰好与 x 轴相切于一点。
The discriminant can also be used to find unknown parameters. If the equation kx² – 4x + 2 = 0 has two distinct real roots, then b² – 4ac > 0. Substituting: (-4)² – 4(k)(2) > 0, so 16 – 8k > 0, giving k < 2. Note k ≠ 0 since the equation must remain quadratic.
判别式还可用于求未知参数。若方程 kx² – 4x + 2 = 0 有两个不等实根，则 b² – 4ac > 0。代入：(-4)² – 4(k)(2) > 0，即 16 – 8k > 0，得 k < 2。注意 k ≠ 0，因为方程须保持为二次方程。
6. The Graph of y = ax² + bx + c | 二次函数图像
The graph of a quadratic function is a parabola. When a > 0, the parabola opens upward (U-shaped), and when a < 0, it opens downward (n-shaped). The vertex is the minimum point when a > 0 and the maximum point when a < 0.
二次函数的图像是抛物线。当 a > 0 时，抛物线开口向上（U 形）；当 a < 0 时，开口向下（∩ 形）。顶点在 a > 0 时为最低点，在 a < 0 时为最高点。
The x-coordinate of the vertex is given by x = -b / 2a. Substituting this value back into the function gives the y-coordinate. Alternatively, rewrite in completed square form a(x + p)² + q, and the vertex is (-p, q).
顶点的 x 坐标为 x = -b / 2a。将该值代回函数即可得到 y 坐标。另一种方法是改写为配方法形式 a(x + p)² + q，顶点即为 (-p, q)。
To sketch the graph of y = x² – 2x – 3, first find the roots by factorisation: (x – 3)(x + 1) = 0, so x = 3 and x = -1. The vertex x-coordinate is -b/2a = 2/2 = 1, and y = 1 – 2 – 3 = -4, so the vertex is (1, -4). The y-intercept is c = -3.
要画出 y = x² – 2x – 3 的草图，首先用因式分解求根：(x – 3)(x + 1) = 0，得 x = 3 和 x = -1。顶点 x 坐标为 -b/2a = 2/2 = 1，y = 1 – 2 – 3 = -4，故顶点为 (1, -4)。y 截距为 c = -3。
7. Transformations of Graphs | 图像变换
IGCSE requires understanding how the graph of y = f(x) changes under translations. The graph of y = f(x) + k is the graph of y = f(x) shifted vertically by k units. The graph of y = f(x + k) is shifted horizontally by -k units (left if k > 0).
IGCSE 要求理解 y = f(x) 的图像在平移下的变化。y = f(x) + k 的图像是 y = f(x) 的图像垂直平移 k 个单位。y = f(x + k) 的图像水平平移 -k 个单位（k > 0 时向左平移）。
For a quadratic, the completed square form makes transformations transparent. The graph of y = (x – 2)² is the graph of y = x² shifted 2 units to the right. The graph of y = x² + 3 is shifted 3 units upward.
对于二次函数，配方法形式使变换一目了然。y = (x – 2)² 的图像是 y = x² 向右平移 2 个单位的图像。y = x² + 3 的图像是向上平移 3 个单位的图像。
Combining shifts: y = (x + 1)² – 4 represents y = x² shifted 1 unit left and 4 units down, with vertex at (-1, -4). Note the sign: inside the bracket, x + 1 means left, and the constant -4 means down.
组合平移：y = (x + 1)² – 4 表示 y = x² 向左平移 1 个单位再向下平移 4 个单位，顶点为 (-1, -4)。注意符号：括号内 x + 1 表示向左，常数项 -4 表示向下。
8. Quadratic Inequalities | 二次不等式
Solving quadratic inequalities requires sketching or considering the sign intervals of the quadratic expression. Solve x² – 5x + 6 > 0 by first factorising: (x – 2)(x – 3) > 0.
解二次不等式需要画出草图或考虑二次表达式的符号区间。解 x² – 5x + 6 > 0，先因式分解：(x – 2)(x – 3) > 0。
The critical values are x = 2 and x = 3. Test intervals: for x < 2, both factors are negative, so the product is positive. For 2 < x < 3, one factor is positive and the other negative, so the product is negative. For x > 3, both are positive, so the product is positive. Therefore the solution is x < 2 or x > 3.
临界值为 x = 2 和 x = 3。测试区间：当 x < 2 时，两个因式均为负，乘积为正。当 2 < x < 3 时，一个因式为正、另一个为负，乘积为负。当 x > 3 时，两者均为正，乘积为正。因此解为 x < 2 或 x > 3。
For the inequality (x – 2)(x – 3) ≤ 0, the solution is 2 ≤ x ≤ 3. Remember: when the coefficient of x² is negative, it is easier to multiply through by -1 first, reversing the inequality sign.
对于不等式 (x – 2)(x – 3) ≤ 0，解为 2 ≤ x ≤ 3。记住：当 x² 系数为负时，可先在整个不等式两边乘以 -1，同时反转不等号。
9. Word Problems | 应用题
Many IGCSE questions test the ability to translate a real-world scenario into a quadratic equation. A rectangle has length 4 cm longer than its width, and the area is 60 cm². Let the width be x. Then the length is x + 4, and the area equation is x(x + 4) = 60.
许多 IGCSE 题目考查将实际情境转化为二次方程的能力。一个矩形的长比宽长 4 cm，面积为 60 cm²。设宽为 x，则长为 x + 4，面积方程为 x(x + 4) = 60。
Expanding and rearranging gives x² + 4x – 60 = 0. Factorising: (x + 10)(x – 6) = 0. So x = -10 or x = 6. Since a width cannot be negative, x = 6. The rectangle is 6 cm wide and 10 cm long. Always reject extraneous negative solutions in geometric contexts.
展开并整理得 x² + 4x – 60 = 0。因式分解：(x + 10)(x – 6) = 0。因此 x = -10 或 x = 6。宽度不能为负数，故 x = 6。矩形宽 6 cm，长 10 cm。在几何情境中务必捨去无意义的负数解。
Another common type involves projectile motion. A ball’s height after t seconds is h = -5t² + 20t + 1. To find when the ball hits the ground, set h = 0: -5t² + 20t + 1 = 0. Multiply by -1 and apply the quadratic formula to obtain the positive root.
另一类常见问题涉及抛体运动。球在 t 秒后的高度为 h = -5t² + 20t + 1。要求球何时落地，令 h = 0：-5t² + 20t + 1 = 0。两边乘以 -1 并应用求根公式取正根。
10. Simultaneous Equations with Quadratics | 含二次方程的联立方程组
A linear equation and a quadratic equation can be solved simultaneously by substitution. Solve y = 3x + 1 together with y = x² + x – 5. Set the expressions equal: 3x + 1 = x² + x – 5.
一次方程与二次方程可通过代入法联立求解。联立 y = 3x + 1 和 y = x² + x – 5。令两式相等：3x + 1 = x² + x – 5。
Rearrange to x² – 2x – 6 = 0. Using the quadratic formula with a = 1, b = -2, c = -6:
移项得 x² – 2x – 6 = 0。使用求根公式，其中 a = 1，b = -2，c = -6：
x = (2 ± √(4 + 24)) / 2 = (2 ± √28) / 2 = 1 ± √7
Substitute each x value back into y = 3x + 1 to obtain the corresponding y values: if x = 1 + √7, then y = 4 + 3√7; if x = 1 – √7, then y = 4 – 3√7. The intersection points are (1 + √7, 4 + 3√7) and (1 – √7, 4 – 3√7).
将每个 x 值代回 y = 3x + 1 求得对应的 y 值：若 x = 1 + √7，则 y = 4 + 3√7；若 x = 1 – √7，则 y = 4 – 3√7。交点坐标为 (1 + √7, 4 + 3√7) 和 (1 – √7, 4 – 3√7)。
Geometrically, solving simultaneously finds the intersection points of the line and the parabola. The discriminant of the resulting quadratic tells us whether the line intersects (Δ > 0), touches (Δ = 0), or misses the parabola (Δ < 0).
从几何角度来看，联立求解即找到直线与抛物线的交点。所得二次方程的判别式告诉我们直线与抛物线相交（Δ > 0）、相切（Δ = 0）还是相离（Δ < 0）。
11. Common Mistakes and Exam Tips | 常见错误与考试技巧
A frequent error is failing to rearrange the equation into standard form before identifying a, b and c. Always check the equation is written as ax² + bx + c = 0 before applying the formula. Another common mistake is losing solutions when dividing both sides by x — if x = 0 is a solution, it will be lost.
一个常见错误是在确定 a、b、c 之前未将方程整理为标准形式。使用公式前务必确认方程为 ax² + bx + c = 0。另一个常见错误是在等式两边同时除以 x 时丢失解——如果 x = 0 是解，则会因此丢失。
When factorising, always check by expanding. When using the quadratic formula, be careful with negative signs inside the square root. For instance, b = -4 gives b² = 16, not -16. Also, pay attention to the order of operations in the denominator 2a.
因式分解后务必展开验证。使用求根公式时，注意根号内的负号。例如 b = -4 时 b² = 16，而非 -16。同时注意分母 2a 的运算顺序。
In graph sketching questions, always show the roots, the vertex and the y-intercept clearly. If a graph does not cross the x-axis, state this explicitly and show the discriminant is negative. Include the line of symmetry x = -b/2a on your sketch.
在画图题中，务必清楚标出根、顶点和 y 截距。若图像不与 x 轴相交，需明确说明并展示判别式为负。草图中还应画出对称轴 x = -b/2a。
For word problems, define the variable clearly, write down the equation before solving, and check whether solutions are reasonable in the context. Present your final answer in complete sentences with correct units.
对于应用题，清晰定义变量，先写出方程再求解，并检查解在情境中是否合理。最终答案应以完整句子呈现并注明正确单位。
12. Practice Summary | 练习总结
Mastery of quadratic equations and functions requires consistent practice across all the methods shown above. Begin with factorisation questions, then move to the quadratic formula, then complete the square. Practise sketching graphs and interpreting discriminants in equal measure.
熟练掌握二次方程与函数需要围绕上述方法进行持续练习。从因式分解题开始，再练习求根公式，然后是配方法。同等重视画图与判别式的解释。
Finally, attempt mixed exam-style questions that combine multiple skills — for example, a question that asks you to solve a quadratic, state the discriminant, sketch the graph, and interpret the results in a real-world context. This is how the IGCSE examination typically integrates topics.
最后，尝试综合多个技能的整体风格的考题——例如，一道题同时要求解二次方程、说明判别式、画出图像并在实际情境中解释结果。这是 IGCSE 考试整合知识点所采用的典型方式。
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