📚 Solving Quadratic Equations and Functions: A Complete IGCSE Guide | 二次方程与函数：IGCSE完整指南
Quadratic equations and functions are central to the IGCSE mathematics syllabus. They appear in algebra, graph sketching, problem solving, and even coordinate geometry. This guide provides a structured, bilingual revision route through every key idea, from standard form to discriminant, from solving methods to real-world applications.
二次方程与函数是 IGCSE 数学大纲的核心内容。它们出现在代数、图像描绘、问题求解、甚至坐标几何中。本指南提供一条系统化的中英双语复习路径，涵盖从标准形式到判别式、从求解方法到实际应用的所有关键概念。
1. Standard Form and Basic Terms | 标准形式与基本概念
A quadratic equation is a polynomial equation of degree 2. Its standard form is ax² + bx + c = 0, where a, b, and c are constants and a is not equal to 0. The graph of a quadratic function y = ax² + bx + c is called a parabola.
二次方程是次数为 2 的多项式方程。它的标准形式是 ax² + bx + c = 0，其中 a、b、c 为常数，且 a 不等于 0。二次函数 y = ax² + bx + c 的图像称为抛物线。
Because the coefficient of x² must not be zero, we exclude the case where the equation becomes linear. You should quickly identify the values of a, b, and c before applying any solution method.
由于 x² 的系数不能为零，我们排除了方程退化成一次方程的情况。在应用任何求解方法之前，你应当快速辨识出 a、b、c 的值。
Teaching tip | 教学提示：Make sure students can rearrange an equation into standard form before solving. For example, 3x² + 2 = 7x must first become 3x² − 7x + 2 = 0.
教学提示：确保学生在求解前能够将方程整理为标准形式。例如，3x² + 2 = 7x 必须先改写为 3x² − 7x + 2 = 0。
2. Solving by Factorisation | 因式分解法
If a quadratic expression can be written as a product of two linear factors, we can use the zero product property: if pq = 0, then p = 0 or q = 0. This gives two possible roots.
如果一个二次表达式可以写成两个一次因式的乘积，我们就可以利用零积性质：若 pq = 0，则 p = 0 或 q = 0。这样就会得到两个可能的根。
Consider x² − 5x + 6 = 0. We look for two numbers that multiply to 6 and add to −5. These are −2 and −3, so x² − 5x + 6 = (x − 2)(x − 3) = 0. Therefore x = 2 or x = 3.
考虑 x² − 5x + 6 = 0。我们要找两个数，它们的乘积为 6，和为 −5。这两个数是 −2 和 −3，因此 x² − 5x + 6 = (x − 2)(x − 3) = 0。所以 x = 2 或 x = 3。
Special patterns include the difference of two squares, such as x² − 9 = (x − 3)(x + 3), and perfect squares, such as x² + 6x + 9 = (x + 3)². Always check whether the coefficient of x² is not 1 before using simple factor pair searches.
特殊模式包括平方差公式，如 x² − 9 = (x − 3)(x + 3)，以及完全平方式，如 x² + 6x + 9 = (x + 3)²。在寻找简单因式对之前，务必检查 x² 的系数是否为 1。
Teaching tip | 教学提示：Encourage students to expand their factorised answer to verify correctness. This builds confidence and catches sign errors early.
教学提示：鼓励学生展开因式分解后的结果来验证是否正确。这有助于建立信心，也能尽早发现符号错误。
3. The Quadratic Formula | 求根公式
The quadratic formula solves every quadratic equation, including those that cannot be factorised easily. For ax² + bx + c = 0, the formula is:
求根公式可以求解所有二次方程，包括那些不易因式分解的方程。对于 ax² + bx + c = 0，公式如下：
x = (−b ± √(b² − 4ac)) / (2a)
Take the equation 2x² + 3x − 2 = 0. Here a = 2, b = 3, and c = −2. Substitute these values carefully:
以方程 2x² + 3x − 2 = 0 为例。这里 a = 2，b = 3，c = −2。小心代入这些数值：
x = (−3 ± √(9 − 4 × 2 × (−2))) / (2 × 2) = (−3 ± √25) / 4 = (−3 ± 5) / 4
Thus x = (2)/4 = 0.5 or x = (−8)/4 = −2. The discriminant, b² − 4ac, determines how many roots exist, which we will examine next.
因此 x = 2/4 = 0.5 或 x = −8/4 = −2。判别式 b² − 4ac 决定了根的数量，我们接下来会讨论这一点。
Teaching tip | 教学提示：Tell students to write out a, b, c before substituting. This prevents careless errors, especially with negative values.
教学提示：告诉学生在代入前先写出 a、b、c。这可以防止粗心错误，尤其当涉及负数时。
4. Completing the Square | 平方完成法
Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. This is especially useful for finding the vertex of a parabola and solving equations without using the quadratic formula.
平方完成法将 ax² + bx + c 改写为 a(x + p)² + q 的形式。这对于求抛物线顶点以及在不使用求根公式的情况下解方程特别有用。
For x² + 6x + 1 = 0, first halve the coefficient of x to get 3. Then write x² + 6x = (x + 3)² − 9. So the equation becomes (x + 3)² − 9 + 1 = 0, or (x + 3)² − 8 = 0. Solving gives (x + 3)² = 8, so x = −3 ± √8 = −3 ± 2√2.
对于 x² + 6x + 1 = 0，先将 x 的系数除以 2 得到 3。然后写出 x² + 6x = (x + 3)² − 9。于是方程变为 (x + 3)² − 9 + 1 = 0，即 (x + 3)² − 8 = 0。求解得 (x + 3)² = 8，所以 x = −3 ± √8 = −3 ± 2√2。
This form also reveals the minimum or maximum value of a quadratic function. For example, y = (x − 2)² + 5 has a minimum value of 5 when x = 2.
这种形式还能揭示二次函数的最小值或最大值。例如，y = (x − 2)² + 5 在 x = 2 时取得最小值 5。
Teaching tip | 教学提示：Have students practise completing the square with a leading coefficient of 1 before moving to harder cases like 2x² + 8x + 3.
教学提示：让学生先练习 x² 系数为 1 的情况，然后再处理更难的题目，例如 2x² + 8x + 3。
5. The Discriminant and the Nature of Roots | 判别式与根的性质
The discriminant is the expression Δ = b² − 4ac. It tells us how many real roots a quadratic equation has without solving it fully.
判别式是表达式 Δ = b² − 4ac。它不需要完全解方程就能告诉我们二次方程有多少个实数根。
| Discriminant Δ | Nature of roots | Graph interpretation |
| Δ > 0 | Two distinct real roots | Parabola crosses the x-axis twice |
| Δ = 0 | One repeated real root | Parabola touches the x-axis once |
| Δ < 0 | No real roots | Parabola does not meet the x-axis |
判别式 Δ
根的性质
图像解释
Δ > 0
两个不同实数根
抛物线与 x 轴有两个交点
Δ = 0
一个重根
抛物线与 x 轴相切
Δ < 0
没有实数根
抛物线与 x 轴没有交点
For example, x² − 2x + 5 = 0 has Δ = 4 − 20 = −16, so it has no real roots. This explains why the graph of y = x² − 2x + 5 never touches the x-axis.
例如，x² − 2x + 5 = 0 的判别式 Δ = 4 − 20 = −16，因此它没有实数根。这解释了为什么 y = x² − 2x + 5 的图像从不接触 x 轴。
6. Features of the Quadratic Graph | 二次函数图像的特征
The graph of y = ax² + bx + c is always a parabola. 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 的图像始终是抛物线。如果 a > 0，抛物线开口向上，存在最小值点；如果 a < 0，抛物线开口向下，存在最大值点。
The axis of symmetry is the vertical line x = −b/(2a). The vertex lies on this axis, so its x-coordinate is −b/(2a). The y-coordinate can be found by substituting this x-value into the original equation.
对称轴是竖直直线 x = −b/(2a)。顶点位于这条对称轴上，因此其 x 坐标为 −b/(2a)。y 坐标可以通过把这个 x 值代入原方程求得。
The y-intercept is simply c, since y = c when x = 0. The x-intercepts, if they exist, are the real roots of ax² + bx + c = 0.
y 截距就是 c，因为当 x = 0 时 y = c。x 截距如果存在，就是 ax² + bx + c = 0 的实数根。
7. Sketching a Parabola | 绘制抛物线草图
A clear sketch does not need to be perfect, but it must show the key features. Follow this order: find the y-intercept, find the x-intercepts or use the discriminant, locate the axis of symmetry and the vertex, and then draw a smooth curve.
一张清晰的草图不需要完美，但必须展示关键特征。按以下顺序操作：求 y 截距，求 x 截距或使用判别式，确定对称轴和顶点，然后绘制平滑曲线。
To sketch y = x² − 2x − 3, first write y = (x − 3)(x + 1), so the x-intercepts are 3 and −1. The axis of symmetry is halfway between them, x = 1. Substituting gives y = 1 − 2 − 3 = −4, so the vertex is (1, −4). The y-intercept is −3.
为了绘制 y = x² − 2x − 3 的草图，先写出 y = (x − 3)(x + 1)，因此 x 截距为 3 和 −1。对称轴位于它们中间，即 x = 1。代入得 y = 1 − 2 − 3 = −4，所以顶点为 (1, −4)。y 截距为 −3。
Plot these four points and remember that the parabola is symmetrical about x = 1. The sketch is complete once the curve passes through all of them.
标出这四个点，并记住抛物线关于 x = 1 对称。当曲线穿过所有这些点时，草图就完成了。
8. Solving Equations and Inequalities by Graphs | 用图像解方程与不等式
Solving a quadratic equation graphically means finding the x-coordinates where the parabola crosses the x-axis. In an exam, these solutions may be approximate if the graph is drawn by hand.
用图像法解二次方程，就是找抛物线与 x 轴交点的 x 坐标。在考试中，如果是手绘图像，这些解可能是近似值。
For example, if the graph of y = x² − 2x − 3 crosses the x-axis at x = −1 and x = 3, then the solutions to x² − 2x − 3 = 0 are x = −1 and x = 3. The inequality x² − 2x − 3 > 0 corresponds to the parts of the graph above the x-axis, so x < −1 or x > 3.
例如，如果 y = x² − 2x − 3 的图像与 x 轴在 x = −1 和 x = 3 处相交，那么 x² − 2x − 3 = 0 的解就是 x = −1 和 x = 3。不等式 x² − 2x − 3 > 0 对应图像中位于 x 轴上方的部分，因此 x < −1 或 x > 3。
Graphical methods also help with simultaneous equations. Solving y = x² − 2x and y = x + 4 together means finding the intersection points of the parabola and the straight line.
图像法也有助于求解联立方程。联立求解 y = x² − 2x 与 y = x + 4，就是找抛物线与直线的交点。
9. Real-World Applications | 实际应用
Quadratic equations model many real situations. A projectile height can be written as h = −5t² + 20t + 10, where h is height in metres and t is time in seconds.
二次方程可以模拟很多现实情境。抛射体的高度可以写成 h = −5t² + 20t + 10，其中 h 是以米为单位的高度，t 是以秒为单位的时间。
To find when the object hits the ground, set h = 0 and solve −5t² + 20t + 10 = 0. Using the quadratic formula with a = −5, b = 20, and c = 10 gives t = (−20 ± √(400 + 200)) / (−10). The positive solution is t ≈ 4.45 seconds.
要找到物体何时落地，令 h = 0，解 −5t² + 20t + 10 = 0。使用求根公式，a = −5，b = 20，c = 10，得 t = (−20 ± √(400 + 200)) / (−10)。正数解约为 t ≈ 4.45 秒。
Area problems often lead to quadratics. If a rectangular field has length x + 4 and width x, and its area is 45 m², then x(x + 4) = 45, or x² + 4x − 45 = 0. Factorising gives (x + 9)(x − 5) = 0, so x = 5.
面积问题常常会导出二次方程。如果一块矩形田地的长为 x + 4，宽为 x，面积为 45 m²，那么 x(x + 4) = 45，即 x² + 4x − 45 = 0。因式分解得 (x + 9)(x − 5) = 0，所以 x = 5。
Always check whether a negative root makes sense in context. Here x = −9 is rejected because a width cannot be negative.
去检验负根是否符合实际情境。这里 x = −9 被舍去，因为宽不能为负数。
10. Intersection of a Quadratic and a Line | 二次函数与直线的交点
To find where a quadratic curve y = ax² + bx + c intersects a straight line y = mx + k, set the equations equal to each other and solve the resulting quadratic equation.
为了求二次曲线 y = ax² + bx + c 与直线 y = mx + k 的交点，令两个方程相等，然后求解由此产生的二次方程。
Suppose y = x² − 2x − 3 and y = x − 1. Then x² − 2x − 3 = x − 1, which simplifies to x² − 3x − 2 = 0. The discriminant is 9 + 8 = 17, which is positive, so there are two distinct intersection points.
假设 y = x² − 2x − 3 与 y = x − 1。那么 x² − 2x − 3 = x − 1，化简得 x² − 3x − 2 = 0。判别式为 9 + 8 = 17，为正数，所以有两个不同的交点。
The discriminant also tells
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