Mastering Quadratic Equations for IGCSE Mathematics | IGCSE 数学：二次方程完全攻略

📚 Mastering Quadratic Equations for IGCSE Mathematics | IGCSE 数学：二次方程完全攻略

Quadratic equations appear throughout the IGCSE Mathematics syllabus, from basic factorisation in the Core paper to completing the square and hidden quadratics in the Extended paper. A strong grasp of their structure, solution methods, and graphical behaviour will help you solve a wide range of problems quickly and accurately.

二次方程贯穿 IGCSE 数学课程的始终，从核心卷的基础因式分解，到扩展卷的配方法与隐藏二次型。牢固掌握二次方程的结构、解法及其图像特征，能帮助你快速、准确地解决大量题目。

1. Recognising a Quadratic Equation | 识别二次方程

A quadratic equation in one variable has the general form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The condition a ≠ 0 is essential because if a = 0 the equation becomes linear, not quadratic. Common examples include x² – 5x + 6 = 0, 2x² + 3x – 2 = 0 and 4x² – 9 = 0.

一元二次方程的一般形式为 ax² + bx + c = 0，其中 a、b、c 为常数且 a ≠ 0。a ≠ 0 这一条件至关重要，因为若 a = 0，方程就变为一次方程而非二次方程。常见例子有 x² – 5x + 6 = 0、2x² + 3x – 2 = 0 和 4x² – 9 = 0。

In IGCSE questions, the quadratic may not be given in standard form at first. You may need to expand brackets or move all terms to one side before identifying a, b and c correctly.

在 IGCSE 题目中，二次方程起初不一定给出标准形式。你可能需要先展开括号或将所有项移到同一边，才能正确识别 a、b 和 c。

For example, x(x – 3) = 4 becomes x² – 3x – 4 = 0 after expansion and rearrangement. Only then can you see that a = 1, b = -3 and c = -4.

例如，x(x – 3) = 4 经过展开和移项后变为 x² – 3x – 4 = 0。此时才能看出 a = 1，b = -3，c = -4。

2. Solving by Factorisation | 因式分解法

When a quadratic expression can be written as a product of two linear factors, we can set each factor equal to zero. For x² – 5x + 6 = 0, factorise to (x – 2)(x – 3) = 0, giving x = 2 or x = 3. Always check that the product of the constants gives c and their sum gives b.

当二次式可以写成两个一次因式的乘积时，我们可以令每个因式等于零。对于 x² – 5x + 6 = 0，因式分解为 (x – 2)(x – 3) = 0，得到 x = 2 或 x = 3。务必检验常数项的乘积等于 c，常数项之和等于 b。

To factorise x² + bx + c, look for two numbers whose product is c and whose sum is b. For x² + 7x + 12, the numbers are 3 and 4, so the factors are (x + 3)(x + 4).

要对 x² + bx + c 进行因式分解，需寻找两个数，使它们的乘积为 c，和为 b。对于 x² + 7x + 12，这两个数是 3 和 4，因此因式分解为 (x + 3)(x + 4)。

Once factorised, use the zero-product rule: if AB = 0, then A = 0 or B = 0. This gives the roots directly. Always check each solution by substituting it into the original equation.

因式分解后，使用零乘积法则：若 AB = 0，则 A = 0 或 B = 0。由此可直接得到根。始终将每个解代回原方程进行检验。

3. Difference of Two Squares | 平方差公式

A quadratic of the form a²x² – b² can be factorised as (ax + b)(ax – b). For example, 4x² – 9 = (2x + 3)(2x – 3), so 4x² – 9 = 0 gives x = -3/2 or x = 3/2.

形如 a²x² – b² 的二次式可以利用平方差公式分解为 (ax + b)(ax – b)。例如 4x² – 9 = (2x + 3)(2x – 3)，因此 4x² – 9 = 0 的解为 x = -3/2 或 x = 3/2。

This method is faster than trying to find two numbers that add to zero. Recognising the pattern is key: both terms must be perfect squares and there must be a minus sign between them.

这种方法比尝试寻找两个相加为零的数更快。识别这一模式很关键：两项都必须是完全平方数，并且它们之间必须是减号。

Do not attempt to apply the difference of two squares to a sum such as x² + 9. That expression does not factorise over the real numbers unless a question explicitly asks for complex roots, which is outside IGCSE.

不要把平方差公式用于 x² + 9 这样的和。除非题目明确要求复数根（超出 IGCSE 范围），否则该式在实数范围内不能因式分解。

4. Completing the Square | 配方法

Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. For x² + 6x + 2, take half of 6, square it: (6/2)² = 9. Write x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7. This form reveals the vertex of the graph and is useful when a quadratic cannot be factorised easily.

配方法将 ax² + bx + c 改写为 a(x + p)² + q 的形式。以 x² + 6x + 2 为例，取 6 的一半再平方：(6/2)² = 9。于是 x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7。这种形式能直接显示图像顶点，并且在二次式不易因式分解时非常有用。

To solve by completing the square, isolate the squared bracket. From (x + 3)² – 7 = 0, we get (x + 3)² = 7, so x + 3 = ±√7, giving x = -3 ± √7.

要用配方法解方程，需要将平方项单独放在一边。由 (x + 3)² – 7 = 0，得到 (x + 3)² = 7，于是 x + 3 = ±√7，最终 x = -3 ± √7。

When the coefficient of x² is not 1, factor it out first. For 2x² + 8x + 5, write 2(x² + 4x) + 5 = 2[(x + 2)² – 4] + 5 = 2(x + 2)² – 8 + 5 = 2(x + 2)² – 3.

当 x² 的系数不是 1 时，需要先将其提出来。对于 2x² + 8x + 5，可写为 2(x² + 4x) + 5 = 2[(x + 2)² – 4] + 5 = 2(x + 2)² – 8 + 5 = 2(x + 2)² – 3。

5. Quadratic Formula | 求根公式

For any quadratic equation ax² + bx + c = 0, the solutions are given by the quadratic formula. This method always works, provided a ≠ 0, and it is especially useful when factorisation is difficult.

对于任意二次方程 ax² + bx + c = 0，其解可由求根公式给出。只要 a ≠ 0，该方法始终有效，在因式分解困难时尤其有用。

x = (-b ± √(b² – 4ac)) / 2a

For 2x² + 3x – 2 = 0, identify a = 2, b = 3 and c = -2. Substituting gives x = (-3 ± √(9 + 16)) / 4 = (-3 ± 5) / 4, so x = 1/2 or x = -2.

对于 2x² + 3x – 2 = 0，确定 a = 2，b = 3，c = -2。代入公式得 x = (-3 ± √(9 + 16)) / 4 = (-3 ± 5) / 4，因此 x = 1/2 或 x = -2。

The quadratic formula should be quoted accurately in the exam. Keep the numerator bracketed if there is any risk of sign errors, and simplify the square root before dividing.

考试中应准确写出求根公式。如果容易出现符号错误，可将分子加上括号，并在除以 2a 前先化简平方根。

6. The Discriminant | 判别式

The expression D = b² – 4ac is called the discriminant. It tells us how many real roots a quadratic has: D > 0 gives two distinct real roots, D = 0 gives one repeated real root, and D < 0 gives no real roots.

表达式 D = b² – 4ac 称为判别式。它说明二次方程有多少个实根：D > 0 有两个不同实根，D = 0 有一个重根，D < 0 没有实根。

This is especially useful for questions that ask whether a line intersects a curve. For example, if the discriminant of the resulting quadratic is negative, the line and the curve do not meet.

这在判断直线与曲线是否相交的问题中尤为有用。例如，如果由两者联立所得二次方程的判别式为负，则直线与曲线没有交点。

You can also use the discriminant to show that a quadratic is always positive or always negative. If a > 0 and D < 0, the graph lies entirely above the x-axis, so the quadratic has no real roots and is positive for all x.

你还可以用判别式证明一个二次式恒为正或恒为负。如果 a > 0 且 D < 0，图像完全位于 x 轴上方，因此该二次方程没有实根，且对所有 x 都为正。

7. Graphs of Quadratic Functions | 二次函数图像

The graph of y = ax² + bx + c is 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. The line of symmetry is x = -b / 2a, and the roots are the x-intercepts when y = 0.

y = ax² + bx + c 的图像是一条抛物线。若 a > 0，抛物线开口向上并且有最低点；若 a < 0，开口向下并且有最高点。对称轴为 x = -b / 2a，方程的根是 y = 0 时图像与 x 轴交点的横坐标。

The turning point can be found by completing the square. In the form y = a(x + p)² + q, the vertex is at (-p, q). For y = (x + 3)² – 7, the vertex is (-3, -7).

顶点可以通过配方法求出。在 y = a(x + p)² + q 的形式中，顶点坐标为 (-p, q)。对于 y = (x + 3)² – 7，顶点为 (-3, -7)。

Sketching a parabola only requires a few key features: the shape determined by a, the y-intercept at (0, c), the roots if they exist, and the vertex. These features are usually enough to answer graph-based questions.

绘制抛物线只需把握几个关键特征：由 a 决定开口方向、y 轴截距 (0, c)、存在的根以及顶点。这些特征通常足以解答图像类题目。

8. Solving by Graphing | 图像法

To solve a quadratic equation graphically, plot y = ax² + bx + c and read the x-coordinates where the graph crosses the x-axis. If the graph just touches the x-axis, there is one repeated root; if it does not cross the x-axis, there are no real roots.

要用图像法求解二次方程，可先画出 y = ax² + bx + c，然后读取图像与 x 轴交点的横坐标。如果图像刚好接触 x 轴，则有一个重根；如果图像不与 x 轴相交，则没有实根。

Graphical methods are often used in problems involving simultaneous equations, where one equation is quadratic and the other is linear. The points of intersection of the line and the parabola give the solutions.

图像法常用于涉及联立方程的问题中，其中一个方程是二次方程，另一个是一次方程。直线与抛物线的交点坐标即为方程组的解。

When reading solutions from a graph, give the best possible estimate unless the roots are exact. The examiner expects reasonable accuracy, not unattainable precision from a hand-drawn sketch.

从图像上读取解时，除非根是精确值，否则应给出最合理的估计。考官期望的是合理的准确性，而不是手工草图不可能达到的精度。

9. Applications and Word Problems | 应用题与文字题

Many IGCSE problems ask you to form a quadratic from a real-life situation such as area, projectile motion, or profit. Define the unknown carefully, write an equation in standard form, solve it, and reject any solution that does not make sense in context, such as a negative length or time.

许多 IGCSE 题目要求根据实际情境建立二次方程，例如面积、抛体运动或利润问题。要仔细设未知数，将方程写成标准形式，求解后舍去在背景下不合理的解，如负的长度或时间。

Example: the product of two consecutive positive integers is 72. Let the smaller be n, so n(n + 1) = 72. This becomes n² + n – 72 = 0, which factorises as (n + 9)(n – 8) = 0, giving n = 8 because n = -9 is not positive.

示例：两个连续正整数的乘积为 72。设较小的数为 n，则 n(n + 1) = 72。方程化为 n² + n – 72 = 0，因式分解为 (n + 9)(n – 8) = 0，因此 n = 8，因为 n = -9 不是正数。

Always translate words into algebra step by step. Phrases such as ‘the area is 48 cm²’ often lead to a product of expressions, while ‘the sum of the squares is 25’ leads to a squared term.

始终将文字逐步翻译成代数式。诸如“面积为 48 cm²”这样的表述通常会导出两个式子的乘积，而“平方和为 25”则会导出平方项。

10. Common Mistakes | 常见错误

Watch out for these common errors: forgetting the ± sign in the quadratic formula, missing the a ≠ 0 condition, incorrect factorisation signs, and mixing up b and b² – 4ac. Always substitute your solutions back into the original equation to check them.

要警惕以下常见错误：在使用求根公式时漏掉 ± 号，遗漏 a ≠ 0 的条件，因式分解时符号出错，以及混淆 b 与 b² – 4ac。务必把解代回原方程进行检验。

Forgetting to set each factor to zero after factorising.
Factorising后忘记令每个因式等于零。
Losing a root when dividing both sides of an equation by x.
在方程两边同时除以 x 时丢失一个根。
Using the quadratic formula with incorrect values of a, b and c.
使用求根公式时 a、b、c 的值代入错误。

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Mastering Quadratic Equations for IGCSE Mathematics | IGCSE 数学：二次方程完全攻略