IGCSE Maths G-2 Student Book 600: Quadratic Equations | IGCSE 数学 G-2 学生用书 600：二次方程

📚 IGCSE Maths G-2 Student Book 600: Quadratic Equations | IGCSE 数学 G-2 学生用书 600：二次方程

Quadratic equations are a central topic in the IGCSE Mathematics syllabus. In this G-2 Student Book 600 unit, we focus on the standard form, solving methods, graphs and real-life applications of quadratic equations. A confident command of these techniques is essential for success in both Core and Extended papers.

二次方程是 IGCSE 数学大纲中的核心主题。在本次 G-2 学生用书 600 单元中，我们重点学习二次方程的标准形式、求解方法、图像以及实际应用。熟练掌握这些技巧对于在 Core 和 Extended 试卷中取得成功至关重要。

1. Understanding Quadratic Equations | 理解二次方程

A quadratic equation in one variable is any equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of x is 2, which is why it is called quadratic. If a = 0, the equation becomes linear, not quadratic.

一元二次方程是可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 是常数，且 a ≠ 0。x 的最高次数是 2，因此称为二次方程。如果 a = 0，方程就变成了一次方程，而不是二次方程。

Quadratic equations can have two real solutions, one repeated real solution, or no real solutions depending on the value of the discriminant. Recognising the standard form is the first step before choosing a solving strategy.

根据判别式的值，二次方程可以有两个实数解、一个重复实数解或没有实数解。在选择求解策略之前，识别标准形式是第一步。

2. Solving by Factorising | 因式分解法

Factorising is often the fastest method when the quadratic expression can be written as a product of two linear brackets. For example, x² + 5x + 6 = 0 can be factorised as (x + 2)(x + 3) = 0. Since the product is zero, either x + 2 = 0 or x + 3 = 0, giving x = -2 or x = -3.

当二次式可以写成两个一次括号的乘积时，因式分解通常是最快的方法。例如，x² + 5x + 6 = 0 可以因式分解为 (x + 2)(x + 3) = 0。由于乘积为零，所以 x + 2 = 0 或 x + 3 = 0，得到 x = -2 或 x = -3。

Always rearrange the equation to the form ax² + bx + c = 0 before factorising.
Always rearrange the equation to the form ax² + bx + c = 0 before factorising. 中文：在因式分解前，一定要把方程整理成 ax² + bx + c = 0 的形式。
Check your brackets by expanding them back to ensure the middle term is correct.
Check your brackets by expanding them back to ensure the middle term is correct. 中文：展开括号进行检验，确保中间项正确。

x² + 7x + 12 = 0 → (x + 3)(x + 4) = 0 → x = -3 or x = -4

3. Solving by Completing the Square | 配方法

Completing the square transforms a quadratic into the form (x + p)² = q, which can then be solved by taking square roots. For example, x² + 6x + 5 = 0 can be rewritten as (x + 3)² – 4 = 0, so (x + 3)² = 4. Taking the square root gives x + 3 = ±2, so x = -1 or x = -5.

配方法将二次方程转化为 (x + p)² = q 的形式，然后通过开平方求解。例如，x² + 6x + 5 = 0 可以改写为 (x + 3)² – 4 = 0，因此 (x + 3)² = 4。开平方得到 x + 3 = ±2，所以 x = -1 或 x = -5。

This method is especially useful when the quadratic does not factorise easily or when you need to find the turning point of a parabola. Remember to halve the coefficient of x, square it, and adjust the constant term.

当二次式不容易因式分解，或者需要求抛物线的顶点时，这种方法尤其有用。记住取 x 系数的一半，平方，然后调整常数项。

x² + bx + c = 0 → (x + b/2)² – (b/2)² + c = 0

4. The Quadratic Formula | 二次公式

The quadratic formula can solve any quadratic equation, even when factorising is difficult or impossible. The formula is x = (-b ± √(b² – 4ac)) / (2a), where a, b and c are the coefficients from ax² + bx + c = 0. Always substitute carefully to avoid sign errors.

二次公式可以求解任何二次方程，即使因式分解困难或不可能。公式为 x = (-b ± √(b² – 4ac)) / (2a)，其中 a、b、c 是来自 ax² + bx + c = 0 的系数。代入时要仔细，以避免符号错误。

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

例如，在 2x² + 3x – 2 = 0 中，a = 2，b = 3，c = -2。代入公式得到 x = (-3 ± √(9 + 16)) / 4 = (-3 ± 5) / 4，所以 x = 1/2 或 x = -2。

x = (-b ± √(b² – 4ac)) / (2a)

5. 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 y-intercept is at (0, c), and the x-intercepts are the real roots of ax² + bx + c = 0.

y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上，有最低点；如果 a < 0，抛物线开口向下，有最高点。y 轴截距在 (0, c)，x 轴截距是 ax² + bx + c = 0 的实数根。

The turning point, or vertex, can be found by completing the square. For y = (x – h)² + k, the vertex is at (h, k). Drawing a smooth curve through the intercepts and vertex helps you visualise the solutions and the symmetry of the graph.

顶点可以通过配方法求得。对于 y = (x – h)² + k，顶点位于 (h, k)。通过截距和顶点画出光滑曲线，有助于直观理解方程的解以及图像的对称性。

y = x² – 4x + 3	y = (x – 2)² – 1	Vertex: (2, -1)
y = -x² + 2x + 3	y = -(x – 1)² + 4	Vertex: (1, 4)

6. The Discriminant | 判别式

The discriminant is the expression Δ = b² – 4ac inside the quadratic formula. It tells you the nature of the roots without solving the equation fully. If Δ > 0, there are two distinct real roots. If Δ = 0, there is one repeated real root. If Δ < 0, there are no real roots, but two complex roots at advanced level.

判别式是二次公式中的表达式 Δ = b² – 4ac。它可以在不完整求解方程的情况下判断根的性质。如果 Δ > 0，则有两个不同的实数根；如果 Δ = 0，则有一个重复实数根；如果 Δ < 0，则没有实数根，但在更高层次中有两个复数根。

In IGCSE, questions often ask you to find the value of k for which a quadratic has equal roots or no real roots. Set Δ = 0 for equal roots, Δ > 0 for two real roots, and Δ < 0 for no real roots.

在 IGCSE 中，题目经常要求你找出使二次方程有相等实根或没有实根的 k 值。有相等实根时令 Δ = 0，有两个实根时令 Δ > 0，没有实根时令 Δ < 0。

Δ = b² – 4ac

7. Roots, Intercepts and Turning Point | 根、截距与顶点

The roots of a quadratic equation are the x-values where the parabola crosses the x-axis. The y-intercept is the value of y when x = 0, which is simply c in y = ax² + bx + c. The turning point lies on the axis of symmetry at x = -b / (2a), and its y-coordinate can be found by substitution or by completing the square.

二次方程的根是抛物线与 x 轴交点的 x 值。y 轴截距是当 x = 0 时 y 的值，在 y = ax² + bx + c 中就是 c。顶点位于对称轴 x = -b / (2a) 上，其 y 坐标可以通过代入或配方法求得。

For y = x² – 6x + 8, the roots are x = 2 and x = 4, the y-intercept is 8, and the axis of symmetry is x = 3. Substituting x = 3 gives y = 9 – 18 + 8 = -1, so the turning point is (3, -1).

对于 y = x² – 6x + 8，根为 x = 2 和 x = 4，y 轴截距为 8，对称轴为 x = 3。代入 x = 3 得到 y = 9 – 18 + 8 = -1，所以顶点为 (3, -1)。

8. Word Problems Involving Quadratics | 二次方程应用题

Many real-life problems lead to quadratic equations, such as finding the dimensions of a rectangle with a fixed area, projectile motion, or profit maximisation. Identify the unknown, form a quadratic equation from the given conditions, solve it, and then interpret the solution in context.

许多实际问题都会引出二次方程，例如求固定面积矩形的尺寸、抛体运动或利润最大化。先确定未知数，根据已知条件建立二次方程，求解后在具体情境中解释解的意义。

For example, a rectangle has length 3 cm more than its width, and its area is 40 cm². Let the width be x cm, so the length is (x + 3) cm. The equation is x(x + 3) = 40, or x² + 3x – 40 = 0. Factorising gives (x + 8)(x – 5) = 0, so x = 5 cm is the valid positive solution.

例如，一个矩形的长比宽多 3 cm，面积为 40 cm²。设宽为 x cm，则长为 (x + 3) cm。方程为 x(x + 3) = 40，即 x² + 3x – 40 = 0。因式分解得到 (x + 8)(x – 5) = 0，因此 x = 5 cm 是有效的正解。

9. Common Mistakes | 常见错误

One common mistake is forgetting to set the equation to zero before factorising. Another is losing the negative sign when substituting into the quadratic formula. Students also sometimes divide both sides by x, which can eliminate a valid root, such as x = 0.

一个常见错误是在因式分解前忘记把方程整理为零。另一个是在代入二次公式时丢失负号。学生有时还会两边除以 x，这可能会消去一个有效根，例如 x = 0。

Do not divide by x unless you are certain x ≠ 0.
Do not divide by x unless you are certain x ≠ 0. 中文：除非确定 x ≠ 0，否则不要两边除以 x。
Always write the formula with brackets around -b to avoid sign errors.
Always write the formula with brackets around -b to avoid sign errors. 中文：写出公式时始终在 -b 外加括号，以避免符号错误。

When completing the square, remember that adding and subtracting the same value inside the bracket keeps the equation balanced. Check your final roots by substituting them back into the original equation.

配方法时，记住在括号内加上并减去同一个值可以保持方程平衡。将最终得到的根代回原方程进行检验。

10. Exam-Style Practice | 真题风格练习

Try solving 3x² – 5x – 2 = 0 using the quadratic formula. First identify a = 3, b = -5, c = -2. Then x = (5 ± √(25 + 24)) / 6 = (5 ± 7) / 6, so x = 2 or x = -1/3. Always write the final answer as a simplified fraction or decimal as required.

尝试用二次公式求解 3x² – 5x – 2 = 0。首先确定 a = 3，b = -5，c = -2。于是 x = (5 ± √(25 + 24)) / 6 = (5 ± 7) / 6，所以 x = 2 或 x = -1/3。按照题目要求，将最终答案写成最简分数或小数。

Now try this exam-style problem: Find the values of k for which x² + kx + 9 = 0 has one repeated real root. For a repeated root, Δ = b² – 4ac = k² – 36 = 0, so k = ±6. This type of discriminant question is very common in IGCSE papers.

现在尝试这道真题风格的问题：求使 x² + kx + 9 = 0 有一个重复实数根的 k 值。因为重复实根要求 Δ = b² – 4ac = k² – 36 = 0，所以 k = ±6。这类判别式问题在 IGCSE 试卷中非常常见。

Practise a mix of factorising, completing the square, the quadratic formula, and graph interpretation. Time yourself on past paper questions to build speed and accuracy before the real exam.

混合练习因式分解法、配方法、二次公式和图像解读。在真实考试前，用往年真题进行计时练习，以提高速度和准确性。

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IGCSE Maths G-2 Student Book 600: Quadratic Equations | IGCSE 数学 G-2 学生用书 600：二次方程