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Mastering Quadratic Equations for IGCSE | 掌握IGCSE二次方程

2026-08-30 05:05 · 本文部分内容由AI辅助创作

📚 Mastering Quadratic Equations for IGCSE | 掌握IGCSE二次方程

Quadratic equations appear in almost every IGCSE Mathematics paper. This article breaks down the essential methods and tips you need to solve them confidently and accurately.

二次方程几乎出现在每份IGCSE数学试卷中。本文为你拆解必备的解法与技巧,帮助你自信、准确地解题。


1. What Is a Quadratic Equation? | 什么是二次方程

that can be written in the form ax² + bx + c = 0, where a, b and c are numbers and a ≠ 0.

二次方程是任何可以写成 ax² + bx + c = 0 形式的方程,其中 a、b、c 是数字,且 a ≠ 0。

The highest power of x is 2, which is why it is called "quadratic". The graph of a quadratic equation is a parabola.

x 的最高次幂是 2,因此称为“二次”。二次方程的图像是一条抛物线。


2. Standard Form and Key Terms | 标准形式与关键术语

t a is the quadratic coefficient, b is the linear coefficient, and c is the constant term.

二次方程的标准形式是 ax² + bx + c = 0。其中 a 是二次项系数,b 是一次项系数,c 是常数项。

For example, in 2x² − 3x + 5 = 0, we have a = 2, b = −3 and c = 5.

例如,在 2x² − 3x + 5 = 0 中,a = 2,b = −3,c = 5。

A root (or solution) of a quadratic equation is a value of x that makes the equation equal to 0.

二次方程的根(或解)是使方程等于 0 的 x 值。


3. Solving by Factorisation | 用因式分解求解

If a quadratic expression can be factorised, we can solve it by setting each factor equal to zero.

如果一个二次表达式可以因式分解,我们可以令每个因式等于零来求解。

Solve x² − 5x + 6 = 0 by factorisation.

用因式分解解 x² − 5x + 6 = 0。

x² − 5x + 6 = (x − 2)(x − 3) = 0

So x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.

因此 x − 2 = 0 或 x − 3 = 0,得到 x = 2 或 x = 3。

Always rearrange the equation into the form ax² + bx + c = 0 before factorising.

因式分解前务必先将方程整理成 ax² + bx + c = 0 的形式。


4. The Quadratic Formula | 二次公式

any quadratic equation ax² + bx + c = 0, the solutions are given by the quadratic formula:

对于任意二次方程 ax² + bx + c = 0,其解由二次公式给出:

x = (−b ± √b² − 4ac) / (2a)

This formula works for all quadratics, including those that cannot be factorised easily.

该公式适用于所有二次方程,包括那些不易因式分解的方程。

Example: Solve x² + 3x − 2 = 0 using the formula.

例:用公式解 x² + 3x − 2 = 0。

Here a = 1, b = 3, c = −2. Substitute into the formula:

这里 a = 1,b = 3,c = −2。代入公式:

x = (−3 ± √(3² − 4×1×(−2))) / (2×1) = (−3 ± √17) / 2

So x ≈ 0.562 or x ≈ −3.562.

因此 x ≈ 0.562 或 x ≈ −3.562。


5. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + p)² + q.

配方法将二次式改写为 a(x + p)² + q 的形式。

Start with x² + 6x + 5. Take half of 6, which is 3, and write:

从 x² + 6x + 5 开始。取 6 的一半,即 3,写为:

x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4

To solve (x + 3)² − 4 = 0, we get (x + 3)² = 4, so x + 3 = ±2, giving x = −1 or x = −5.

解 (x + 3)² − 4 = 0,得 (x + 3)² = 4,所以 x + 3 = ±2,即 x = −1 或 x = −5。

When the coefficient of x² is not 1, factor it out first before completing the square.

当 x² 的系数不为 1 时,先将该系数提取出来,再进行配方。


6. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root: b² − 4ac.

判别式是二次公式根号下的部分:b² − 4ac。

If b² − 4ac > 0, the equation has two distinct real roots.

若 b² − 4ac > 0,方程有两个不同实数根。

If b² − 4ac = 0, the equation has exactly one real root (a repeated root).

若 b² − 4ac = 0,方程恰有一个实数根(重根)。

If b² − 4ac < 0, the equation has no real roots.

若 b² − 4ac < 0,方程没有实数根。

x² + 2x + 5 = 0, the discriminant is 2² − 4×1×5 = −16 < 0, so there are no real solutions.

例:对于 x² + 2x + 5 = 0,判别式 = 2² − 4×1×5 = −16 < 0,因此没有实数解。


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7. Sketching Graphs of Quadratics | 二次函数图像

y = ax² + bx + c is a parabola. If a > 0, it opens upwards; if a < 0, it opens downwards.

y = ax² + bx + c 的图像是一条抛物线。若 a > 0,开口向上;若 a < 0,开口向下。

The roots of the equation are the x-intercepts of the graph.

方程的根就是图像与 x 轴的交点。

by x = −b / (2a). The y-coordinate can be found by substituting this x into the equation.

顶点的 x 坐标为 x = −b / (2a)。将该 x 值代入方程即可求得 y 坐标。

The y-intercept is the value of c, which is the point (0, c).

y 轴截距为 c,即点 (0, c)。


8. Applications: Word Problems | 应用题

Quadratic equations often arise in geometry, physics and motion problems.

二次方程经常出现在几何、物理和运动问题中。

Example: A rectangle has length (x + 3) cm and width (x − 1) cm. Its area is 12 cm². Find x.

例:一个长方形的长为 (x + 3) cm,宽为 (x − 1) cm,面积为 12 cm²。求 x。

(x + 3)(x − 1) = 12 ⟹ x² + 2x − 3 = 12 ⟹ x² + 2x − 15 = 0

Factorise: (x + 5)(x − 3) = 0. Since length cannot be negative, x = 3.

因式分解:(x + 5)(x − 3) = 0。由于长度不能为负,所以 x = 3。

Always check that your answers make sense in the original context.

务必检查答案是否满足原题的实际意义。


9. Common Mistakes and Exam Tips | 常见错误与考试技巧

One common mistake is forgetting to rearrange the equation to zero before factorising.

一个常见错误是因式分解前忘记把方程整理为等于零的形式。

Another mistake is applying the quadratic formula with the wrong sign for b.

另一个错误是在使用二次公式时把 b 的符号弄错。

Learn to recognise which method is fastest: factorisation for simple cases, formula for standard use, completing the square for finding the vertex.

学会判断哪种方法最快:简单情形用因式分解,常规情况用公式,求顶点时用配方法。


10. Practice Questions | 练习题

Try these problems on your own, then check your answers.

先独立尝试以下题目,再核对答案。

1. Solve x² − 7x + 12 = 0.

1. 解 x² − 7x + 12 = 0。

2. Solve 2x² + 5x − 3 = 0 using the quadratic formula.

2. 用二次公式解 2x² + 5x − 3 = 0。

3. Express x² + 8x + 3 in the form (x + p)² + q.

3. 将 x² + 8x + 3 写成 (x + p)² + q 的形式。

4. For what value of k does x² + kx + 9 have exactly one real root?

4. k 为何值时,x² + kx + 9 恰有一个实数根?

Answers: 1. x = 3 or x = 4. 2. x = 0.5 or x = −3. 3. (x + 4)² − 13. 4. k = ±6.

答案:1. x = 3 或 x = 4。2. x = 0.5 或 x = −3。3. (x + 4)² − 13。4. k = ±6。


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