Solving Quadratic Equations | 二次方程求解

📚 Solving Quadratic Equations | 二次方程求解

Quadratic equations are among the most important topics in IGCSE Mathematics. They appear in algebra, coordinate geometry, problem-solving and graph questions, and a strong command of them is essential for a high grade. In this revision guide, we will cover the standard form, three core solving methods, the discriminant, graphical interpretation, common mistakes, and exam strategies.

二次方程是 IGCSE 数学中最重要的考点之一，广泛出现在代数、坐标几何、应用题和图像题中。熟练掌握二次方程是取得高分的关键。本复习指南将系统讲解标准形式、三种核心解法、判别式、图像意义、常见错误以及考试策略。

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

a, b and c are constants and a ≠ 0. If a = 0, the equation becomes linear, not quadratic.

二次方程是最高次数为 2 的多项式方程，即变量的最高次幂为 2。它的一般形式为 ax² + bx + c = 0，其中 a、b、c 为常数，且 a ≠ 0。若 a = 0，则方程退化为一次方程，而不是二次方程。

ations. The coefficient a is called the leading coefficient, and it must not be zero for the equation to be genuinely quadratic.

例如，x² − 4x + 3 = 0 是二次方程，因为 x 的最高次数为 2。而 2x + 1 = 0（一次方程）或 x³ − 1 = 0（三次方程）都不是二次方程。系数 a 称为首项系数，若 a 为零，方程就不再是真正的二次方程。

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2. Standard Form and Identifying Coefficients | 标准形式与系数识别

he same side so that one side equals zero, then collecting like terms. Once the equation is in standard form, you can identify a, b and c directly.

在求解任何二次方程之前，都要先将方程整理成标准形式 ax² + bx + c = 0，即把所有项移到同一边，使另一边为零，然后合并同类项。整理成标准形式后，就可以直接读出 a、b、c 的值。

lose attention to signs; a very common error is to copy the wrong sign for b when rearranging.

示例：整理方程 2x² + 3 = 7x − 1。两边同时减去 7x 并加上 1，得 2x² − 7x + 4 = 0。因此 a = 2，b = −7，c = 4。要特别注意符号，一个常见错误就是在移项时把 b 的符号抄错。

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3. Solving by Factorisation | 因式分解法求解

if we can rewrite ax² + bx + c as (x + m)(x + n), the roots are simply x = −m and x = −n.

当二次式可以写成两个一次因式的乘积时，因式分解是最快的解法。它依赖于零乘积性质：若 p × q = 0，则 p = 0 或 q = 0。因此，如果我们把 ax² + bx + c 改写成 (x + m)(x + n)，根就是 x = −m 和 x = −n。

Follow these steps:

解题步骤如下：

• Write the equation in the form ax² + bx + c = 0.
  将方程写成 ax² + bx + c = 0 的形式。
• Factorise the quadratic into two brackets.
  将二次式分解为两个括号相乘的形式。
• Set each bracket equal to zero and solve the linear equations.
  令每个括号等于零，并解所得的一次方程。

y to +6 and add to −5 are −2 and −3, so we write (x − 2)(x − 3) = 0. Hence x = 2 or x = 3.

示例：解方程 x² − 5x + 6 = 0。乘积为 +6 且和为 −5 的两个数是 −2 和 −3，因此写成 (x − 2)(x − 3) = 0。所以 x = 2 或 x = 3。

x² − 5x + 6 = 0 ⇒ (x − 2)(x − 3) = 0 ⇒ x = 2 or x = 3

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4. Completing the Square | 配方法求解

Completing the square rewrites a quadratic as a perfect square plus a constant. For an expression in the form x² + bx, add and subtract (b/2)². This method is useful when factorisation is not obvious, and it is also essential for finding the turning point of a quadratic graph.

配方法将二次式改写成一个完全平方加上一个常数。对于形如 x² + bx 的表达式，需要加上并减去 (b/2)²。当因式分解不明显时，配方法非常有用；同时它也是求二次函数图像顶点坐标的必要工具。

• If a ≠ 1, divide both sides of the equation by a so that the coefficient of x² is 1.
  若 a ≠ 1，先将方程两边同除以 a，使 x² 的系数为 1。
• Move the constant term to the right-hand side.
  将常数项移到等号右边。
• Add (b/2)² to both sides of the equation.
  在方程两边同时加上 (b/2)²。
• Write the left side as (x + b/2)² and then take the square root of both sides.
  将左边写成 (x + b/2)²，再对两边开平方。

9 to both sides: x² + 6x + 9 = 7, which gives (x + 3)² = 7. Taking square roots: x + 3 = ±√7, so x = −3 ± √7.

示例：解方程 x² + 6x + 2 = 0。这里 b = 6，所以 (b/2)² = 9。两边加 9，得 x² + 6x + 9 = 7，即 (x + 3)² = 7。开平方得 x + 3 = ±√7，因此 x = −3 ± √7。

x² + 6x + 2 = 0 ⇒ (x + 3)² = 7 ⇒ x = −3 ± √7

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5. The Quadratic Formula | 求根公式法求解

eneral form ax² + bx + c = 0. You should memorise it and be able to substitute accurately.

求根公式适用于所有二次方程，包括那些难以直接因式分解的情形。它通过对一般形式 ax² + bx + c = 0 配方推导而来。你需要牢记该公式，并能准确代入计算。

x = (−b ± √b² − 4ac) / (2a)

t: b² − 4ac = 49 − 24 = 25. Then substitute: x = (7 ± 5) / 4. This gives x = 3 or x = 1/2.

示例：用公式解方程 2x² − 7x + 3 = 0。这里 a = 2，b = −7，c = 3。先计算判别式：b² − 4ac = 49 − 24 = 25。再代入公式：x = (7 ± 5) / 4，得到 x = 3 或 x = 1/2。

x = (−(−7) ± √25) / (2 × 2) = (7 ± 5) / 4 ⇒ x = 3 or x = 1/2

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6. The Discriminant | 判别式及其应用

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