Solving Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 解二次方程：因式分解、配方法与求根公式

📚 Solving Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 解二次方程：因式分解、配方法与求根公式

Quadratic equations are a central topic in IGCSE Mathematics. They appear in pure algebra, in coordinate geometry when finding intersections, and in real-life problems such as area, speed and projectile motion. Mastering factorising, completing the square and the quadratic formula will help you solve problems accurately and choose the quickest method in an exam.

二次方程是 IGCSE 数学的核心内容之一。它出现在纯代数、坐标几何中求交点，以及面积、速度和抛体运动等实际应用问题中。掌握因式分解法、配方法和求根公式法，能帮助你准确解题，并在考试中选择最快的方法。

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

d a ≠ 0. If a were equal to zero, the x² term would disappear and the equation would become linear rather than quadratic. The highest power of x is 2, which is why the graph of a quadratic function is a parabola.

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 为常数，且 a ≠ 0。如果 a 等于零，x² 项就会消失，方程就变成了一次方程而不是二次方程。x 的最高次数是 2，因此二次函数的图像是一条抛物线。

ax² + bx + c = 0, a ≠ 0

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2. Why Standard Form Matters | 为什么标准形式很重要

s essential because factorising and the quadratic formula both rely on correct values of a, b and c.

在解二次方程之前，一定要先将其整理成标准形式 ax² + bx + c = 0。把所有项移到方程的一边并完全化简。例如，3x² = 5x − 2 应先化为 3x² − 5x + 2 = 0。这一步非常关键，因为因式分解法和求根公式法都依赖于正确的 a、b、c 值。

3x² = 5x − 2 → 3x² − 5x + 2 = 0

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3. Solving by Factorising | 因式分解法

) = 0. Since the product is zero, at least one factor must be zero, giving x = 3 or x = 4.

当二次式可以整齐地分解为两个一次因式时，因式分解法通常是最快的方法。对于 x² − 7x + 12 = 0，我们需要找到两个数，它们的乘积为 +12，并且和为 −7。这两个数是 −3 和 −4，因此方程变为 (x − 3)(x − 4) = 0。由于乘积为零，至少有一个因式为零，所以 x = 3 或 x = 4。

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

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4. Solving by Completing the Square | 配方法

ite the expression as (x + 4)² − 16 + 9 = 0. This simplifies to (x + 4)² = 7, so x + 4 = ±√7. The solutions are x = −4 ± √7.

配方法把二次式改写为 a(x + p)² + q 的形式。这种方法在求抛物线的顶点或精确的无理数根时特别有用。对于 x² + 8x + 9 = 0，取 x 系数的一半为 4，平方得到 16，将表达式改写为 (x + 4)² − 16 + 9 = 0。化简得 (x + 4)² = 7，因此 x + 4 = ±√7。解为 x = −4 ± √7。

x² + 8x + 9 = 0
(x + 4)² − 16 + 9 = 0
(x + 4)² = 7
x = −4 ± √7

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5. Solving by the Quadratic Formula | 求根公式法

solve any quadratic equation directly. It states that if ax² + bx + c = 0, then x = (−b ± √b² − 4ac ² − 3x − 5 = 0, we have a = 2, b = −3 and c = −5. Substituting these values gives x = (3 ± √49) / 4, so x = 10/4 = 2.5 or x = −4/4 = −1.

求根公式法可以直接求解任何二次方程。公式为：若 ax² + bx + c = 0，则 x = (−b ± √b² − 4ac) / 2a。对于 2x² − 3x − 5 = 0，a = 2，b = −3，c = −5。代入这些值得到 x = (3 ± √49) / 4，所以 x = 10/4 = 2.5 或 x = −4/4 = −1。

x = (−b ± √b² − 4ac) / 2a
2x² − 3x − 5 = 0 → x = 2.5 or x = −1

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6. The Discriminant and Nature of Roots | 判别式与根的性质

ated real root, sometimes called a double root. If Δ < 0, there are no real roots, although two complex roots exist in advanced mathematics.

求根公式中平方根内的式子 b² − 4ac 叫做判别式，通常记作 Δ。它可以在不解方程的情况下判断根的性质。若 Δ > 0，方程有两个不同的实根；若 Δ = 0，方程有一个重根，也称作二重根；若 Δ < 0，方程没有实根，但在高等数学中存在两个复数根。

Δ = b² − 4ac	Discriminant | 判别式
Δ > 0	Two distinct real roots | 两个不同实根
Δ = 0	One repeated real root | 一个重根
Δ < 0	No real roots | 无实根

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7. Choosing the Best Method | 选择最佳方法

r an exact surd answer, completing the square is useful. If the coefficients are large, awkward or do not factorise easily, the quadratic formula is the most reliable option.

在 IGCSE 考试中，选择合适的方法可以节省时间并减少错误。如果二次式有简单的整数因式，因式分解法通常最快。如果方程已经是 x² + bx + c = 0 的形式，并且需要求顶点或精确的无理数根，配方法很有用。如果系数较大、较复杂或不容易因式分解，求根公式法是最可靠的选择。

• Factorising: quickest for simple integer roots | 因式分解法：有简单整数根时最快
• Completing the square: best for vertex and surd form | 配方法：适合求顶点和根式形式
• Quadratic formula: works for every quadratic equation | 求根公式法：适用于所有二次方程

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8. Common Mistakes to Avoid | 常见错误

ys remember the ± sign when taking the square root in completing the square or the quadratic formula.

许多分数是因为可以避免的错误而丢失的。在确定 a、b、c 之前，不要忘记把方程整理成标准形式。除非你确定 x ≠ 0，否则不要将方程两边同时除以 x，因为这样可能会丢掉 x

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