📚 Completing the Square and the Quadratic Formula | 配方法与二次公式
In IGCSE Mathematics, quadratic equations appear constantly in algebra, graphs, and problem-solving. This revision guide explains two powerful algebraic methods: completing the square and the quadratic formula, showing how they are linked and when to use each one.
在 IGCSE 数学中，二次方程在代数、图像和实际问题中频繁出现。本复习指南讲解两种重要的代数方法——配方法与二次公式，说明它们之间的联系以及各自的适用场景。
1. Quadratic Equations: Standard Form | 二次方程的标准式
A quadratic equation in one variable is any equation that can be written in the standard form ax² + bx + c = 0 0. The highest power of x is 2, which is why the graph of y = ax² + bx + c is a parabola.
一元二次方程是指可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。x 的最高次数是 2，因此 y = ax² + bx + c 的图像是一条抛物线。
ax² + bx + c = 0, a ≠ 0
Solutions of the equation are called roots, and they correspond to the x-intercepts of the parabola. A quadratic equation can have two distinct real roots, one repeated real root, or no real roots.
方程的解称为根，它们对应于抛物线与 x 轴的交点。二次方程可能有两个不相等的实根、一个重根或没有实数根。
2. Why Completing the Square Works | 配方法的原理
Completing the square rewrites a quadratic expression of the form x² + bx + c into the form (x + p)² + q. Since a squared term is always non-negative, this form immediately reveals the minimum or maximum value of the expression and the coordinates of the turning point.
配方法将形如 x² + bx + c 的二次式改写为 (x + p)² + q 的形式。由于平方项总是非负的，这种形式可以立即揭示表达式的最小值或最大值以及顶点坐标。
². To complete the square for x² + bx, we add and subtract (b/2)², because the middle term 2d must equal b.
关键恒等式来自完全平方的展开：(x + d)² = x² + 2dx + d²。为了对 x² + bx 配方，我们需要加上并减去 (b/2)²，因为中间项 2d 必须等于 b。
x² + bx = (x + b/2)² − (b/2)²
3. Step-by-Step Completing the Square | 配方法的步骤
For a quadratic expression with leading coefficient 1, follow these steps: write x² + bx + c, find half of b, square it, add and subtract that square, then factor the perfect square and simplify the constants.
对于首项系数为 1 的二次式，可按照以下步骤进行：写出 x² + bx + c，求出 b 的一半并平方，加上再减去这个平方数，然后因式分解完全平方项并化简常数项。
- Step 1: Identify b from x² + bx + c. 步骤一：从 x² + bx + c 中找出 b。
- Step 2: Compute (b/2)². 步骤二：计算 (b/2)²。
- Step 3: Rewrite as x² + bx + (b/2)² − (b/2)² + c. 步骤三：改写为 x² + bx + (b/2)² − (b/2)² + c。
- Step 4: Factor the first three terms as (x + b/2)² and combine constants. 步骤四：将前三项因式分解为 (x + b/2)² 并合并常数。
If the leading coefficient is not 1, first factor it out from the x² and x terms, then complete the square inside the brackets.
如果首项系数不是 1，应先将 x² 项和 x 项的系数提取出来，再在括号内进行配方。
4. Worked Example: Leading Coefficient 1 | 首项系数为1的例题
Half of 6 is 3, and 3² = 9, so we add and subtract 9: x² + 6x + 9 − 9 + 5 = (x + 3)² − 4.
对 x² + 6x + 5 进行配方。6 的一半是 3，3² = 9，因此加上并减去 9：x² + 6x + 9 − 9 + 5 = (x + 3)² − 4。
x² + 6x + 5 = (x + 3)² − 4
e x² + 6x + 5 = 0 using this form, set (x + 3)² − 4 = 0, so (x + 3)² = 4, giving x + 3 = ±√4, hence x = −3 ± 2, so x = −1 or x = −5.
这表明 y = x² + 6x + 5 的顶点是 (−3, −4)。要用这种形式解方程 x² + 6x + 5 = 0，令 (x + 3)² − 4 = 0，得到 (x + 3)² = 4，因此 x + 3 = ±√4，从而 x = −3 ± 2，即 x = −1 或 x = −5。
5. Worked Example: Leading Coefficient Not 1 | 首项系数不为1的例题
omes (x + 2)² − 4. Multiply by 2: 2[(x + 2)² − 4] + 3 = 2(x + 2)² − 8 + 3 = 2(x + 2)² − 5.
对 2x² + 8x + 3 配方。首先从前两项中提取 2：2(x² + 4x) + 3。4 的一半是 2，2² = 4，因此 x² + 4x 变为 (x + 2)² − 4。乘以 2：2[(x + 2)² − 4] + 3 = 2(x + 2)² − 8 + 3 = 2(x + 2)² − 5。
2x² + 8x + 3 = 2(x + 2)² − 5
ation 2x² + 8x + 3 = 0, this becomes 2(x + 2)² − 5 = 0, so (x + 2)² = 5/2, giving x = −2 ± √5/2.
对于方程 2x² + 8x + 3 = 0，可写成 2(x + 2)² − 5 = 0，因此 (x + 2)² = 5/2，得到 x = −2 ± √5/2。
6. Deriving the Quadratic Formula | 推导二次公式
The quadratic formula is obtained by completing the square
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