📚 Quadratics: Completing the Square and the Quadratic Formula | 二次函数：配方法与求根公式
Quadratic equations appear throughout IGCSE Mathematics, from geometry and physics problems to graph sketching and optimisation. Two of the most powerful algebraic tools for solving any quadratic are completing the square and the quadratic formula. This article explains how both methods work, how they are connected, and when to use each one effectively in an exam.
二次方程贯穿 IGCSE 数学的许多内容，包括几何、物理问题、函数图像和优化问题。配方法和求根公式是求解任意二次方程的两个核心代数工具。本文将解释这两种方法的原理、它们之间的联系，以及在考试中如何高效选用。
1. Quadratic Equations in Standard Form | 二次方程的标准形式
ot quadratic. The values of x that satisfy the equation are called the roots or solutions.
二次方程是可以写成 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。如果 a = 0，方程就变成一次方程，而不是二次方程。满足方程的 x 值称为根或解。
ax² + bx + c = 0, a ≠ 0
ents, a quadratic may be solved by factorising, completing the square, using the quadratic formula, or graphically.
例如，2x² - 5x - 3 = 0 中 a = 2，b = -5，c = -3。根据系数的不同，二次方程可以通过因式分解、配方法、求根公式或图像法求解。
2. Why Completing the Square Matters | 为什么配方法很重要
Completing the square rewrites a quadratic expression in the form (x + p)² + q. This form immediately reveals the turning point of a quadratic graph and helps solve equations that cannot be factorised easily.
配方法将二次表达式改写为 (x + p)² + q 的形式。这种形式可以立即显示二次函数图像的顶点，并帮助求解不易因式分解的方程。
It is also the method used to derive the quadratic formula, so understanding completing the square gives a deeper understanding of where the formula comes from.
它也是推导求根公式的方法，因此理解配方法可以更深入地理解求根公式的来源。
In IGCSE exams, completing the square is often tested both as a manipulation skill and as a way to find maximum or minimum values.
在 IGCSE 考试中，配方法既作为代数变形技能考查，也作为求最大值或最小值的方法考查。
3. Perfect Square Trinomials | 完全平方三项式
Notice that the constant term 9 is the square of half the coefficient of x: (6 ÷ 2)² = 9.
完全平方三项式是一个二项式平方的结果。例如，(x + 3)² = x² + 6x + 9。注意常数项 9 是 x 项系数一半的平方：(6 ÷ 2)² = 9。
Similarly, (x - 4)² = x² - 8x + 16, and here half of -8 is -4, whose square is 16.
类似地，(x - 4)² = x² - 8x + 16，这里 -8 的一半是 -4，其平方为 16。
(x + a)² = x² + 2ax + a²
Recognising this pattern is essential because completing the square works by constructing a perfect square trinomial from the x² and x terms.
识别这一模式非常重要，因为配方法的核心就是利用 x² 项和 x 项构造一个完全平方三项式。
4. Algorithm for Completing the Square | 配方法的步骤
For a monic quadratic, x² + bx + c, we can complete the square in four short steps. First, write x² + bx as the square of a binomial, using half of b. Second, subtract the square of that half to keep the expression equal. Third, add the original constant c. Fourth, simplify the constant terms.
对于首项系数为 1 的二次式 x² + bx + c，我们可以用四个简短步骤完成配方。第一，将 x² + bx 写成二项式的平方，使用 b 的一半。第二，减去该一半的平方以保持表达式相等。第三，加上原来的常数项 c。第四，化简常数项。
x² + bx + c = (x + b/2)² - (b/2)² + c
If the coefficient of x² is not 1, factor it out from the first two terms before applying the same process inside the bracket.
如果 x² 的系数不是 1，需要先从 x² 项和 x 项中提出该系数，再在括号内进行同样的配方过程。
Many students lose marks by forgetting to multiply the subtracted square by the factor when it is outside the bracket.
许多学生失分的原因是忘记在括号外有系数时，被减去的平方需要乘以该系数。
5. Worked Example: Leading Coefficient 1 | 例题：首项系数为 1
Complete the square for y = x² + 6x + 5. Half of 6 is 3, so we write:
将 y = x² + 6x + 5 配方。6 的一半是 3，因此写作：
y = (x + 3)² - 9 + 5
Simplify the constant terms to get:
化简常数项得到：
y = (x + 3)² - 4
alue of y is -4, and it occurs when x = -3. The vertex of the graph is therefore (-3, -4).
这告诉我们 y 的最小值为 -4，且当 x = -3 时取得。因此图像的顶点为 (-3, -4)。
his form, set (x + 3)² - 4 = 0, then (x + 3)² = 4, so x + 3 = ±2, giving x = -1 or x = -5.
使用该形式解方程 x² + 6x + 5 = 0，令 (x + 3)² - 4 = 0，则 (x + 3)² = 4，所以 x + 3 = ±2，得到 x = -1 或 x = -5。
6. Worked Example: Leading Coefficient Not 1 | 例题：首项系数不为 1
Complete the square for y = 2x² - 8x + 3. First factor 2 from the first two terms:
将 y = 2x² - 8x + 3 配方。首先从前两项提出因数 2：
y = 2(x² - 4x) + 3
Inside the bracket, half of -4 is -2, so add and subtract 4:
在括号内，-4 的一半是 -2，因此加上并减去 4：
y = 2[(x - 2)² - 4] + 3
Distribute the 2 and simplify:
展开 2 并化简：
y = 2(x - 2)² - 8 + 3 = 2(x - 2)² - 5
s positive, the parabola opens upward and the function has a minimum value of -5 at x = 2.
顶点为 (2, -5)。由于 a = 2 为正，抛物线开口向上，函数在 x = 2 处取得最小值 -5。
This format is especially useful for graph sketching because it gives the vertex without differentiation or a table of values.
这种形式对画函数图像特别有用，因为它无需求导或列表就能直接给出顶点。
7. Deriving the Quadratic Formula | 推导求根公式
re on the general quadratic ax² + bx + c = 0, where a ≠ 0. Start by dividing through by a:
求根公式可以通过对一般二次方程 ax² + bx + c = 0（a ≠ 0）进行配方推导。首先两边同时除以 a：
x² + (b/a)x + c/a = 0
Move the constant term to the right, then complete the square on the left:
将常数项移到右边，然后对左边进行配方：
x² + (b/a)x = -c/a
Half of b/a is b/(2a), and its square is b²/(4a²). Add this to both sides:
b/a 的一半是 b/(2a)，其平方为 b²/(4a²)。两边同时加上这个平方：
x² + (b/a)x + b²/(4a²) = b²/(4a²) - c/a
The left side is a perfect square:
左边是一个完全平方：
(x + b/(2a))² = (b² - 4ac)/(4a²)
Taking the square root of both sides and solving for x gives the quadratic formula:
两边开平方并解出 x，得到求根公式：
x = [-b ± √b² - 4ac] / (2a)
Notice that this derivation uses the same half-and-square pattern from completing the square. Many exam boards expect students to recognise this connection rather than just memorise the formula.
注意该推导使用了与配方法相同的“一半再平方”模式。许多考试局希望学生理解这一联系，而不仅仅是记住公式。
8. Using the Quadratic Formula | 使用求根公式
n standard form ax² + bx + c = 0. Identify a, b and c carefully, including negative signs.
使用求根公式时，首先将方程写成标准形式 ax² + bx + c = 0，并仔细识别 a、b、c，包括负号。
For 3x² + 5x - 2 = 0, we have a = 3, b = 5 and c = -2. Substitute into the formula:
对于 3x² + 5x - 2 = 0，我们有 a = 3，b = 5，c = -2。代入公式：
x = [-5 ± √(5² - 4 × 3 × (-2))] / (2 × 3)
Simplify the discriminant: 25 + 24 = 49, so √49 = 7. This gives:
化简判别式：25 + 24 = 49，因此 √49 = 7。于是：
x = (-5 ± 7) / 6
The two roots are x = 1/3 and x = -2.
两个根为 x = 1
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