📚 Quadratic Equations: Factorisation, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与求根公式
tants and a ≠ 0. Solving quadratic equations means finding the values of x that make the equation true, and IGCSE candidates are expected to be fluent in factorisation, completing the square, the quadratic formula and graphical interpretation.
二次方程贯穿 IGCSE 数学考纲，从纯代数运算到几何、测量和建模都会出现。任何可化为 ax² + bx + c = 0 形式的方程都是二次方程，其中 a、b、c 为常数且 a ≠ 0。解二次方程就是找出使等式成立的 x 值，IGCSE 考生需要熟练掌握因式分解法、配方法、求根公式以及图像解释。
1. Standard Form and Coefficients | 标准形式与系数
The standard form of a quadratic equation is:
二次方程的标准形式为：
ax² + bx + c = 0, where a ≠ 0
xample, 3x² − 7 = 0 is a quadratic with b = 0, and 5x² + 2x = 0 is a quadratic with c = 0.
这里 a 是 x² 的系数，b 是 x 的系数，c 是常数项。a ≠ 0 这一条件至关重要，因为如果 a 等于零，方程就变为一次方程，而不是二次方程。不过，b 和 c 可以为零。例如，3x² − 7 = 0 是 b = 0 的二次方程，5x² + 2x = 0 是 c = 0 的二次方程。
Before choosing a solution method, always rearrange the equation into standard form. Move all terms to one side so that the other side equals zero. This step keeps coefficients clear and reduces sign errors later.
在选择解法之前，一定要把方程整理成标准形式。将所有项移到一边，使另一边等于零。这一步可以使系数清晰，并减少后续的符号错误。
2. Solving by Factorisation | 因式分解法
If the quadratic expression factorises over integers, factorisation is usually the fastest method. For a monic quadratic of the form x² + bx + c, find two numbers whose product is c and whose sum is b. These two numbers allow you to write the expression as (x + p)(x + q). Then use the zero product property: if the product is zero, at least one factor must be zero.
如果二次式可以在整数范围内因式分解，因式分解法通常是最快的方法。对于形如 x² + bx + c 的首一二次式，找到两个数，使它们的乘积为 c，和为 b。利用这两个数可把表达式写成 (x + p)(x + q)。然后使用零积性质：如果乘积为零，则至少有一个因式为零。
+ 5x + 6 = (x + 2)(x + 3) = 0. Therefore x + 2 = 0 or x + 3 = 0, giving x = −2 or x = −3.
例子：解 x² + 5x + 6 = 0。两个数是 2 和 3，因为 2 × 3 = 6 且 2 + 3 = 5。所以 x² + 5x + 6 = (x + 2)(x + 3) = 0。因此 x + 2 = 0 或 x + 3 = 0，得到 x = −2 或 x = −3。
factor: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0. The solutions are x = −1/2 or x = −3.
当 a ≠ 1 时，使用 ac 法。找到两个数，其乘积为 ac，和为 b，将中项拆开，再分组分解。对于 2x² + 7x + 3，ac = 6。6 和 1 满足 6 × 1 = 6 且 6 + 1 = 7。将 7x 拆成 6x + x，然后分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0。解为 x = −1/2 或 x = −3。
3. Special Factorisations | 特殊因式分解
Perfect square trinomials factor as a² + 2ab + b² = (a + b)² or a² − 2ab + b² = (a − b)².
如果你能识别标准模式，一些二次方程可以快速求解。平方差公式为 a² − b² = (a + b)(a − b)。完全平方三项式可分解为 a² + 2ab + b² = (a + b)² 或 a² − 2ab + b² = (a − b)²。
= −3. Example: solve x² + 6x + 9 = 0. This is (x + 3)² = 0, so x = −3 is a repeated root.
例子：解 x² − 9 = 0。把它看作 x² − 3²，所以 (x + 3)(x − 3) = 0。因此 x = 3 或 x = −3。例子：解 x² + 6x + 9 = 0。它是 (x + 3)² = 0，所以 x = −3 是二重根。
These patterns also appear when an equation is not immediately in standard form. Always expand and simplify first, then look for a common factor before applying any special factorisation.
这些模式也会出现在方程尚未化为标准形式的情形中。应先展开并化简，然后在应用任何特殊因式分解之前寻找公因式。
4. Solving by Completing the Square | 配方法
Completing the square transforms x² + bx into a perfect square form. The method is useful when factorisation is difficult and it is also the basis for the quadratic formula. Follow these steps: move the constant term to the right-hand side; take half of the coefficient of x, square it, and add it to both sides; write the left side as a squared binomial; take the square root of both sides; solve for x.
配方法将 x² + bx 转化为完全平方形式。当因式分解困难时，这种方法很有用，它也是求根公式的基础。步骤如下：把常数项移到右边；取 x 系数的一半，平方后加到两边；将左边写成二项式的平方；两边开平方；解出 x。
x² + bx + (b/2)² = (x + b/2)²
9 = 4. This gives (x + 3)² = 4. Taking square roots gives x + 3 = ±2, so x = −1 or x = −5.
例子：解 x² + 6x + 5 = 0。把 5 移项：x² + 6x = −5。6 的一半是 3，3² = 9，所以两边加 9：x² + 6x + 9 = 4。于是 (x + 3)² = 4。两边开平方得 x + 3 = ±2，所以 x = −1 或 x = −5。
0 before completing the square. This normalisation keeps the coefficient of x² equal to 1.
如果 a ≠ 1，先将每一项除以 a。例如，2x² + 8x + 6 = 0 变为 x² + 4x + 3 = 0，然后再配方。这样的标准化处理可使 x² 的系数等于 1。
5. The Quadratic Formula | 求根公式
ion ax² + bx + c = 0. It gives the solutions directly without factorising. The formula is:
求根公式是通过对一般方程 ax² + bx + c = 0 配方得到的。它无需因式分解即可直接给出解。公式为：
x = (-b ± √b² - 4ac) / (2a)
solve 2x² − 4x − 3 = 0. Here a = 2, b = −4, c = −3. Substitute into the formula: x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2.
这个公式非常强大，因为它适用于任何二次方程，包括根为无理数的方程。例子：解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。代入公式：x = (4 ± √16 + 24) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2。
When using the formula, write down the values of a, b and c carefully, especially when coefficients are negative. Do not simplify away the denominator incorrectly; the entire numerator is divided by 2a.
使用公式时，要仔细写出 a、b、c 的值，尤其是系数为负数时。不要错误地约去分母；整个分子都要除以 2a。
6. The Discriminant | 判别式
formula: Δ = b² − 4ac. It determines the nature of the roots without solving the equation.
判别式是求根公式中平方根号下的表达式：Δ = b² − 4ac。它无需解方程即可判断根的性质。
| Discriminant Δ = b² - 4ac | Nature of roots |
|---|---|
| Δ > 0 | Two distinct real roots |
| Δ = 0 | One repeated real root (two equal roots) |
| Δ < 0 | No real roots |
you to find the value of k for which a quadratic has equal roots, so set Δ = 0 and solve.
例如，x² − 4x + 4 = 0 的判别式 Δ = (−4)² − 4 × 1 × 4 = 0，所以它有一个二重根。方程 x² + x + 1 = 0 的判别式 Δ = 1 − 4 = −3，所以它没有实数根。IGCSE 题目经常要求找出使二次方程有等根的 k 值，此时令 Δ = 0 并求解。
7. Graphical Interpretation | 图像意义
of ax² + bx + c = 0 are the x-intercepts of the graph, where the parabola cuts the x-axis.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上，有一个最低点；如果 a < 0，抛物线开口向下，有一个最高点。方程 ax² + bx + c = 0 的解就是图像与 x 轴交点的横坐标，即抛物线与 x 轴相交的位置。
x-intercepts: x = (-b ± √b² - 4ac) / (2a)
a), which is also the axis of symmetry. The y-intercept is c. The discriminant tells how many times the parabola crosses the x-axis
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