📚 Quadratic Equations: Solving by Factorising, the Quadratic Formula and Completing the Square | 二次方程：因式分解法、求根公式与配方法
m a x² + b x + c = 0, where a, b and c are constants and a ≠ 0. This guide explains the three main algebraic solution methods - factorising, completing the square and the quadratic formula - together with the discriminant, the link between roots and coefficients, the shape of the parabola, and the exam strategies that turn a long calculation into a quick, reliable mark.
二次方程是 IGCSE、GCSE 与 A Level 代数的核心骨架。二次方程是任何可以写成标准形式 a x² + b x + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。本指南系统讲解三种主要代数解法——因式分解法、配方法与求根公式，并涵盖判别式、根与系数的关系、抛物线图像，以及把冗长计算变成快速得分的考试策略。
1. Standard Form and Key Terms | 标准形式与关键术语
re constants with a ≠ 0. That condition on a is essential: if a were zero, the x² term would disappear and the equation would collapse into a linear equation with a single solution.
二次方程是任何可以整理成标准形式 a x² + b x + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。对 a 的这个限制非常关键：如果 a 等于零，x² 项就会消失，方程退化为只有一个解的一次方程。
a x² + b x + c = 0, a ≠ 0
The coefficient a is the number multiplying x², b is the number multiplying x, and c is the constant term. Always rearrange the equation so that one side equals zero before you do anything else - expand any brackets, collect like terms, and multiply through by -1 if a is negative so that the x² coefficient becomes positive. Getting this step right prevents most sign errors later.
系数 a 是 x² 前的乘数，b 是 x 前的乘数，c 是常数项。动手解题之前，务必先把方程整理成一边等于零：展开所有括号、合并同类项；若 a 为负数，可整体乘以 -1 使 x² 系数变为正数。把这一步做对，可以避免后面绝大多数的符号错误。
crosses or touches the x-axis, which is why the algebraic and graphical views of quadratics always agree.
二次方程的解有三个常见名称：根、零点或 x 轴截距。它们是抛物线 y = a x² + b x + c 与 x 轴相交或相切处的 x 值，这也是二次方程的代数视角与图像视角总是彼此吻合的原因。
- Root / zero | 根 / 零点: a value of x that makes the equation true. | 使方程成立的 x 值。
- Coefficient | 系数: the numerical multiplier of a term, such as a and b. | 某一项的数值乘数，如 a 与 b。
- Constant term | 常数项: the term with no x, namely c. | 不含 x 的项，即 c。
- Discriminant | 判别式: the expression b² − 4ac, which reveals how many real roots exist. | 表达式 b² − 4ac，用来判断实数根的个数。
- Monic quadratic | 首一二次式: a quadratic in which a = 1. | a = 1 的二次式。
2. Solving by Factorising | 因式分解法
Factorising is usually the fastest method when the quadratic has simple integer or fractional roots. The idea is to rewrite the quadratic expression as a product of two linear brackets, then use the fact that a product is zero only when one of its factors is zero.
当二次方程有简单的整数根或分数根时，因式分解法通常是最快的解法。思路是把二次式改写为两个一次因式的乘积，然后利用"乘积为零则至少一个因式为零"这一事实来求解。
The method follows a fixed routine that examiners expect to see written down, because the working carries method marks even if the final answer slips.
这套方法有固定的书写流程，阅卷人期望看到这些步骤，因为即使最终答案出错，过程仍可获得方法分。
- Step 1: Write the equation in the form a x² + b x + c = 0. | 步骤 1：把方程写成 a x² + b x + c = 0 的形式。
- Step 2: Factorise the left-hand side into (p x + q)(r x + s). | 步骤 2：把左边分解为 (p x + q)(r x + s)。
- Step 3: Set each bracket equal to zero: p x + q = 0 and r x + s = 0. | 步骤 3：令每个括号等于零：p x + q = 0 与 r x + s = 0。
- Step 4: Solve each linear equation and state both roots clearly. | 步骤 4：解出两个一次方程，清楚写出两个根。
- Step 5: Substitute back to check, especially when the roots are negative or fractional. | 步骤 5：代回原式检验，尤其是根为负数或分数时。
6 and add to −5 are −2 and −3, so x² − 5x + 6 = (x − 2)(x − 3) = 0, giving x = 2 or x = 3.
例 1：解 x² − 5x + 6 = 0。相乘得 +6、相加得 −5 的两个数是 −2 与 −3，因此 x² − 5x + 6 = (x − 2)(x − 3) = 0，得 x = 2 或 x = 3。
rs: 2x² + 10x − 3x − 15 = 2x(x + 5) − 3(x + 5) = (x + 5)(2x − 3). Hence x = −5 or x = 3/2.
例 2：解 2x² + 7x − 15 = 0。把中间项拆成 10x − 3x，再分组分解：2x² + 10x − 3x − 15 = 2x(x + 5) − 3(x + 5) = (x + 5)(2x − 3)。因此 x = −5 或 x = 3/2。
divide both sides by x here - that would silently destroy the root x = 0 and lose a mark.
例 3：解 3x² = 12x。整理得 3x² − 12x = 0，即 3x(x − 4) = 0，所以 x = 0 或 x = 4。此处千万不要两边同除以 x——那会悄悄抹掉根 x = 0 并丢掉分数。
3. Key Factorising Patterns You Must Recognise | 必须掌握的因式分解模式
Most exam quadratics are built from a small set of patterns. Recognising the pattern instantly saves minutes and reduces the risk of arithmetic slips.
考试中的二次式大多由少数几种模式构成。一眼认出模式能节省大量时间并降低算错的风险。
- Common factor | 公因式: a x² + a b x = a x(x + b). Always take out the highest common factor first. | a x² + a b x = a x(x + b)。永远先提取最大公因式。
- Difference of two squares | 平方差: a² − b² = (a − b)(a + b). | a² − b² = (a − b)(a + b)。
- Perfect square trinomial | 完全平方三项式: a² ± 2ab + b² = (a ± b)². | a² ± 2ab + b² = (a ± b)²。
- Monic trinomial | 首一三项式: x² + (p + q)x + p q = (x + p)(x + q). | x² + (p + q)x + p q = (x + p)(x + q)。
- Non-monic trinomial | 非首一三项式: split the middle term, then factorise in pairs. | 非首一三项式：拆中间项，再分组分解。
a² − b² = (a − b)(a + b)
For a monic quadratic, you are searching for two numbers whose product is c and whose sum is b. If no such integer pair exists, the quadratic does not factorise over the integers and you should switch immediately to the formula or completing the square.
对于首一二次式，你要找的是乘积为 c、和为 b 的两个数。如果不存在这样的整数对，说明该二次式在整数范围内无法分解，此时应立即改用求根公式或配方法。
3)(x + 3) does. Sign matters enormously: the difference of two squares needs a minus sign between the squared terms.
注意 x² + 9 在实数范围内无法分解，而 x² − 9 = (x − 3)(x + 3) 可以。符号极其重要：平方差公式要求两个平方项之间是减号。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q or a(x + p)² + q. It works for every quadratic, even those that do not factorise, and it reveals the vertex of the parabola directly.
配方法把二次式改写为 (x + p)² + q 或 a(x + p)² + q 的形式。它对所有二次式都适用，即使无法因式分解的也可以，并且能直接读出抛物线的顶点。
x + b/2)² − (b/2)². You add and subtract the square of half the coefficient of x, so the value of the expression never changes.
核心恒等式很简单：x² + b x = (x + b/2)² − (b/2)²。你加上再减去 x 系数一半的平方，表达式的值始终不变。
x² + b x + c = (x + b/2)² + c − b²/4
(x + 3)² − 9 + 1 = (x + 3)² − 8. Solving (x + 3)² − 8 = 0 gives (x + 3)² = 8, so x = −3 ± √8 = −3 ± 2√2.
例：对 x² + 6x + 1 配方。6 的一半是 3，故 x² + 6x = (x + 3)² − 9，于是原式 = (x + 3)² − 9 + 1 = (x + 3)² − 8。解 (x + 3)² − 8 = 0 得 (x + 3)² = 8，所以 x = −3 ± √8 = −3 ± 2√2。
= a(x + b/2a)² + c − b²/4a. For example, 2x² − 8x + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3.
当 a ≠ 1 时，先把 a 从含 x 的项中提出。一般地，a x² + b x + c = a(x + b/2a)² + c − b²/4a。例如 2x² − 8x + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3。
- Step 1: If a ≠ 1, factor a out of the first two terms. | 步骤 1：若 a ≠ 1，从前两项提出 a。
- Step 2: Halve the coefficient of x inside the bracket and write (x + p)². | 步骤 2：把括号内 x 的系数除以 2，写出 (x + p)²。
- Step 3: Subtract p² (multiplied by any factor taken out) to balance. | 步骤 3：减去 p²（乘以已提出的系数）以保持平衡。
- Step 4: Simplify the constant and, if solving, take square roots of both sides. | 步骤 4：化简常数；若在解方程，则两边开平方。
5. Deriving the Quadratic Formula | 推导求根公式
The quadratic formula is not magic - it is exactly completing the square applied to the general equation. Understanding the derivation means you can reproduce it if you ever forget it.
求根公式并非凭空而来——它正是对一般方程做配方的结果。理解推导过程意味着即使一时忘记公式，你也能自己推出来。
a x² + b x + c = 0
x² + (b/a)x + c/a = 0
(x + b/2a)² − b²/4a² + c/a = 0
(x + b/2a)² = (b² − 4ac) / 4a²
x + b/2a = ± √b² − 4ac / 2a
x = (−b ± √b² − 4ac) / 2a
Divide through by a, complete the square on the left, move the constant to the right, take square roots of both sides, then subtract b/2a. Every step is reversible, which is why the formula gives all real roots of any quadratic equation.
两边同除以 a，对左边配方，把常数移到右边，两边开平方，再减去 b/2a。每一步都可逆，这正是该公式能给出任意二次方程全部实数根的原因。
The ± symbol is what produces two roots. It means "plus or minus", so you calculate the expression once with a plus sign and once with a minus sign.
± 符号正是产生两个根的原因。它表示"正或负"，因此你需要用加号算一次、用减号算一次。
6. Solving with the Quadratic Formula | 使用求根公式解题
The formula is the universal fallback. If a quadratic does not factorise nicely, or if the question involves surds or three significant figures, reach for the formula.
求根公式是万能退路。如果二次式不能漂亮地因式分解，或者题目涉及根式、三位有效数字，就直接用公式。
- Step 1: Rearrange into a x² + b x + c = 0 and identify a, b, c with their signs. | 步骤 1：整理成 a x² + b x + c = 0，连同符号写出 a、b、c。
- Step 2: Substitute carefully, putting negative values in brackets. | 步骤 2：仔细代入，负数要加括号。
- Step 3: Work out the discriminant b² − 4ac separately before the square root. | 步骤 3：开方之前先单独算出判别式 b² − 4ac。
- Step 4: Simplify the surd if the answer must be exact, or round to the required accuracy. | 步骤 4：若要求精确值则化简根式，否则按要求的精度四舍五入。
- Step 5: Write both roots, not just the positive one. | 步骤 5：两个根都要写，不能只写正根。
² − 4x − 3 = 0. Here a = 2, b = −4, c = −3, so b² − 4ac = 16 + 24 = 40. Therefore x = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2, which is approximately x ≈ 2.58 or x ≈ −0.58 to two decimal places.
例题：解 2x² − 4x − 3 = 0。此处 a = 2，b = −4，c = −3，故 b² − 4ac = 16 + 24 = 40。因此 x = (4 ± √40) / 4 = (4 ± 2√10) / 4 = (2 ± √10) / 2，约为 x ≈ 2.58 或 x ≈ −0.58（保留两位小数）。
stitution line in full, with brackets, is the single best habit for avoiding these errors.
常见陷阱是漏掉符号：b = −4 时 −b 变成 +4；a = 2、c = −3 时 −4ac 变成 +24。把代入过程完整写出并加上括号，是避免这类错误最有效的习惯。
7. The Discriminant and the Nature of Roots | 判别式与根的性质
nt. Its sign tells you how many real roots the equation has before you do any further work.
根号下的表达式 Δ = b² − 4ac 称为判别式。它的符号在继续计算之前就能告诉你方程有几个实数根。
Δ = b² − 4ac
| Discriminant | 判别式 | Nature of roots | 根的性质 | Graph | 图像 |
| Δ > 0 | Two distinct real roots | 两个不相等的实根 | Crosses the x-axis twice | 与 x 轴相交两次 |
| Δ = 0 | One repeated real root | 一个重根 | Touches the x-axis at the vertex | 在顶点处与 x 轴相切 |
| Δ < 0 | No real roots | 无实根 | Does not meet the x-axis | 与 x 轴不相交 |
When a, b and c are integers and Δ is a perfect square, both roots are rational, which means the quadratic would have factorised. If
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