📚 T-3-1083: Quadratic Equations - Complete Revision Guide | T-3-1083：二次方程完全复习指南
Quadratic equations are one of the most important topics in the IGCSE Mathematics syllabus, and they continue to appear throughout A-Level algebra, coordinate geometry, mechanics and modelling. A quadratic equation can have two real solutions, one repeated solution or no real solution at all, and which case happens depends entirely on the coefficients. This article gives you a complete revision guide, moving from standard form to factorising, completing the square, the quadratic formula, the discriminant, graphs, inequalities, word problems and exam-style applications, with the examiner traps marked along the way.
二次方程是 IGCSE 数学大纲中最重要的主题之一，并且在 A-Level 的代数、坐标几何、力学和数学建模中持续出现。二次方程根据系数的不同，可能有两个实数解、一个重复解，或完全没有实数解。本文为你提供一份完整的复习指南，从标准形式出发，依次讲因式分解、配方法、求根公式、判别式、图像、不等式、应用题以及考试风格的实际应用，并标出考官常设的陷阱。
1. Standard Form of a Quadratic Equation | 二次方程的标准形式
, because if a = 0 the x² term disappears and the equation becomes linear, with only one solution. Any equation that can be rearranged into this shape counts as quadratic, even if it does not look like one at first glance.
二次方程是任何可以写成 a x² + b x + c = 0 形式的方程，其中 a、b、c 是常数，且 a ≠ 0。a ≠ 0 这个条件非常重要，因为如果 a = 0，x² 项就消失了，方程变成一次方程，只会有一个解。任何能整理成这种形状的方程都算二次方程，即使它第一眼看上去并不像。
ongs to the coefficient: b is −5, not 5. Writing the equation in standard form before solving is a key first step, and it prevents most careless errors later.
例如，在 2x² − 5x + 3 = 0 中，a = 2，b = −5，c = 3。注意符号属于系数本身：b 是 −5，而不是 5。解题前先把方程写成标准形式是关键的第一步，它能避免之后大多数粗心错误。
ctly. For instance, x² = 4x − 3 becomes x² − 4x + 3 = 0, and 3x(x − 2) = 9 becomes 3x² − 6x − 9 = 0, which can be divided by 3 to give x² − 2x − 3 = 0.
如果方程不是标准形式，就把所有项移到方程的一边，令另一边等于 0。这样可以更容易地正确确定 a、b 和 c。例如，x² = 4x − 3 可以整理为 x² − 4x + 3 = 0；3x(x − 2) = 9 展开为 3x² − 6x − 9 = 0，再除以 3 得到 x² − 2x − 3 = 0。
a x² + b x + c = 0, a ≠ 0
- A quadratic never has more than two real solutions. 二次方程最多只有两个实数解。
- Always expand brackets and collect like terms before reading off a, b and c. 先展开括号并合并同类项，再读出 a、b、c。
- Keep fractions and surds exact until the final answer line. 分数和根式保持精确形式，直到最后一行答案再化简。
2. Solving by Factorising | 因式分解法
rs multiply to give zero, then at least one of them must be zero. In symbols, if p q = 0 then p = 0 or q = 0. So once the quadratic is written as a product of two brackets, each bracket can be set equal to zero in turn.
当二次方程有简单的整数根时，因式分解通常是最快的方法。这个方法依赖于零乘积性质：如果两个数相乘等于零，那么其中至少有一个是零。用符号表示就是，如果 p q = 0，那么 p = 0 或 q = 0。所以一旦把二次式写成两个括号的乘积，就可以依次令每个括号等于零。
d product 6, namely −2 and −3, so x² − 5x + 6 = (x − 2)(x − 3) = 0, giving x = 2 or x = 3.
对于首项系数为 1 的二次式 x² + b x + c，要找两个数，它们的和是 b，乘积是 c。例如 x² − 5x + 6 需要两个数，和为 −5、积为 6，即 −2 和 −3，所以 x² − 5x + 6 = (x − 2)(x − 3) = 0，得到 x = 2 或 x = 3。
m = −3, n = −1 gives (2x − 3)(x − 1), and indeed −2 − 3 = −5. Setting each bracket to zero gives x = 3/2 or x = 1.
当 a 不等于 1 时，试错过程稍长，但逻辑完全相同。对于 2x² − 5x + 3，我们需要 (2x + m)(x + n)，满足 m n = 3 且 2n + m = −5。试 m = −3、n = −1 得到 (2x − 3)(x − 1)，确实有 −2 − 3 = −5。令每个括号等于零，得到 x = 3/2 或 x = 1。
quares, p² − q² = (p − q)(p + q), handles equations such as x² − 9 = 0 directly, giving x = 3 or x = −3.
有两个特殊模式可以节省大量时间。首先要提取公因式：3x² − 12x = 3x(x − 4) = 0，得到 x = 0 或 x = 4。平方差公式 p² − q² = (p − q)(p + q) 可以直接处理像 x² − 9 = 0 这样的方程，得到 x = 3 或 x = −3。
- Never divide both sides by x, because you will lose the solution x = 0. 绝对不要两边同除以 x，否则会丢掉 x = 0 这个解。
- Check by expanding your brackets back to the original quadratic. 把括号展开回原二次式来检验。
- If no integer pair works, switch to the formula rather than guessing. 如果没有整数对可用，就改用求根公式，不要继续猜。
3. Completing the Square | 配方法
only reliable hand method when the roots are irrational. The basic identity is x² + b x = (x + b/2)² − (b/2)², because expanding gives back x² + b x once the subtracted square is included.
配方法把二次式改写为 a(x + p)² + q 的形式。它是求抛物线顶点（转折点）的标准方法，也是当根是无理数时唯一可靠的手算方法。基本恒等式是 x² + b x = (x + b/2)² − (b/2)²，因为展开后加上被减去的平方项，正好还原成 x² + b x。
ation becomes (x + 3)² − 9 + 1 = 0, that is (x + 3)² = 8. Taking square roots, x + 3 = ± 2√2, so x = −3 + 2√2 or x = −3 − 2√2. Rounded, these are approximately −0.17 and −5.83.
例题：用配方法解 x² + 6x + 1 = 0。6 的一半是 3，所以 x² + 6x = (x + 3)² − 9。方程变为 (x + 3)² − 9 + 1 = 0，即 (x + 3)² = 8。开平方得 x + 3 = ± 2√2，所以 x = −3 + 2√2 或 x = −3 − 2√2。取近似值约为 −0.17 和 −5.83。
is therefore (2, −3), and since a = 2 is positive it is a minimum. The general vertex form is a(x + b/2a)² + (c − b²/4a).
当 a 不等于 1 时，只从前两项中提取 a。对于 2x² − 8x + 5，写成 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3。因此顶点是 (2, −3)，由于 a = 2 为正，这是一个最小值。一般的顶点形式是 a(x + b/2a)² + (c − b²/4a)。
- Completed-square form instantly gives the minimum or maximum value. 配方法形式可以直接给出最小值或最大值。
- If a > 0 the parabola opens upwards and has a minimum; if a < 0 it opens downwards and has a maximum. 若 a > 0，抛物线开口向上且有最小值；若 a < 0，开口向下且有最大值。
- Always write the half-coefficient as a fraction if b is odd, for example (x + 1/2)². 若 b 是奇数，半系数要写成分数，例如 (x + 1/2)²。
4. The Quadratic Formula | 求根公式
0 with a ≠ 0, the solutions are given by the expression below. The ± sign is what produces two answers from a single formula.
求根公式可以解出任何二次方程，包括那些在整数范围内无法因式分解的方程。它表明，对于 a x² + b x + c = 0（a ≠ 0），解由下面的表达式给出。± 符号正是从同一个公式产生两个答案的原因。
x = (−b ± √b² − 4ac) ÷ (2a)
the square gives (x + b/2a)² = (b² − 4ac)/(4a²), and taking square roots with the ± symbol gives x + b/2a = ± √b² − 4ac ÷ (2a), which rearranges to the formula above. Understanding this derivation means you never need to memorise the formula blindly.
这个公式并非凭空而来，它是对一般方程配方得到的。把 a x² + b x + c = 0 除以 a 得 x² + (b/a)x + c/a = 0。配方后得 (x + b/2a)² = (b² − 4ac)/(4a²)，带上 ± 符号开平方得 x + b/2a = ± √b² − 4ac ÷ (2a)，整理即得上式。理解这个推导过程，你就不必死记公式。
4/4 = 1. Substituting both values back into the original equation confirms that both work.
例题：用公式解 2x² − 5x + 3 = 0。这里 a = 2，b = −5，c = 3，所以 b² − 4ac = 25 − 24 = 1。于是 x = (5 ± 1) ÷ 4，得到 x = 6/4 = 3/2 或 x = 4/4 = 1。把两个值代回原方程检验，都成立。
- Write the formula down before substituting, and use brackets around negative numbers. 先写出公式再代入，负数要加括号。
- If the answer is irrational, simplify the surd and keep it exact unless told to round. 若答案是无理数，要化简根式，除非题目要求取近似，否则保留精确形式。
- Give answers to the required accuracy, usually 3 significant figures. 按题目要求的精度作答，通常是 3 位有效数字。
5. The Discriminant | 判别式
nt. It tells you how many real roots the equation has without solving it, and it is one of the most frequently examined ideas at both IGCSE and A-Level. The sign of the discriminant is what matters, not its exact value.
根号下的表达式 Δ = b² − 4ac 称为判别式。它可以在不解方程的情况下告诉你方程有多少个实数根，这是 IGCSE 和 A-Level 都最常考的概念之一。真正重要的是判别式的符号，而不是它的具体数值。
| Discriminant 判别式 | Number of real roots 实数根个数 | Graph behaviour 图像表现 |
| b² − 4ac > 0 | Two distinct real roots 两个不同实根 | Curve cuts the x-axis twice 曲线与 x 轴交于两点 |
| b² − 4ac = 0 | One repeated real root 一个重根 | Curve touches the x-axis 曲线与 x 轴相切 |
| b² − 4ac < 0 | No real roots 没有实根 | Curve misses the x-axis 曲线与 x 轴不相交 |
; 0, so it has no real roots and the parabola stays above the x-axis for every value of x.
例如，x² + 4x + 4 的 Δ = 16 − 16 = 0，所以它是一个完全平方 (x + 2)²，有重根 x = −2。相反，x² + x + 1 的 Δ = 1 − 4 = −3 < 0，因此没有实根，抛物线对所有 x 都位于 x 轴上方。
- "Show that the equation has equal roots" almost always means set Δ = 0 and solve for the unknown constant. "证明方程有等根"几乎总是意味着令 Δ = 0 并解出未知常数。
- "Find the range of k for which the equation has two distinct roots" means solve Δ > 0 as an inequality. "求方程有两个不同实根时 k 的取值范围"就是解不等式 Δ > 0。
- Do not confuse the discriminant with the formula itself; Δ is only the part inside the square root. 不要混淆判别式与公式本身；Δ 只是根号内的部分。
6. Roots, x-Intercepts and the Graph | 根、x 轴交点与图像
, and the vertex sits on that line. The roots of the equation a x² + b x + c = 0 are exactly the x-coordinates where the curve crosses, touches or misses the x-axis.
y = a x² + b x + c 的图像是一条抛物线。它的 y 轴截距在 (0, c)，因为令 x = 0 得 y = c。它的对称轴是竖直线 x = −b/2a，顶点就在这条线上。方程 a x² + b x + c = 0 的根，正是曲线与 x 轴相交、相切或不相交处的 x 坐标。
and the vertex using x = −b/2a with the corresponding y value. For y = x² − 4x + 3, a = 1 so it opens upwards, the y-intercept is 3, the roots are x = 1 and x = 3, and the axis of symmetry is x = 2, giving a vertex at (2, −1).
要快速画出抛物线草图，按顺序找四件事：a 的符号决定开口向上还是向下；y 轴截距 c；根（如果存在）；以及用 x = −b/2a 及其对应的 y 值求出的顶点。对于 y = x² − 4x + 3，a = 1 开口向上，y 轴截距为 3，根为 x = 1 和 x = 3，对称轴为 x = 2，顶点为 (2, −1)。
Δ = 0, two crossing points mean Δ > 0, and no intersection means Δ < 0. This single idea links algebra and geometry and appears in many exam papers.
当直线 y = m x + k 与抛物线相交时，代入消元形成二次方程，再用判别式判断交点个数。相切意味着 Δ = 0，两个交点意味着 Δ > 0，没有交点意味着 Δ < 0。这一个想法把代数与几何联系起来，在很多试卷中都会出现。
- The vertex form a(x + p)² + q gives the turning point (−p, q) immediately. 顶点形式 a(x + p)² + q 可以直接给出转折点 (−p, q)。
- Use completing the square or x = −b/2a; both give the same axis of symmetry. 配方法与 x = −b/2a 都可以求对称轴，结果相同。
- Always label the intercepts and the vertex on a sketch to earn method marks. 画草图时标出截距和顶点，以取得方法分。
7. Quadratic Inequalities | 二次不等式
A quadratic inequality asks for a range of values rather than exact roots. The safe method has three steps: solve the corresponding equation to find the critical values, sketch the parabola, then read off the intervals where the curve lies above or below the x-axis. Never simply write the two roots joined by "and" or "or" without checking the sketch.
二次不等式要求的是一个取值范围，而不是精确的根。安全的方法有三步：先解对应的方程求出临界值，再画出抛物线草图，然后读出曲线位于 x 轴上方或下方的区间。千万不要不检查草图就简单地把两个根用 "and" 或 "or" 连起来。
rds, so it is above the x-axis on the outside of the two roots. The solution is therefore x < −2 or x > 3. If the question had asked for x² − x − 6 < 0, the answer would be −2 < x < 3, the interval between the roots.
例题：解 x² − x −
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