📚 Solving Quadratic Equations | 解二次方程
Quadratic equations sit at the heart of IGCSE and A Level algebra: the same three techniques - factorisation, completing the square and the quadratic formula - reappear in coordinate geometry, graph work, inequalities, sequences and modelling questions. This revision guide works through every method step by step, explains what the discriminant tells you, shows how examiners word problem questions, and finishes with a checklist of the errors that cost the most marks.
二次方程是 IGCSE 与 A Level 代数的核心：因式分解、配方法与二次公式这三种技巧，会反复出现在坐标几何、图像、不等式、数列与建模题中。本复习指南将逐步讲解每一种方法，说明判别式所传达的信息，展示考官在应用题中的措辞方式，并在最后给出最容易失分的错误清单。
1. What is a Quadratic Equation? | 什么是二次方程？
x + c = 0, where x is the unknown variable and a, b, c are constants with a ≠ 0. The word 'quadratic' comes from the Latin 'quadratus' is exactly why the condition a ≠ 0 is always stated. A quadratic equation has at most two real solutions, also called roots or zeros, and these are the values of x that make the expression equal to zero.
二次方程是次数为二的多项式方程，也就是说未知数的最高次幂为 2。它的一般形式是 a x² + b x + c = 0，其中 x 为未知数，a、b、c 为常数，且 a ≠ 0。"二次"一词源于拉丁文 quadratus，意为"平方"，因为未知数被平方了。如果 a 等于 0，x² 项就会消失，方程退化为一次方程——这正是必须写明 a ≠ 0 的原因。二次方程最多有两个实数解，也称为根或零点，它们就是使该表达式等于零的 x 取值。
anged into the standard form. The graph of the function y = a x² + b x + c is a parabola: it opens upwards when a > 0 and downwards when a < 0, and it crosses the x-axis at the real roots of the corresponding equation.
二次方程的例子包括 x² − 5x + 6 = 0、3x² + 4x = 0、x² = 16 以及 2x² − 7x + 3 = 0。注意其中有些看起来并不完整，但它们都可以整理成标准形式。函数 y = a x² + b x + c 的图像是一条抛物线：当 a > 0 时开口向上，当 a < 0 时开口向下，它与 x 轴的交点即为对应方程的实数根。
2. Standard Form and Key Terms | 标准形式与关键术语
ient of x² (the quadratic term), b is the coefficient of x (the linear term), and c is the constant term. For instance, 3x² − 12x + 9 = 0 is already in standard form, while 2x² = 5x − 4 must be rewritten as 2x² − 5x + 4 = 0. Similarly, (x + 1)(x − 4) = 6 must first be expanded and rearranged into x² − 3x − 10 = 0, because the right-hand side is not zero.
在运用任何方法之前，务必先整理方程，使其中一边等于零。在标准形式 a x² + b x + c = 0 中，a 是 x² 的系数（二次项系数），b 是 x 的系数（一次项系数），c 是常数项。例如，3x² − 12x + 9 = 0 已经是标准形式，而 2x² = 5x − 4 需要重写为 2x² − 5x + 4 = 0。同理，(x + 1)(x − 4) = 6 必须先展开并整理为 x² − 3x − 10 = 0，因为右边并不等于零。
Rewriting is not just cosmetic: the quadratic formula and the discriminant both read a, b and c directly from the standard form, so an un-rearranged equation will produce the wrong answer. When a coefficient is negative, many students prefer to multiply the whole equation by −1 to make a positive, which does not change the roots but makes factorisation and substitution much cleaner.
整理形式并不只是形式上的要求：二次公式与判别式都直接从标准形式中读取 a、b、c，因此没有整理好的方程必然算错。当系数为负时，很多同学习惯把整个方程乘以 −1，让 a 变为正数，这样不会改变方程的根，却能让因式分解和代入计算清爽许多。
3. Solving by Factorisation | 因式分解法
wo numbers multiply to give zero, then at least one of them must be zero. So if (x − p)(x − q) = 0, then x − p = 0 or x − q = 0, giving x = p or x = q. Write the equation in standard form, factorise the left-hand side, then set each bracket equal to zero and solve the resulting linear equations.
因式分解通常是最快的方法，但只有当一个方程的根是有理数时才会顺利奏效。其原理是零积性质：如果两个数相乘等于零，那么其中至少有一个为零。因此若 (x − p)(x − q) = 0，则 x − p = 0 或 x − q = 0，得到 x = p 或 x = q。先把方程化为标准形式，对左边进行因式分解，再令每个括号等于零，解出相应的一次方程即可。
nd +2. Hence (x − 3)(x + 2) = 0, so x = 3 or x = −2. Special patterns are worth recognising instantly: a difference of two squares such as x² − 9 = 0 factorises as (x − 3)(x + 3), giving x = ±3, and a perfect square such as x² + 6x + 9 = 0 becomes (x + 3)² = 0, giving the repeated root x = −3.
当 a = 1 时，要寻找两个数，使它们相乘得 c、相加得 b。对于 x² − x − 6 = 0，我们需要两个数相乘得 −6、相加得 −1，即 −3 与 +2。因此 (x − 3)(x + 2) = 0，所以 x = 3 或 x = −2。有些特殊结构应当一眼认出：平方差如 x² − 9 = 0 可分解为 (x − 3)(x + 3)，得 x = ±3；完全平方式如 x² + 6x + 9 = 0 变为 (x + 3)² = 0，得到重根 x = −3。
write the middle term as 6x − x, then factorise in pairs: 2x² + 6x − x − 3 = 2x(x + 3) − 1(x + 3) = (2x − 1)(x + 3). Therefore 2x − 1 = 0 or x + 3 = 0, so x = ½ or x = −3. Always substitute your answers back into the original equation as a one-line check.
当 a ≠ 1 时，需要拆分中间项。以 2x² + 5x − 3 = 0 为例：a × c = 2 × (−3) = −6，需要两个数相乘得 −6、相加得 +5，即 +6 与 −1。把中间项改写为 6x − x，再分组分解：2x² + 6x − x − 3 = 2x(x + 3) − 1(x + 3) = (2x − 1)(x + 3)。因此 2x − 1 = 0 或 x + 3 = 0，即 x = ½ 或 x = −3。最后一定要把答案代回原方程做一行验算。
4. Solving by Completing the Square | 配方法
d. For a monic quadratic x² + b x + c = 0, halve the coefficient of x, write (x + b/2)², and subtract the square of that half. The identity is x² + b x = (x + b/2)² − (b/2)². Then move the constant to the right-hand side and take square roots of both sides, remembering the ± sign.
配方法把二次式改写为"完全平方加常数"的形式，即使根是无理数或根式也能直接看出。对于首项系数为 1 的二次方程 x² + b x + c = 0，把一次项系数除以 2，写成 (x + b/2)²，再减去这个一半的平方。恒等式为 x² + b x = (x + b/2)² − (b/2)²。之后把常数移到等号右边，对两边开平方，别忘了 ± 号。
x² + 6x + 2 = 0 → (x + 3)² − 9 + 2 = 0 → (x + 3)² = 7 → x = −3 ± √7
re inside the bracket: 2[(x − 2)² − 4] + 3 = 0, which simplifies to 2(x − 2)² − 5 = 0. Hence (x − 2)² = 5/2, so x = 2 ± √5/2, which can be written as x = 2 ± √10/2 ≈ 3.58 or 0.42.
当 a ≠ 1 时，先从前两项中提取 a，注意不要把常数项也提进去。对于 2x² − 8x + 3 = 0，写成 2(x² − 4x) + 3 = 0，再在括号内配方：2[(x − 2)² − 4] + 3 = 0，化简得 2(x − 2)² − 5 = 0。因此 (x − 2)² = 5/2，即 x = 2 ± √5/2，也可写作 x = 2 ± √10/2 ≈ 3.58 或 0.42。
s a minimum point, and when a < 0 it is a maximum point. Examiners frequently ask for the turning point and the line of symmetry x = h in the same question.
配方法也是求抛物线顶点的标准方法。写成 y = a(x − h)² + k 后即可看出顶点为 (h, k)：当 a > 0 时为最低点，当 a < 0 时为最高点。考官常常在同一道题中既要求写出顶点，也要求写出对称轴 x = h。
5. Deriving the Quadratic Formula | 推导二次公式
d simplify. Following the algebra once makes the formula far easier to remember and to trust under exam pressure. Start by dividing: x² + (b/a)x + c/a = 0.
二次公式其实就是把配方法用于一般形式 a x² + b x + c = 0。将每一项除以 a，对结果配方，再化简。亲自推一遍这个代数过程，会让你在考场上更容易记住并信任这个公式。先除以 a：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 ± √b² − 4ac) / 2a
er the coefficients, provided a ≠ 0. Learn it exactly as written: the whole numerator −b ± √b² − 4ac is divided by 2a, not just the square root part. A common slip is to write −b² instead of (−b)², or to forget that the subtraction applies only inside the square root. Write the formula down at the top of your working before substituting, and quote it in the answer to secure the method mark even if an arithmetic error follows.
只要 a ≠ 0，无论系数如何，这个公式都成立。要一字不差地记住它：整个分子 −b ± √b² − 4ac 都要除以 2a，而不只是根号那部分。常见错误是把 (−b)² 写成 −b²，或者误以为减号在根号外面。建议先把公式写在解题开头，再代入数值，并在答案中引用公式，这样即便后面算错，也能拿到方法分。
6. Using the Quadratic Formula in Practice | 二次公式的实际应用
2x² − 4x − 3 = 0, so a = 2, b = −4 and c = −3. The discriminant is b² − 4ac = (−4)² − 4 × 2 × (−3) = 16 + 24 = 40, which is positive, so there are two distinct real roots. Then x = (4 ± √40) / 4, and since √40 = 2√10, this simplifies to x = (2 ± √10) / 2 ≈ 2.58 or −0.58.
代入计算最稳妥的做法是按固定顺序进行：先从标准形式中读出 a、b、c 并写下来，再代入公式。以 2x² − 4x − 3 = 0 为例，a = 2，b = −4，c = −3。判别式 b² − 4ac = (−4)² − 4 × 2 × (−3) = 16 + 24 = 40，为正，所以有两个不同的实根。于是 x = (4 ± √40) / 4，由于 √40 = 2√10，可化简为 x = (2 ± √10) / 2 ≈ 2.58 或 −0.58。
itution line, the value of the discriminant and the final simplification separately - method marks are awarded for each visible step. When the answer is meant to be exact, simplify the surd fully and cancel any common factor; leaving (4 ± √40) / 4 unsimplified may lose the final accuracy mark.
除非题目要求精确的根式形式，否则应按试卷说明保留 3 位有效数字或 2 位小数。要把代入式、判别式的取值和最终化简分步写清楚——每个可见步骤都有方法分。若答案要求精确值，必须把根式化到最简并约去公因数；把 (4 ± √40) / 4 直接留着不化简，可能会丢掉最后的精度分。
7. The Discriminant and the Number of Roots | 判别式与根的个数
nt, and it tells you how many real roots the equation has before you solve it. Because a square root of a negative number is not real, the sign of the discriminant completely determines the nature of the roots. This is one of the most heavily examined ideas in the topic.
根号下的表达式 Δ = b² − 4ac 称为判别式，它能在求解之前就告诉你方程有多少个实根。由于负数的平方根不是实数，判别式的符号完全决定了根的性质。这是本专题中考查频率最高的知识点之一。
| Discriminant / 判别式 | Nature of roots / 根的性质 | Graph / 图像 |
| Δ > 0 | Two distinct real roots / 两个不同实根 | Crosses x-axis twice / 与 x 轴相交两次 |
| Δ = 0 | One repeated real root / 一个重根 | Touches x-axis / 与 x 轴相切 |
| Δ < 0 | No real roots / 无实根 | Does not meet x-axis / 与 x 轴不相交 |
Typical exam wording includes 'find the values of k for which the equation has two equal roots' or 'show that the equation has no real solutions' t; 0, and for two distinct roots set Δ > 0. Remember that a perfect square discriminant also tells you the quadratic factorises nicely over the integers.
常见的考题措辞包括"求使方程有两个相等实根的 k 值"或"证明该方程没有实数解"。两者都通过构造判别式的不等式来解决：重根令 Δ = 0，无实根令 Δ < 0，两个不同实根令 Δ > 0。还要注意，若判别式是完全平方数，则该二次式在整数范围内可以顺利因式分解。
8. Solving Quadratic Equations by Graphing | 图像法解二次方程
x² + b x + c. Reading roots off a graph is an acceptable method when the question provides a grid or an accurate sketch, and it is invaluable for checking whether an algebraic answer is sensible. Draw a smooth parabola through the plotted points, then read where it meets the x-axis.
a x² + b x + c = 0 的实根，正是曲线 y = a x² + b x + c 与 x 轴交点的横坐标。当题目给出网格或较准确的草图时，读图求根是被认可的方法，而且它对检验代数答案是否合理极有帮助。先描点画出光滑的抛物线，再读出它与 x 轴的交点。
2x = x + 1 and draw the straight line y = x + 1; the x-coordinates of the intersections are the solutions. The vertex of the parabola lies on the line of symmetry x = −b/2a, and the y-value there is the minimum (a > 0) or maximum (a < 0) value of the function.
图像法还能求解并非直接二次的方程。若已画出曲线 y = x² − 2x，要图解 x² − 3x − 1 = 0，可整理为 x² − 2x = x + 1，再画出直线 y = x + 1，交点的横坐标即为解。抛物线的顶点位于对称轴 x = −b/2a 上，该处的纵坐标就是函数的最小值（a > 0）或最大值（a < 0）。
9. Word Problems and Modelling | 应用题与建模
Word problems hide a quadratic inside a context: areas, projectile heights, profit functions or consecutive-number puzzles. The routine is always the same. Define a variable in words, write an equation from the given information, rearrange it to standard form, solve it, and then apply common sense to the answers.
应用题会把二次方程隐藏在具体情境中：面积、抛体高度、利润函数或连续整数谜题。解题流程始终一致：用文字明确设出变量，
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