📚 Solving Quadratic Equations | 解二次方程
Quadratic equations appear throughout the IGCSE and A-Level Mathematics syllabus, from algebraic manipulation to coordinate geometry, graph sketching and real-world modelling. This revision guide explains how to recognise a quadratic equation, how to rewrite it in standard form, and how to solve it by factorising, by completing the square and by the quadratic formula. It also covers the discriminant, the graphical meaning of roots, quadratic word problems, and the simultaneous-equation questions that mix a straight line with a curve. Work through the paired English and Chinese notes, then test yourself on the practice checklist at the end.
二次方程在 IGCSE 与 A-Level 数学大纲中贯穿始终，从代数运算到坐标几何、图像绘制以及实际建模都有出现。本复习指南将介绍如何识别二次方程、如何把它整理成标准形式，以及如何用因式分解法、配方法和求根公式求解。文章还会讲解判别式、根在图像上的意义、二次方程应用题，以及直线与曲线联立的方程组题型。请按中英对照笔记学习，并在最后的练习清单中自我检测。
1. What is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is any equation that can be written in the general form shown below, where a, b and c are constants and a ≠ 0. The highest power of the variable x is 2, which is why the equation is called 'quadratic'. If a were equal to zero, the x² term would vanish and the equation would become linear, not quadratic.
一元二次方程是可以写成下面一般形式的方程，其中 a、b、c 是常数且 a ≠ 0。变量 x 的最高次数是 2，因此它被称为二次方程。如果 a 等于 0，x² 项就会消失，方程变成一次方程，不再是二次方程。
a x² + b x + c = 0, a ≠ 0
ged as x² − 9 = 0. However, 3x + 4 = 0 is linear, and x³ − 2x = 0 is cubic, not quadratic.
例如，2x² − 5x + 3 = 0 是二次方程，其中 a = 2，b = −5，c = 3。方程 x² = 9 也是二次方程，因为它可以整理为 x² − 9 = 0。但 3x + 4 = 0 是一次方程，x³ − 2x = 0 是三次方程，都不属于二次方程。
A quadratic equation can have two distinct real solutions, one repeated real solution, or no real solutions at all. The number of real solutions depends on the value of the discriminant, which is introduced later in this guide. Note the word 'solutions' is used interchangeably with 'roots' and, in graph questions, with 'x-intercepts'.
二次方程可以有两个不相等的实数解、一个重复的实数解，或者没有实数解。实数解的个数取决于判别式的值，本指南稍后会介绍判别式。要注意的是，"解"与"根"在这里是同义的，在图像题中它们还等同于曲线与 x 轴的交点横坐标。
2. Standard Form and Identifying Coefficients | 标准形式与系数识别
de so that the other side equals zero, and arrange the terms in descending powers of x. Then identify a, b and c carefully, including their signs. Writing the coefficients down explicitly before you start is the single most effective way to avoid careless errors.
解二次方程之前，要先把方程写成标准形式：a x² + b x + c = 0。把所有项移到一边，令另一边等于 0，并按 x 的降幂排列。然后仔细识别 a、b、c，包括它们的符号。动手计算之前先把三个系数明确写出来，是避免粗心错误最有效的办法。
and c = −5. Misreading c as 5 instead of −5 is a very common source of mistakes, and it changes the answers completely.
例如，把 5 − 3x = 2x² 整理为 −2x² − 3x + 5 = 0，或者乘以 −1 后等价地写成 2x² + 3x − 5 = 0。在第二个版本中，a = 2，b = 3，c = −5。把 c 误读为 5 而不是 −5 是常见错误来源，而且会让答案完全改变。
5 − 3x = 2x² → 2x² + 3x − 5 = 0 (a = 2, b = 3, c = −5)
e x² + 6x = 10, then rearranged to x² + 6x − 10 = 0. This makes the coefficients easier to work with. Similarly, brackets should always be expanded before coefficients are read off, since a product such as (x + 1)(x − 4) = 0 must be converted into x² − 3x − 4 = 0 if you intend to use a = 1, b = −3, c = −4.
如果方程中含有分数，要先乘以最小公分母。例如，x²/2 + 3x = 5 可以先乘以 2，得到 x² + 6x = 10，再整理为 x² + 6x − 10 = 0。这样系数会更容易处理。同样，遇到括号时应当先展开再读系数，因为像 (x + 1)(x − 4) = 0 这样的乘积式，若要用 a = 1、b = −3、c = −4，必须先展开成 x² − 3x − 4 = 0。
3. Solving by Factorising | 因式分解法
roduct of two numbers is zero, then at least one of them must be zero. In symbols, if p q = 0 then p = 0 or q = 0. Applied to a factorised quadratic, this turns one equation into two much simpler linear equations.
当二次式能整齐地分解为整数系数的因式时，因式分解法是最快的方法。它的依据是"零乘积性质"：如果两个数的乘积等于零，那么其中至少有一个数为零。用符号表示就是：若 p q = 0，则 p = 0 或 q = 0。把它用在已分解的二次式上，就能把一个方程拆成两个更简单的一次方程。
if (x − p)(x − q) = 0 then x = p or x = q
−3 = 6 and −2 + (−3) = −5. Hence x = 2 or x = 3. Always expand your brackets back mentally to check that the middle term and the constant term are correct.
要分解 x² + b x + c，需要找到两个数，它们相乘等于 c，相加等于 b。例如，x² − 5x + 6 = 0 可分解为 (x − 2)(x − 3) = 0，因为 −2 × −3 = 6，且 −2 + (−3) = −5。因此 x = 2 或 x = 3。分解完成后，最好在脑中把括号重新展开一次，检查一次项和常数项是否正确。
ing x = −1/2 or x = −3. A useful technique is to multiply a and c to get 6, then find two numbers multiplying to 6 and adding to 7, namely 1 and 6, and split the middle term before grouping. Remember that a factorised answer is only valid if the original equation was first put equal to zero.
当 x² 的系数不是 1 时，试算会稍微复杂一些。对于 2x² + 7x + 3 = 0，分解形式是 (2x + 1)(x + 3) = 0，得到 x = −1/2 或 x = −3。一个实用技巧是先把 a 与 c 相乘得到 6，再找两个数使它们相乘等于 6、相加等于 7，即 1 和 6，然后拆分中间项并分组提取公因式。要记住，只有在原方程已经等于 0 的前提下，因式分解的结论才成立。
4. Common Factors and the Difference of Two Squares | 公因式与平方差
tor of x, take it out first. The equation 3x² − 12x = 0 becomes 3x(x − 4) = 0, so x = 0 or x = 4. Never divide both sides by x, because that silently throws away the root x = 0.
有两类特殊结构在考试中反复出现，几乎可以一眼认出。第一类是公因式：如果每一项都含有 x 这个因式，就先把它提出来。方程 3x² − 12x = 0 可写成 3x(x − 4) = 0，因此 x = 0 或 x = 4。千万不要两边同时除以 x，因为那样会悄悄丢掉 x = 0 这个根。
refore x = ±3. It also applies to expressions like 4x² − 25 = 0, giving (2x + 5)(2x − 5) = 0 and x = ±5/2. Recognising this pattern saves a great deal of time compared with using the formula.
第二类是平方差：a² − b² = (a + b)(a − b)。它可以处理像 x² − 9 = 0 这样的方程，分解为 (x + 3)(x − 3) = 0，因此 x = ±3。它也适用于 4x² − 25 = 0 这类式子，分解为 (2x + 5)(2x − 5) = 0，得到 x = ±5/2。与使用求根公式相比，识别出这一结构能节省大量时间。
a² − b² = (a + b)(a − b)
eal numbers and has no real roots. If a question asks for the exact number of solutions, this is one situation where the answer is zero.
要注意，两个平方之和，例如 x² + 9 = 0，在实数范围内无法分解，也没有实数根。如果题目问实数解的个数，这就是答案为"零个"的一种情形。
5. Completing the Square | 配方法
Completing the square rewrites a quadratic in the form (x + p)² + q, which reveals the minimum or maximum value of the expression and the coordinates of the vertex. The method also solves equations that do not factorise, and it is the standard route to the quadratic formula itself. For x² + b x + c, take half of b, square it, add it and subtract it again.
配方法把二次式改写成 (x + p)² + q 的形式，从而直接显示出表达式的最小值或最大值，以及抛物线的顶点坐标。这种方法还能求解无法因式分解的方程，并且正是推导求根公式的标准途径。对于 x² + b x + c，取 b 的一半，平方后加上再减去。
x² + b x + c = (x + b/2)² − (b/2)² + c
3)² − 9 − 10 = 0, that is (x + 3)² = 19. Taking square roots of both sides gives x + 3 = ±√19, so x = −3 + √19 or x = −3 − √19 ese are surds, so a decimal answer such as x ≈ 1.36 or x ≈ −7.36 is only an approximation.
以 x² + 6x − 10 = 0 为例。6 的一半是 3，3² = 9，所以 x² + 6x = (x + 3)² − 9。方程变为 (x + 3)² − 9 − 10 = 0，即 (x + 3)² = 19。两边开平方得到 x + 3 = ±√19，因此 x = −3 + √19 或 x = −3 − √19。这些是无理数，所以像 x ≈ 1.36 或 x ≈ −7.36 这样的答案只是近似值。
3. When the coefficient of x² is not 1, factor it out first: for 2x² − 8x + 3, write 2(x² − 4x) + 3, complete the square inside the bracket to get 2[(x − 2)² − 4] + 3, and simplify to 2(x − 2)² − 5. Always take the leading coefficient out before halving b.
配方后的形式 (x + 3)² − 19 还告诉你，曲线 y = x² + 6x − 10 的最低点是 (−3, −19)，对称轴是 x = −3。当 x² 的系数不是 1 时，要先把它提出来：对于 2x² − 8x + 3，写成 2(x² − 4x) + 3，在括号内配方得到 2[(x − 2)² − 4] + 3，化简为 2(x − 2)² − 5。一定要先把首项系数提出来，再去取 b 的一半。
6. The Quadratic Formula | 求根公式
The quadratic formula solves any quadratic equation, whether or not it factorises, and it should be memorised exactly. Substituting the values of a, b and c from standard form gives the two roots directly. The formula is printed on most formula sheets, but relying on the sheet wastes time in an exam.
求根公式可以求解任何二次方程，无论它能否因式分解，因此必须准确记忆。把标准形式中的 a、b、c 代入公式，就能直接得到两个根。虽然大多数公式表上都会印出这个公式，但每次翻表查看会浪费考试时间。
x = (−b ± √b² − 4ac) / (2a)
3x − 5 = 0, where a = 2, b = 3 and c = −5. The discriminant is b² − 4ac = 9 + 40 = 49, and √49 back into the original equation, and factorising confirms the answer: (2x + 5)(x − 1) = 0.
以 2x² + 3x − 5 = 0 为例，其中 a = 2，b = 3，c = −5。判别式 b² − 4ac = 9 + 40 = 49，而 √49 = 7。因此 x = (−3 ± 7) / 4，得到 x = 4/4 = 1 或 x = −10/4 = −2.5。把两个根代回原方程都可以检验，而因式分解 (2x + 5)(x − 1) = 0 也证实了答案。
r a given number of decimal places. For x² − 4x + 1 = 0, a = 1, b = −4, c = 1, so x = (4 ± √12) / 2 = (4 ± 2√3) / 2 = 2 ± √3. Simplify surds whenever possible, and always write both roots clearly separated by 'or'.
当判别式不是完全平方数时，除非题目要求保留若干位小数，否则应保留根号形式。对于 x² − 4x + 1 = 0，a = 1，b = −4，c = 1，所以 x = (4 ± √12) / 2 = (4 ± 2√3) / 2 = 2 ± √3。能化简的根号一定要化简，并且始终把两个根写清楚，用"或"分开。
7. The Discriminant and the Nature of Roots | 判别式与根的性质
The expression under the square root sign in the quadratic formula is called the discriminant, and it determines how many real roots the equation has without the need to solve it. This is a frequent exam question, often phrased as 'find the number of real solutions' or 'show that the equation has two distinct roots'.
求根公式中根号下的表达式称为判别式，它无需真正求解就能决定方程有多少个实数根。这是考试中的高频题型，常见问法包括"求实数解的个数"或"证明该方程有两个不相等的实根"。
Δ = b² − 4ac
| Value of Δ | 判别式的值 | Nature of roots | 根的性质 | Graph | 图像 |
| Δ > 0 大于零 |
Two distinct real roots 两个不相等的实根 |
Curve cuts the x-axis twice 曲线与 x 轴交于两点 |
| Δ = 0 等于零 |
One repeated real root 一个重复的实根（重根） |
Curve touches the x-axis once 曲线与 x 轴相切于一点 |
| Δ < 0 小于零 |
No real roots 没有实根 |
Curve does not meet the x-axis 曲线与 x 轴不相交 |
al roots. Questions sometimes give a quadratic in terms of an unknown constant k and ask for the range of k values that produce two distinct roots; set Δ > 0, solve the resulting inequality, and state the answer in inequality form.
例如，方程 x² − 6x + 9 = 0 的 Δ = 36 − 36 = 0，因此它有一个重根，即 x = 3。相反，x² + x + 1 = 0 的 Δ = 1 − 4 = −3 < 0，所以没有实根。有些题目会给出含未知常数 k 的二次方程，要求求出使方程有两个不相等的实根时 k 的取值范围；方法是令 Δ > 0，解出所得不等式，并用不等式形式写出答案。
8. Graphical Interpretation of Roots | 根的图像解释
. Its roots are exactly the x-coordinates of the points where the curve crosses or touches the x-axis, which is why solving a quadratic equation and finding x-intercepts are the same task. Recognising this link allows you to check algebraic answers against a sketch and to answer 'how many solutions' questions without calculating.
函数 y = a x² + b x + c 的图像是一条抛物线。它的根正好就是曲线与 x 轴相交或相切处的横坐标，因此"解二次方程"和"求 x 轴截距"本质上是同一件事。认识到这层联系，你就能用草图检验代数答案，也能在不计算的情况下回答"有多少个解"的问题。
try x = −b/(2a), and the y-coordinate of the vertex is the minimum or maximum value of the quadratic expression. Completing the square gives the same information in the form (x + p)² + q, where the vertex is (−p, q).
若 a > 0，抛物线开口向上，有最低点；若 a < 0，抛物线开口向下，有最高点。顶点位于对称轴 x = −b/(2a) 上，顶点的纵坐标就是该二次式的最大值或最小值。配方法给出的 (x + p)² + q 形式提供了同样的信息，此时顶点为 (−p, q)。
q). If the curve passes through (0, c) as well, substitute that point to find k. Remember
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