📚 Quadratic Equations: Factorising, Completing the Square and the Quadratic Formula | 二次方程：因式分解、配方法与求根公式
A quadratic equation is one of the most important topics in the IGCSE Mathematics syllabus. It appears in pure algebra questions, in graph sketching, in problem solving and in applications such as area, speed and simple projectile motion. This article covers the standard form, factorising, completing the square, the quadratic formula, the discriminant, graph interpretation, word problems and exam technique. You should aim to solve any quadratic equation by at least two methods and check your answers by substitution.
二次方程是 IGCSE 数学大纲中最重要的主题之一。它出现在纯代数题、图像绘制、问题解决以及面积、速度和简单抛体运动等应用中。本文涵盖标准形式、因式分解、配方法、求根公式、判别式、图像解读、文字题以及考试技巧。你应当至少会用两种方法解任何二次方程，并通过代入检验答案。
1. What is a Quadratic Equation? | 什么是二次方程？
0, the equation becomes linear rather than quadratic, so the condition a ≠ 0 is essential.
二次方程是次数为 2 的多项式方程。它的一般形式是 ax² + bx + c = 0，其中 a、b、c 是常数且 a ≠ 0。未知数（通常写作 x）的最高次数是 2。如果 a = 0，方程就变成一次方程而不是二次方程，因此条件 a ≠ 0 是至关重要的。
c equations. The equation 3x + 7 = 0 is not quadratic because the highest power of x is 1.
例如，x² − 5x + 6 = 0、2x² + 3x − 2 = 0 和 −x² + 4x = 0 都是二次方程。方程 3x + 7 = 0 不是二次方程，因为 x 的最高次数是 1。
Quadratic equations can have two real solutions, one repeated real solution, or no real solutions. In IGCSE level, you usually work with real numbers, but you should recognise that some equations have no real roots. Later we will use the discriminant to decide this without solving fully.
二次方程可以有两个实数解、一个重复实数解或没有实数解。在 IGCSE 阶段，你通常在实数范围内解题，但你应当认识到有些方程没有实根。稍后我们会使用判别式在不解出全部答案的情况下判断根的情况。
2. Standard Form and Key Terms | 标准形式与关键术语
lled the leading coefficient, b is the coefficient of x, and c is the constant term. The unknown is usually x, but you can use any letter such as t or y.
二次方程的标准形式是 ax² + bx + c = 0。数字 a 称为首项系数，b 是 x 的系数，c 是常数项。未知数通常是 x，但你也可以使用任何字母，如 t 或 y。
rearranged to 3x² − 5x − 2 = 0. This makes factorising and using the formula much easier.
在求解之前，你应当总是重新整理方程，使一边为零，并且各项按 x 的降幂排列。例如，3x² = 5x + 2 应整理为 3x² − 5x − 2 = 0。这样因式分解和使用公式都会方便得多。
th types are common in IGCSE questions, and the method of factorising changes slightly for non-monic cases.
首一二次方程是指 a = 1 的方程，例如 x² + 7x + 10 = 0。非首一二次方程是指 a ≠ 1 的方程，例如 2x² − 3x + 1 = 0。这两种类型在 IGCSE 考试中都很常见，非首一情形的因式分解方法会略有不同。
3. Solving by Factorising | 因式分解法
Factorising is often the fastest method when the quadratic expression can be written as a product of two linear factors. The key idea is the zero product property: if the product of two numbers is zero, then at least one of the numbers must be zero.
当二次式可以写成两个一次因式的乘积时，因式分解通常是最快的方法。关键思想是零乘积性质：如果两个数的乘积为零，那么至少其中一个数必须为零。
nd −3, so we write (x − 2)(x − 3) = 0. Then x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.
例如，解方程 x² − 5x + 6 = 0。我们寻找两个数，它们的乘积为 +6，和为 −5。这两个数是 −2 和 −3，因此我们写成 (x − 2)(x − 3) = 0。然后 x − 2 = 0 或 x − 3 = 0，得到 x = 2 或 x = 3。
actor in pairs: 2x(x + 1) + 3(x + 1) = 0, so (2x + 3)(x + 1) = 0. Thus x = −3/2 or x = −1.
对于非首一二次方程，如 2x² + 5x + 3 = 0，你可以使用 ac 方法。将 a 乘以 c：2 × 3 = 6。找到两个数，它们相乘为 6 且相加为 5；它们是 2 和 3。将中间项改写：2x² + 2x + 3x + 3 = 0。分组分解：2x(x + 1) + 3(x + 1) = 0，所以 (2x + 3)(x + 1) = 0。因此 x = −3/2 或 x = −1。
f you have x² + 5x = −6, first add 6 to both sides to get x² + 5x + 6 = 0, then factorise.
在因式分解之前，务必检查方程是否等于零。如果你有 x² + 5x = −6，首先两边加 6，得到 x² + 5x + 6 = 0，然后再因式分解。
4. Solving by Completing the Square | 配方法
Completing the square is a powerful method because it works for every quadratic, even when factorising is difficult or impossible. It also leads directly to the quadratic formula and helps you find the turning point of a quadratic graph.
配方法是一种强大的方法，因为它适用于所有二次方程，即使因式分解困难或不可能时也能使用。它还直接引出求根公式，并帮助你找到二次函数图像的顶点。
of the coefficient of x, which is 3, and square it to get 9. Then x² + 6x = (x + 3)² − 9.
对于首一二次式，关键恒等式是 x² + 2hx + h² = (x + h)²。要为 x² + 6x 配方，取 x 系数的一半，即 3，再平方得到 9。然后 x² + 6x = (x + 3)² − 9。
the square root of both sides: x + 3 = ±2. Therefore x = −3 ± 2, giving x = −1 or x = −5.
例题：用配方法解 x² + 6x + 5 = 0。先将常数移项：x² + 6x = −5。两边加 9：x² + 6x + 9 = 4。左边写成 (x + 3)²，所以 (x + 3)² = 4。两边开平方：x + 3 = ±2。因此 x = −3 ± 2，得到 x = −1 或 x = −5。
le, 2x² + 8x + 5 = 0 becomes x² + 4x + 2.5 = 0. Then complete the square in the usual way.
当首项系数 a 不等于 1 时，先把每一项都除以 a。例如，2x² + 8x + 5 = 0 变为 x² + 4x + 2.5 = 0。然后按通常方法配方。
Completing the square is especially useful when the question asks for the solution in surd form or when you need to find the vertex of a parabola. The completed square form a(x + p)² + q shows the turning point at (−p, q).
当题目要求以根式形式给出解，或者需要找到抛物线的顶点时，配方法特别有用。配方后的形式 a(x + p)² + q 可直接显示顶点为 (−p, q)。
5. The Quadratic Formula | 求根公式
The quadratic formula gives the solution of ax² + bx + c = 0 directly. It states that:
求根公式直接给出 ax² + bx + c = 0 的解。它写作：
x = (−b ± √b² − 4ac) ÷ 2a
the general equation ax² + bx + c = 0, so it is not a separate trick but a general result.
该公式适用于所有实系数二次方程，只要 a ≠ 0。它是由对一般方程 ax² + bx + c = 0 进行配方推导出来的，因此它不是一个孤立的技巧，而是一个普遍结论。
4, c = −3. Substitute into the formula: x = (4 ± √((−4)² − 4 × 2 × (−3))) ÷ (2 × 2) = (4 ± √16 + 24) ÷ 4 = (4 ± √40) ÷ 4 = (4 ± 2√10) ÷ 4 = (2 ± √10) ÷ 2.
例题：解 2x² − 4x − 3 = 0。这里 a = 2，b = −4，c = −3。代入公式：x = (4 ± √((−4)² − 4 × 2 × (−3))) ÷ (2 × 2) = (4 ± √16 + 24) ÷ 4 = (4 ± √40) ÷ 4 = (4 ± 2√10) ÷ 4 = (2 ± √10) ÷ 2。
r b = −4, −b is +4. Also, the entire numerator must be divided by 2a, not just part of it.
注意负号。一个常见错误是当 b 已经是负数时，仍把 −b 写成负数。对于 b = −4，−b 是 +4。另外，整个分子必须除以 2a，而不是只除以一部分。
When the answer is an irrational surd, IGCSE questions often ask for an exact answer in surd form or for approximate decimal values to a given number of significant figures. Use your calculator carefully and only round at the final step.
当答案是无理根式时，IGCSE 题目通常要求以根式形式给出精确答案，或按给定的有效数字给出近似小数值。小心使用计算器，并且只在最后一步取近似值。
6. The Discriminant and Nature of Roots | 判别式与根的性质
on inside the square root in the quadratic formula. It is usually written as Δ = b² − 4ac.
二次方程 ax² + bx + c = 0 的判别式是求根公式中平方根符号内的表达式。它通常写作 Δ = b² − 4ac。
The value of the discriminant tells you the nature of the roots without solving the equation:
判别式的值可以在不求解方程的情况下告诉你根的性质：
Δ > 0: the equation has two distinct real roots.
Δ = 0: the equation has one repeated real root, sometimes called two equal roots.
Δ < 0: the equation has no real roots. The roots are complex or imaginary, but they are not part of the IGCSE real number work.
Δ > 0：方程有两个不同的实数根。
Δ = 0：方程有一个重复实数根，有时称为两个相等的根。
Δ < 0：方程没有实数根。根是复数或虚数，但它们不属于 IGCSE 实数范围。
= 5, so Δ = 4² − 4 × 1 × 5 = 16 − 20 = −4. Since Δ < 0, the equation has no real roots.
例题：判断 x² + 4x + 5 = 0 的根的性质。这里 a = 1，b = 4，c = 5，所以 Δ = 4² − 4 × 1 × 5 = 16 − 20 = −4。因为 Δ < 0，该方程没有实数根。
and solve for k. For example, x² + kx + 4 = 0 has equal roots when k² − 16 = 0, so k = ±4.
有些 IGCSE 题目要求你求出使二次方程有两个相等根时 k 的值。在这种情况下，令判别式等于零并解 k。例如，x² + kx + 4 = 0 有相等根，当 k² − 16 = 0，所以 k = ±4。
7. Graphs of Quadratic Functions | 二次函数图像
x² + bx + c, and its graph is a parabola. If a > 0, the parabola opens upwards and has a minimum point. If a < 0, the parabola opens downwards and has a maximum point.
二次函数的形式为 y = ax² + bx + c，它的图像是一条抛物线。如果 a > 0，抛物线开口向上，且有一个最小值点。如果 a < 0，抛物线开口向下，且有一个最大值点。
x² + bx + c = 0, if they exist. These are often called the roots or zeros of the function.
图像与 y 轴的截距是常数项 c，因为当 x = 0 时，y = c。图像与 x 轴的截距是 ax² + bx + c = 0 的实数解（如果存在）。这些通常称为函数的根或零点。
h the vertex, or turning point, of the parabola. The vertex has coordinates (−b ÷ 2a, f(−b ÷ 2a)).
对称轴是竖直线 x = −b ÷ 2a。这条线经过抛物线的顶点，也就是拐点。顶点的坐标为 (−b ÷ 2a, f(−b ÷ 2a))。
y-intercept is 3, and the roots are x = 1 and x = 3 because x² − 4x + 3 = (x − 1)(x − 3).
例如，y = x² − 4x + 3 有 a = 1，b = −4，c = 3。对称轴是 x = 2。代入 x = 2 得到 y = 4 − 8 + 3 = −1，所以顶点是 (2, −1)。y 轴截距为 3，根为 x = 1 和 x = 3，因为 x² − 4x + 3 = (x − 1)(x − 3)。
Sketching a quadratic graph usually requires the y-intercept, the roots if possible, the vertex and the shape. You do not need to plot many points if you know these key features.
画二次函数图像通常需要 y 轴截距、如果可能的话根、顶点以及开口方向。如果你知道这些关键特征，就无需描很多点。
8. Applications and Word Problems | 应用问题与文字题
Quadratic equations appear in many real-life contexts at IGCSE level. Typical problems involve areas of rectangles, consecutive integers, simple projectile paths and optimisation of profit or cost.
在 IGCSE 阶段，二次方程出现在许多实际情境中。典型问题涉及矩形面积、连续整数、简单抛体轨迹以及利润或成本的最优化。
8)(x − 5) = 0. Since x must be positive, x = 5. The width is 5 cm and the length is 8 cm.
例题：一个矩形的长比宽长 3 厘米，面积为 40 平方厘米。设宽为 x 厘米。那么长为 x + 3 厘米。面积方程为 x(x + 3) = 40，所以 x² + 3x − 40 = 0。因式分解：(x + 8)(x − 5) = 0。由于 x 必须为正，所以 x = 5。宽为 5 厘米，长为 8 厘米。
When solving word problems, always define your variable clearly. Write the equation from the information given, solve it, and then check that the solutions make sense in the original situation. Reject negative or zero lengths, times or speeds when they are not physically meaningful.
解文字题时，务必清楚地定义变量。根据所给信息写出方程，求解，然后检查解是否符合原始情境。当长度、时间或速度为负值或零且没有实际意义时，应舍去这些解。
is the initial vertical speed. Solving h = 0 tells you when the object reaches the ground.
抛体问题通常给出形如 h = −5t² + vt + h₀ 的二次式，其中 h 是高度，t 是时间，v 是初始竖直速度。解 h = 0 可以得到物体何时到达地面。
9. Common Mistakes and How to Avoid Them | 常见错误与避免方法
oes not allow you to set each factor to zero. You must first rearrange to x² + 3x − 4 = 0.
一个非常常见的错误是在方程不等于零时进行因式分解。例如，如果 x² + 3x = 4，写成 (x)(x + 3) = 4 并不能让你令每个因式为零。你必须先重新整理为 x² + 3x − 4 = 0。
Sign errors in the quadratic formula are also frequent. If b is negative, −b becomes positive. Also remember that the term is b², so squaring a negative number gives a positive result. The expression is b² − 4ac, not b²
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