📚 T-K-1034 IGCSE Mathematics: Quadratic Equations | T-K-1034 IGCSE 数学：二次方程
A quadratic equation is one of the most important algebraic ideas in IGCSE Mathematics. It appears in topics such as factorisation, functions, graphs, and real-life modelling. In this article, you will learn how to recognise, solve, and interpret quadratic equations step by step.
二次方程是 IGCSE 数学中最重要的代数概念之一。它出现在因式分解、函数、图像和实际建模等主题中。本文将带你逐步学会识别、求解和解释二次方程。
1. Recognising a Quadratic Equation | 识别二次方程
a, b and c are constants and a ≠ 0. If a = 0, the equation becomes linear, not quadratic.
二次方程是指任何可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b 和 c 是常数，且 a ≠ 0。如果 a = 0，方程就变成了一次方程，而不是二次方程。
ratic equations. However, 3x + 1 = 0 is not quadratic because the highest power of x is 1.
例如，2x² + 5x − 3 = 0 和 x² − 7 = 0 都是二次方程。然而，3x + 1 = 0 不是二次方程，因为 x 的最高次数是 1。
In IGCSE questions, you may also see quadratic equations that are not yet in standard form. Always rearrange the equation so one side is zero before solving.
在 IGCSE 题目中，你可能会遇到还没有写成标准形式的二次方程。求解前一定要先移项，让等式的一边等于 0。
2. Standard Form and Coefficients | 标准形式与系数
nd c are called coefficients: a is the coefficient of x², b is the coefficient of x, and c is the constant term.
二次方程的标准形式是 ax² + bx + c = 0。数字 a、b 和 c 被称为系数：a 是 x² 的系数，b 是 x 的系数，c 是常数项。
dratic formula or the discriminant. For 2x² − 4x + 1 = 0, we have a = 2, b = −4 and c = 1.
在使用公式法或判别式之前，正确识别这些系数非常重要。对于 2x² − 4x + 1 = 0，我们有 a = 2，b = −4，c = 1。
you must first rewrite it as 2x² − 4x + 1 = 0 to obtain the correct signs for a, b and c.
如果方程写成 2x² = 4x − 1，你必须先把它改写为 2x² − 4x + 1 = 0，才能得到 a、b 和 c 的正确符号。
3. Solving by Factorising | 因式分解法
ten as a product of two brackets. For example, x² + 5x + 6 = 0 becomes (x + 2)(x + 3) = 0.
当二次方程左边可以写成两个括号的乘积时，因式分解是最快的解法。例如，x² + 5x + 6 = 0 可以写成 (x + 2)(x + 3) = 0。
+ 2 = 0 gives x = −2, and x + 3 = 0 gives x = −3. So the solutions are x = −2 and x = −3.
因式分解后，让每个括号分别等于 0：x + 2 = 0 得到 x = −2，x + 3 = 0 得到 x = −3。因此解是 x = −2 和 x = −3。
6x + 3 = 0. Factor by grouping to get (2x + 1)(x + 3) = 0, which gives x = −1/2 or x = −3.
对于 2x² + 7x + 3 = 0 这样的方程，需要拆分中间项：2x² + x + 6x + 3 = 0。分组分解得到 (2x + 1)(x + 3) = 0，所以 x = −1/2 或 x = −3。
4. Completing the Square | 配方法
Completing the square rewrites a quadratic expression in the form (x + p)² + q, which makes the turning point easy to find. For x² + 6x + 5, take half of 6, which is 3, and write (x + 3)² − 9 + 5.
配方法把二次表达式写成 (x + p)² + q 的形式，这样可以很容易地找到顶点。对于 x² + 6x + 5，取 6 的一半，即 3，写成 (x + 3)² − 9 + 5。
Setting this equal to zero gives (x + 3)² = 4, so x + 3 = ±2. Therefore x = −1 or x = −5.
化简后得到 (x + 3)² − 4。让它等于 0，得到 (x + 3)² = 4，所以 x + 3 = ±2。因此 x = −1 或 x = −5。
When the coefficient of x² is not 1, first factor it out from the x² and x terms. For example, 2x² + 8x + 6 becomes 2(x² + 4x) + 6, then complete the square inside the brackets.
当 x² 的系数不是 1 时，先把它从 x² 项和 x 项中提出来。例如，2x² + 8x + 6 先写成 2(x² + 4x) + 6，然后在括号内完成配方。
5. The Quadratic Formula | 公式法
The quadratic formula is a general method that works for every quadratic equation. The solutions are given by:
公式法是适用于所有二次方程的通用方法。它的解为：
x = [ −b ± √b² − 4ac ] ÷ 2a
d c = −2. Substituting into the formula gives x = [ −5 ± √(5² − 4 × 3 × (−2)) ] ÷ (2 × 3).
对于 3x² + 5x − 2 = 0，确定 a = 3，b = 5，c = −2。代入公式得到 x = [ −5 ± √(5² − 4 × 3 × (−2)) ] ÷ (2 × 3)。
quare root: 25 + 24 = 49, so x = (−5 ± 7) ÷ 6. This gives x = 2/6 = 1/3 or x = −12/6 = −2.
化简根号内部：25 + 24 = 49，所以 x = (−5 ± 7) ÷ 6。得到 x = 2/6 = 1/3 或 x = −12/6 = −2。
Always write the formula as x = [ −b ± √b² − 4ac ] ÷ 2a when using it in an exam. This helps avoid sign errors and shows clear method marks.
考试中使用公式时，要始终写成 x = [ −b ± √b² − 4ac ] ÷ 2a 的形式。这有助于避免符号错误，也能展示清晰的过程分。
6. The Discriminant and Roots | 判别式与根的性质
D = b² − 4ac. It tells you the nature of the roots without actually solving the equation.
二次方程 ax² + bx + c = 0 的判别式定义为 D = b² − 4ac。它可以在不实际求解方程的情况下告诉你根的性质。
exactly one real repeated root. If D < 0, there are no real roots, only complex roots.
如果 D > 0，方程有两个不同的实根。如果 D = 0，方程有一个实重根。如果 D < 0，方程没有实根，只有复数根。
xample, x² − 4x + 4 = 0 has D = (−4)² − 4 × 1 × 4 = 0, so it has one repeated root: x = 2.
例如，x² − 4x + 4 = 0 的判别式 D = (−4)² − 4 × 1 × 4 = 0，所以它有一个重根：x = 2。
t kx² + 2x + 1 = 0 has two real roots. Set up D > 0 and solve the resulting inequality.
IGCSE 题目经常要求求 k 的取值范围，使得 kx² + 2x + 1 = 0 有两个实根。这时要建立 D > 0 的不等式并求解。
7. Graphs of Quadratic Functions | 二次函数图像
. If a > 0, the parabola opens upwards and has a minimum point; if a < 0, it opens downwards and has a maximum point.
y = ax² + bx + c 的图像是一条抛物线。如果 a > 0，抛物线开口向上，有最小值点；如果 a < 0，抛物线开口向下，有最大值点。
l roots of ax² + bx + c = 0. The y-intercept is always at y = c because when x = 0, y = c.
图像与 x 轴的交点就是 ax² + bx + c = 0 的实根。与 y 轴的交点总是在 y = c 处，因为当 x = 0 时，y = c。
. In IGCSE, you may be asked to sketch the graph and label the roots, intercept and turning point.
顶点可以通过配方法或使用 x = −b ÷ 2a 来找到。在 IGCSE 中，你可能需要绘制图像并标出根、截距和顶点。
8. Roots, Vertex and Line of Symmetry | 根、顶点与对称轴
This line passes through the vertex and halfway between the two roots when they are real.
每个二次函数图像都有一条垂直对称轴，方程为 x = −b ÷ 2a。这条线经过顶点，并且当两个根为实数时，它位于两个根的正中间。
re x = 1 and x = 5. The midpoint is x = (1 + 5) ÷ 2 = 3, so the line of symmetry is x = 3.
对于 y = x² − 6x + 5，根是 x = 1 和 x = 5。中点是 x = (1 + 5) ÷ 2 = 3，所以对称轴是 x = 3。
× 3 + 5 = −4, so the vertex is (3, −4). The vertex is the minimum point because a > 0.
把 x = 3 代入方程得到 y = 3² − 6 × 3 + 5 = −4，所以顶点是 (3, −4)。由于 a > 0，这个顶点是最小值点。
9. Applications and Word Problems | 应用与文字题
unctions. The key is to translate the words into an equation of the form ax² + bx + c = 0.
二次方程可以模拟许多实际情况，例如抛体运动、面积问题和利润函数。关键是把文字转化为 ax² + bx + c = 0 形式的方程。
idth is x, and the area is 40, then x(x + 3) = 40. Expand and rearrange: x² + 3x − 40 = 0.
对于一个矩形，如果长是 x + 3，宽是 x，面积是 40，那么 x(x + 3) = 40。展开并移项：x² + 3x − 40 = 0。
, so x = 5 or x = −8. Since a length cannot be negative, x = 5 is the only valid solution.
因式分解得到 (x + 8)(x − 5) = 0，所以 x = 5 或 x = −8。由于长度不能为负，x = 5 是唯一有效的解。
Always check that your answers make sense in the original context. Reject roots that give impossible values such as negative lengths, negative time or negative quantities.
一定要检查答案在原始情境中是否有意义。舍去那些导致长度、时间或数量为负等不可能值的根。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
ual zero before factorising. For example, x² + 3x = 10 must first become x² + 3x − 10 = 0.
一个常见错误是在因式分解之前忘记先把方程移项使右边为 0。例如，x² + 3x = 10 必须先变成 x² + 3x − 10 = 0。
Another mistake is losing the ± sign in the quadratic formula. Always write x = [ −b ± √b² − 4ac ] ÷ 2a, not x = [ −b + √b² − 4ac ] ÷ 2a.
另一个错误是在公式法中丢失 ± 号。一定要写成 x = [ −b ± √b² − 4ac ] ÷ 2a，而不是 x = [ −b + √b² − 4ac ] ÷ 2a。
tituting b. If b is negative, −b becomes positive. For x² − 5x + 6 = 0, b = −5, so −b = 5.
此外，代入 b 时要注意符号。如果 b 是负数，−b 就变成正数。对于 x² − 5x + 6 = 0，b = −5，所以 −b = 5。
In IGCSE exams, show each step clearly: rewrite, identify a, b, c, substitute, simplify, and give both roots. Full method marks are often available even if a small arithmetic error is made.
在 IGCSE 考试中，要清晰地写出每一步：整理方程、确定 a、b、c、代入公式、化简，并给出两个根。即使出现小的算术错误，通常也能拿到过程分。
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