📚 Mastering Quadratic Equations for IGCSE Maths | IGCSE 数学二次方程完全掌握
A quadratic equation is one of the most frequently examined topics in IGCSE Mathematics. You will meet it in algebra, graphs, problem solving and even in applied questions involving area, motion or finance. This article gives you a complete walk-through of the methods you need to solve quadratic equations confidently and accurately.
二次方程是 IGCSE 数学中最常考查的主题之一。你会在代数、图像、解决问题以及涉及面积、运动或金融的应用题中遇到它。本文将带你完整掌握求解二次方程所需的方法，帮助你自信且准确地解题。
1. What Is a Quadratic Equation? | 什么是二次方程？
A quadratic equation is a polynomial equation in which the highest power of the variable is 2. It can always be recognised by the presence of an x² term, while terms with higher powers such as x³ are not allowed.
二次方程是一个多项方程式，其中变量的最高次数为 2。它的显著特点是含有 x² 项，并且不允许出现 x³ 等更高次数的项。
In IGCSE, most quadratic equations use x as the variable, but you may also see equations involving t, y or other letters. The general shape of a quadratic expression is ax² + bx + c, where a, b and c are constants and a ≠ 0.
在 IGCSE 考试中，大多数二次方程以 x 为变量，但你也可能看到含有 t、y 或其他字母的方程。二次表达式的一般形式为 ax² + bx + c，其中 a、b、c 为常数，且 a ≠ 0。
- Examples: x² + 5x + 6 = 0, 2x² - 3x - 5 = 0, 4x² - 9 = 0
- 示例：x² + 5x + 6 = 0、2x² - 3x - 5 = 0、4x² - 9 = 0
2. Standard Form and Key Terms | 标准式与关键术语
Before solving any quadratic equation, you should rewrite it in standard form. The standard form is obtained by moving all terms to one side of the equation so that the other side equals zero.
在求解任何二次方程之前，你应先将方程改写为标准式。标准式是将所有项移到方程的一边，使另一边等于零。
ax² + bx + c = 0
c is the constant term. For example, in 3x² - 7x + 2 = 0, we have a = 3, b = -7 and c = 2.
在这种形式中，a 是 x² 的系数，b 是 x 的系数，c 是常数项。例如，在 3x² - 7x + 2 = 0 中，a = 3，b = -7，c = 2。
If the equation is not in standard form, rearrange it carefully by collecting like terms and using inverse operations. This step is essential because the factorising method and the quadratic formula both require the standard form.
如果方程不是标准式，需要仔细整理，合并同类项并使用逆运算。这一步非常重要，因为因式分解法和求根公式都要求方程为标准式。
3. Solving by Factorising | 因式分解法
Factorising is often the fastest method when the quadratic expression can be written as the product of two linear brackets. After factorising, you apply the zero product property: if the product of two factors is zero, then at least one factor must be zero.
当二次表达式可以写成两个一次因式的乘积时，因式分解法通常是最快的方法。因式分解后，你应用零乘积性质：如果两个因式的乘积为零，那么至少有一个因式必须为零。
nd add to give 7. These numbers are 3 and 4, so the factorised form is (x + 3)(x + 4) = 0.
以方程 x² + 7x + 12 = 0 为例。我们需要找到两个数，它们相乘得 12，相加得 7。这两个数是 3 和 4，因此因式分解形式为 (x + 3)(x + 4) = 0。
x = -4. Always check your solutions by substituting them back into the original equation.
令每个括号分别为零，得到 x + 3 = 0 或 x + 4 = 0，因此解为 x = -3 或 x = -4。始终通过将解代回原方程来检验答案。
ple, 2x² + 5x + 3 = 0 can be factorised as (2x + 3)(x + 1) = 0, giving x = -3/2 or x = -1.
当 a ≠ 1 时，你可能需要分解中间项或使用尝试法。例如，2x² + 5x + 3 = 0 可以因式分解为 (2x + 3)(x + 1) = 0，得到 x = -3/2 或 x = -1。
4. Solving by Completing the Square | 配方法
Completing the square is a powerful algebraic technique that always works, even when the quadratic does not factorise easily. It is also the method used to derive the quadratic formula and to find the turning point of a parabola.
配方法是一种强大的代数技巧，即使二次方程不易因式分解，它也能始终有效。它也是推导求根公式以及求抛物线顶点的方法。
rite x² + bx as (x + b/2)² - (b/2)². This allows you to isolate the squared term and solve by taking square roots.
对于形如 x² + bx + c = 0 的首一二次方程，关键变换是将 x² + bx 写成 (x + b/2)² - (b/2)²。这样可以分离出平方项并通过开平方来求解。
x² + bx = (x + b/2)² - (b/2)²
)² - 9 = 7, so (x + 3)² = 16. Taking square roots gives x + 3 = ±4, hence x = 1 or x = -7.
例如，解方程 x² + 6x - 7 = 0。将其整理为 x² + 6x = 7，然后配方：(x + 3)² - 9 = 7，因此 (x + 3)² = 16。开平方得 x + 3 = ±4，所以 x = 1 或 x = -7。
If a ≠ 1, first divide the whole equation by a before completing the square. This normalisation step is often missed in exams, so write it down clearly to avoid errors.
如果 a ≠ 1，在配方前先对方程整体除以 a。考试中这一步常常被忽略，因此要清楚地写出来以避免错误。
5. The Quadratic Formula | 求根公式
The quadratic formula gives the solutions to any quadratic equation in standard form. It is especially useful when factorising is difficult or when the coefficients are not integers.
求根公式可以求解任何标准形式的二次方程。当因式分解困难或系数不是整数时，它尤其有用。
x = (-b ± √b² - 4ac) ÷ 2a
inus sign. You must identify a, b and c correctly from the standard form ax² + bx + c = 0.
在这个公式中，符号 ± 表示通常有两个解：一个使用加号，一个使用减号。你必须从标准式 ax² + bx + c = 0 中正确识别 a、b 和 c。
-5. Substituting into the formula gives x = (3 ± √((-3)² - 4 × 2 × (-5))) ÷ (2 × 2) = (3 ± √9 + 40) ÷ 4 = (3 ± √49) ÷ 4, so x = (3 + 7) ÷ 4 = 2.5 or x = (3 - 7) ÷ 4 = -1.
例如，解方程 2x² - 3x - 5 = 0。这里 a = 2，b = -3，c = -5。代入公式得到 x = (3 ± √((-3)² - 4 × 2 × (-5))) ÷ (2 × 2) = (3 ± √9 + 40) ÷ 4 = (3 ± √49) ÷ 4，所以 x = (3 + 7) ÷ 4 = 2.5 或 x = (3 - 7) ÷ 4 = -1。
When using the formula, write the substitution step clearly before simplifying. This helps you gain method marks even if a small arithmetic mistake occurs later.
使用公式时，在化简之前清楚地写出代入步骤。这样即使后面出现小的计算错误，也有助于你获得方法分。
6. The Discriminant: Nature of Roots | 判别式与根的性质
The discriminant is the part of the quadratic formula under the square root sign. It tells you how many real solutions a quadratic equation has without actually solving it.
判别式是求根公式中平方根符号下的部分。它可以在不求方程的情况下告诉你二次方程有多少个实数解。
Δ = b² - 4ac
- If b² - 4ac > 0, there are two distinct real roots.
- 如果 b² - 4ac > 0，则有两个不同的实数根。
- If b² - 4ac = 0, there is exactly one real root, sometimes called a repeated root.
- 如果 b² - 4ac = 0，则恰好有一个实数根，有时称为重根。
- If b² - 4ac < 0, there are no real roots; the equation has complex roots at advanced level.
- 如果 b² - 4ac < 0，则没有实数根；在更高阶学习中方程有复数根。
IGCSE questions may ask you to determine the number of solutions or to find values of a parameter, such as k, for which an equation has equal roots. To do this, set the discriminant equal to zero and solve for the parameter.
IGCSE 题目可能会要求你判断解的个数，或求参数的值，例如 k，使得方程有相等实根。要做到这一点，令判别式等于零并求解该参数。
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，抛物线开口向上并有最小值点。如果 a < 0，抛物线开口向下并有最大值点。
refore, solving the equation is the same as finding where the parabola crosses the x-axis.
图像与 x 轴的交点就是二次方程 ax² + bx + c = 0 的实数解。因此，解方程就相当于找出抛物线与 x 轴相交的位置。
k shows that the vertex is (h, k). For example, y = (x - 2)² + 3 has its vertex at (2, 3).
顶点可以通过配方法求得。将函数写成 y = (x - h)² + k 表明顶点为 (h, k)。例如，y = (x - 2)² + 3 的顶点是 (2, 3)。
If the discriminant is positive, the graph crosses the x-axis twice. If it is zero, the graph touches the x-axis at one point. If it is negative, the graph does not cross the x-axis at all.
如果判别式为正，图像与 x 轴相交两次。如果为零，图像与 x 轴相切于一点。如果为负，图像完全不与 x 轴相交。
8. Choosing the Best Method | 选择最佳解法
There are three main algebraic methods for solving quadratic equations in IGCSE: factorising, completing the square and using the quadratic formula. Choosing the best one can save time and reduce mistakes.
IGCSE 中求解二次方程有三种主要的代数方法：因式分解法、配方法和求根公式法。选择最佳方法可以节省时间并减少错误。
| Method | Best used when |
| Factorising | The quadratic has simple integer factors. |
| Completing the square | You need the vertex or the equation is monic. |
| Quadratic formula | Coefficients are awkward or factorising is unclear. |
In an exam, always check whether the quadratic can be factorised quickly. If not, the formula is usually the safest choice. Completing the square is best when a question asks for the turning point or asks you to solve by that specific method.
在考试中，总是先检查二次方程是否可以快速因式分解。如果不能，公式通常是最稳妥的选择。当题目要求求顶点或指定使用配方法时，配方法最合适。
9. Worked IGCSE-Style Example | IGCSE 例题精讲
otice that all terms have a common factor of 3, so dividing through gives x² - 4x + 3 = 0.
让我们用两种不同的方法解方程 3x² - 12x + 9 = 0 并比较答案。首先，注意到所有项都有公因数 3，因此方程两边除以 3 得到 x² - 4x + 3 = 0。
iply to 3 and add to -4. These are -1 and -3, so (x - 1)(x - 3) = 0. Hence x = 1 or x = 3.
方法一：因式分解法。我们需要两个数，相乘得 3，相加得 -4。这两个数是 -1 和 -3，因此 (x - 1)(x - 3) = 0。所以 x = 1 或 x = 3。
Method 2: Quadratic formula. Using a = 3, b = -12 and c = 9, we get x = (12 ± √144 - 108) ÷ 6 = (12 ± √36) ÷ 6 = (12 ± 6) ÷ 6. This gives x = 18 ÷ 6 = 3 or x = 6 ÷ 6 = 1. The solutions agree.
方法二：求根公式法。使用 a = 3，b = -12，c = 9，得到 x = (12 ± √144 - 108) ÷ 6 = (12 ± √36) ÷ 6 = (12 ± 6) ÷ 6。由此得到 x = 18 ÷ 6 = 3 或 x = 6 ÷ 6 = 1。两种方法的解一致。
This example shows that factorising is quicker when possible, but the formula gives the same result reliably. In an IGCSE exam, always show your method even if you can guess the roots.
这个例子说明，在可能的情况下因式分解更快，但公式法同样可靠地给出相同结果。在 IGCSE 考试中，即使你能猜出根，也要展示你的解题过程。
10. Common Pitfalls and Exam Tips | 常见错误与考试提示
Many students lose marks on quadratic equations for avoidable mistakes. The most common error is forgetting to rearrange the equation into standard form before identifying a, b and c.
许多学生在二次方程上因可避免的错误而失分。最常见的错误是在识别 a、b 和 c 之前忘记将方程整理为标准式。
x; instead rearrange to x² - 5x = 0 and factorise as x(x - 5) = 0, giving x = 0 or x = 5.
另一个常见错误是对方程两边同除以 x，而没有考虑 x 可能为零。这会使一个解消失。例如，在 x² = 5x 中，不要约去 x；而是将其整理为 x² - 5x = 0 并因式分解为 x(x - 5) = 0，得到 x = 0 或 x = 5。
Always simplify radicals completely and rationalise denominators if required. Also check whether the question asks for exact answers or answers to a given number of decimal places.
始终将根式完全化简，并在需要时将分母有理化。同时检查题目要求的是精确答案还是给定小数位的答案。
Finally, practise recognising the structure of quadratic equations in word problems. Phrases such as "the product of two consecutive integers" or "the area of a rectangle" often lead to quadratic equations that you can then solve using the methods above.
最后，练习在文字题中识别二次方程的结构。诸如“两个连续整数的乘积”或“矩形的面积”等表述通常会引出二次方程，然后你可以用上述方法求解。
Published by TutorHao | Maths Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图