📚 IGCSE Edexcel Maths Topic 53: Mastering Quadratic Equations | IGCSE爱德思数学专题53：精通二次方程
Quadratic equations form a fundamental part of the IGCSE Edexcel Mathematics syllabus. They appear across a wide range of question styles, from straightforward algebraic manipulation to complex problem-solving involving area, motion and optimisation. A deep understanding of both symbolic techniques (factorising, completing the square, quadratic formula) and graphical interpretation (parabolas, intercepts, vertex) is essential not only for achieving a high grade but also for building a bridge to A Level mathematics.
二次方程是IGCSE爱德思数学课程大纲的基础组成部分。这类方程涵盖广泛的题型，从直接的代数运算到涉及面积、运动和最优化的复杂问题解决。深入理解符号技巧（因式分解、配方法、二次公式）和图像解析（抛物线、截距、顶点）不仅对取得高分至关重要，也为衔接A Level数学搭建桥梁。
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 standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the variable x is 2, which is why it is called quadratic (from the Latin word quadratus meaning square).
二次方程是任何能写成标准形式 ax² + bx + c = 0 的方程，其中 a、b 和 c 是常数且 a ≠ 0。变量 x 的最高次数为 2，因此被称为二次方程。
ked to rearrange an equation into this standard form before solving. For example, 3x² = 5x – 2 must be rewritten as 3x² – 5x + 2 = 0.
在许多考试题目中，你需要先将一个方程整理成这种标准形式再求解。例如，3x² = 5x – 2 必须改写为 3x² – 5x + 2 = 0。
The solutions to a quadratic equation are called its roots. Depending on the equation, there can be two real roots, one repeated real root or no real roots at all.
二次方程的解被称为它的根。根据方程的不同，可能有两个实根、一个重复的实根，或者根本没有实根。
2. Factorising Quadratics | 因式分解二次式
we apply the principle that if the product equals zero, at least one factor must be zero.
因式分解通常是解二次方程最快的方法，前提是二次式可以顺利分解。目标是把 ax² + bx + c 写成两个一次式的乘积：(px + q)(rx + s) = 0，然后应用“积为零则至少一个因式为零”的原理。
and 5 work because 2 × 5 = 10 and 2 + 5 = 7, giving (x + 2)(x + 5) = 0. The roots are x = –2 and x = –5.
对于 a = 1 的简单二次式，我们寻找两个数，它们的乘积等于 c，相加等于 b。例如，因式分解 x² + 7x + 10，数字 2 和 5 满足，因为 2 × 5 = 10 且 2 + 5 = 7，得到 (x + 2)(x + 5) = 0，根为 x = –2 和 x = –5。
When a ≠ 1, you can use the ‘ Rewrite as 2x² + 6x + x + 3, then factorise as 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0.
当 a ≠ 1 时，可使用“ac 方法”。将 a 与 c 相乘，然后找出两个数，它们的乘积等于 ac，相加等于 b。分裂中间项并进行分组分解。对于 2x² + 7x + 3，ac = 6；数字 6 和 1 可行，因为 6 × 1 = 6 且 6 + 1 = 7。重写为 2x² + 6x + x + 3，然后因式分解为 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0。
3. Solving by Completing the Square | 配方法求解
Completing the square is a powerful technique that works for any quadratic equation and is particularly useful when the quadratic does not factorise easily. The idea is to rewrite ax² + bx + c in the form a(x + p)² + q, from which the roots can be found by isolating x.
配方法是一种强大的技巧，适用于任何二次方程，在二次式不易因式分解时尤为有用。其思想是将 ax² + bx + c 重写为 a(x + p)² + q 的形式，然后通过分离 x 来求根。
ratic x² + bx + c, we add and subtract (b/2)² to create a perfect square: x² + bx + (b/2)² – (b/2)² + c = (x + b/2)² + [c – (b/2)²].
对于首一二次式 x² + bx + c，我们加上并减去 (b/2)² 以创建完全平方：x² + bx + (b/2)² – (b/2)² + c = (x + b/2)² + [c – (b/2)²]。
² + 6x + 2 = 0 by completing the square. Half of 6 is 3, 3² = 9. So x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7 = 0. Then (x + 3)² = 7, giving x = –3 ± √7.
示例：用配方法解 x² + 6x + 2 = 0。6 的一半是 3，3² = 9。所以 x² + 6x + 2 = (x + 3)² – 9 + 2 = (x + 3)² – 7 = 0。于是 (x + 3)² = 7，得到 x = –3 ± √7。
If a ≠ 1, first divide every term by a before completing the square. This step ensures the coefficient of x² is 1, making the process straightforward.
如果 a ≠ 1，先在等式两边除以 a，再完成配方。这一步确保 x² 的系数为 1，使过程直接明了。
4. The Quadratic Formula | 二次公式
a direct way to find the roots of any quadratic equation ax² + bx + c = 0. It is given by:
x = [ –b ± √b2 – 4ac ] / (2a)
二次公式由配方法推导而来，为求解任何二次方程 ax² + bx + c = 0 提供了一条直接途径。公式如下：
x = [ –b ± √b2 – 4ac ] / (2a)
When applying the formula, always write down the values of a, b and c from the standard form, taking care to include any negative signs. Substitute carefully and simplify in stages.
应用公式时，务必从标准形式中写下 a、b 和 c 的值，注意包含负号。仔细代入并分步化简。
For example, to solve 3x² – 4x – 5 = 0, identify a = 3, b = –4, c = –5. Then the discriminant b² – 4ac = (–4)² – 4 × 3 × (–5) = 16 + 60 = 76. Hence x = [4 ± √76] / 6, which simplifies to x = [4 ± 2√19] / 6 = (2 ± √19) / 3.
例如，解 3x² – 4x – 5 = 0，确定 a = 3，b = –4，c = –5。判别式 b² – 4ac = (–4)² – 4 × 3 × (–5) = 16 + 60 = 76。因此 x = [4 ± √76] / 6，化简为 x = [4 ± 2√19] / 6 = (2 ± √19) / 3。
The formula is especially helpful when dealing with large or awkward coefficients, or when an exact surd answer is required in the IGCSE exam.
在处理较大或棘手的系数，或者 IGCSE 考试要求保留根号形式的精确答案时，二次公式尤其有用。
5. The Discriminant and Nature of Roots | 判别式与根的性质
The expression under the square root in the quadratic formula, Δ = b² – 4ac, is called the discriminant. It reveals the nature of the roots without actually solving the equation.
二次公式中根号下的表达式 Δ = b² – 4ac 被称为判别式。它能在不求根的情况下揭示根的性质。
| Discriminant (Δ) | Nature of roots | 判别式的值 | 根的性质 |
|---|---|---|---|
| Δ > 0 | Two distinct real roots | Δ > 0 | 两个相异实根 |
| Δ = 0 | One repeated real root (or two equal roots) | Δ = 0 | 一个重复实根（两个相等根） |
| Δ < 0 | No real roots (the roots are complex) | Δ < 0 | 无实根（根是复数） |
n x has two distinct real roots. For instance, if x² + kx + 9 = 0 has equal roots, then b² – 4ac = k² – 36 = 0, so k = ±6.
IGCSE 考题经常检验这一概念，通常会要求你判断方程有多少个实根，或找出使关于 x 的二次式有两个相异实根时 k 的取值范围。例如，若 x² + kx + 9 = 0 有等根，则 b² – 4ac = k² – 36 = 0，因此 k = ±6。
6. Quadratic Graphs and Their Properties | 二次函数图像及其性质
e turning point (vertex) can be found by completing the square or by using the formula x = –b/(2a).
y = ax² + bx + c 的图像是一条抛物线。若 a > 0，抛物线开口向上且有一个最低点；若 a < 0，开口向下且有一个最高点。其转折点（顶点）可通过配方法或公式 x = –b/(2a) 求得。
positive, the parabola cuts the x-axis at two points; if zero, it touches the x-axis; if negative, it does not meet the x-axis at all.
y 截距在 x = 0 处，得到点 (0, c)。x 截距（根）对应于 ax² + bx + c = 0 的解。若判别式为正，抛物线与 x 轴交于两点；若为零，则相切；若为负，则完全不与 x 轴相交。
graph given its equation, marking the intercepts, the vertex and the axis of symmetry x = –b/(2a).
考题经常要求你根据方程画出二次图像草图，并标出截距、顶点和对称轴 x = –b/(2a)。
7. Solving Quadratic Inequalities | 解二次不等式
Quadratic inequalities such as x² – 5x + 6 > 0 or 2x² + 3x – itical values (the roots of the corresponding equation) and then testing intervals on a sketch of the parabola.
诸如 x² – 5x + 6 > 0 或 2x² + 3x – 2 ≤ 0 的二次不等式，可首先求出临界值（相应方程的根），然后在抛物线草图上检验区间来解决。
Step 1: Solve the related equation. For x² – 5x + 6 = 0, factorising gives (x – 2)(x – 3) = 0, so roots are x = 2 and x = 3. Step 2: Sketch the graph of y = x² – the graph to find where the curve is above the x-axis (for > 0) or below (for < 0). For x² – 5x + 6 > 0, the solution is x < 2 or x > 3.
步骤1：解相关方程。对于 x² – 5x + 6 = 0，因式分解得 (x – 2)(x – 3) = 0，因此根为 x = 2 和 x = 3。步骤2：画出 y = x² – 5x + 6 的草图。由于 a > 0，这是一条开口向上的 U 形抛物线，在 x = 2 和 x = 3 处与 x 轴相交。步骤3：利用图像确定曲线在 x 轴上方（> 0）或下方（< 0）的区域。对于 x² – 5x + 6 > 0，解为 x < 2 或 x > 3。
Remember to use solid or open dots when representing the solution on a number line, depending on whether the inequality includes equality.
记住，在数轴上表示解时，根据不等式是否包含等号来决定使用实心点还是空心点。
8. Problem Solving with Quadratic Equations | 二次方程应用解题
Many IGCSE word problems lead to quadratic equations. Common contexts include finding dimensions of a rectangle given area and perimeter, consecutive integer problems, and problems involving areas of composite shapes.
许多IGCSE文字题最终都会化归为二次方程。常见情境包括在已知面积和周长的情况下求矩形边长、连续整数问题，以及涉及组合形状面积的问题。
e width. Let width = w, then length = w + 3. The area is w(w + 3) = 40, leading to w² + 3w – 40 = 0. Factorising gives (w + 8)(w – 5) = 0. Since width must be positive, w = 5 cm.
示例：一个矩形的长比宽多 3 cm。若面积为 40 cm²，求宽。设宽 = w，则长 = w + 3。面积满足 w(w + 3) = 40，得出 w² + 3w – 40 = 0。因式分解得 (w + 8)(w – 5) = 0。因为宽必须为正，故 w = 5 cm。
Always check that your solutions make sense in the context of the problem. Negative or zero lengths often need to be rejected.
务必检查所得解在题目情境下是否合理。负值或零长度通常需要舍弃。
9. Algebraic Manipulation of Quadratic Expressions | 二次式的代数变形
Beyond solving equations, you must be comfortable rearranging quadratic expressions, simplifying rational expressions that involve quadratics, and completing the square to rewrite a quadratic in vertex form.
除了解方程，你还需熟练掌握对二次式进行变形、化简含有二次式的有理表达式，以及通过配方法将二次式改写为顶点形式。
from the first two terms: 2(x² + 4x) + 5. Complete the square inside the bracket: (x + 2)² – 4, then 2((x + 2)² – 4) + 5 = 2(x + 2)² – 8 + 5 = 2(x + 2)² – 3. This reveals a minimum value of –3 when x = –2.
例如，将 2x² + 8x + 5 表示为 a(x + p)² + q 的形式。从首两项提取 x² 的系数：2(x² + 4x) + 5。在括号内配成完全平方：(x + 2)² – 4，然后 2((x + 2)² – 4) + 5 = 2(x + 2)² – 8 + 5 = 2(x + 2)² – 3。这揭示了当 x = –2 时，最小值为 –3。
Being able to convert between standard form and vertex form is extremely useful in optimisation problems and graph sketching.
能够在标准形式和顶点形式之间转换，对最优化问题和图像绘制极为有用。
10. Common Mistakes and How to Avoid Them | 常见错误及避免方法
rising or using the formula. Always rearrange to the standard form ax² + bx + c = 0 first.
许多学生因在因式分解或使用公式前忘记将方程设为零而失分。务必先将其整理成标准
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