IGCSE Edexcel Mathematics #083: Solving Quadratic Equations

📚 IGCSE Edexcel Mathematics #083: Solving Quadratic Equations | 解二次方程

Quadratic equations appear in almost every IGCSE Edexcel Mathematics exam. Mastering this topic is essential because it connects to coordinate geometry, functions, and problem solving. This revision guide explains the key methods, the discriminant, and common pitfalls in a clear step-by-step way.

在 IGCSE Edexcel 数学考试中，二次方程几乎每份试卷都会出现。掌握这一主题至关重要，因为它联系着坐标几何、函数与实际问题。本篇复习指南将逐步清晰地解释核心方法、判别式以及常见易错点。

1. What Is a Quadratic Equation? | 什么是二次方程？

In its simplest form, a quadratic equation is a polynomial equation of degree 2. The highest power of the variable is 2. On the Edexcel IGCSE syllabus, it is usually written in the form ax² + bx + c = 0, where a ≠ 0.

二次方程最简单的形式是最高次项为 2 的多项式方程。在 Edexcel IGCSE 考纲中，通常写作 ax² + bx + c = 0，其中 a ≠ 0。

A quadratic equation may have two real roots, one repeated real root, or no real roots. These possibilities depend on the discriminant, which we will explore later.

二次方程可能有两个实数根、一个重根，或者没有实数根。具体取决于判别式，稍后我们会详细讨论。

2. Standard Form and Key Terms | 标准形式与关键术语

The standard form is: ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. Here, a is the coefficient of x², b is the coefficient of x, and c is the constant term.

标准形式为：ax² + bx + c = 0，其中 a、b、c 为实数，且 a ≠ 0。a 是 x² 项的系数，b 是 x 项的系数，c 是常数项。

For example, in 2x² – 5x + 3 = 0, a = 2, b = -5, c = 3. A root (or solution) is a value of x that makes the equation true.

例如，在 2x² – 5x + 3 = 0 中，a = 2，b = -5，c = 3。根（或解）是使等式成立的 x 值。

Before solving, always rearrange the equation into the standard form. Many marks are lost by forgetting to bring all terms to one side.

在求解之前，务必先将方程整理成标准形式。许多失分都是因为忘记把所有项移到等号一侧。

3. Solving by Factorisation | 因式分解法

Factorisation is the quickest method when it works. Look for two brackets that multiply to give the original quadratic. For a simple case where a = 1, find two numbers that multiply to c and add to b.

因式分解是可行时最快捷的方法。寻找两个括号，使其相乘得到原二次式。当 a = 1 时，找两个数，它们相乘等于 c，相加等于 b。

Example: Solve x² + 7x + 12 = 0. The two numbers are 3 and 4, because 3 × 4 = 12 and 3 + 4 = 7. So (x + 3)(x + 4) = 0. Hence x = -3 or x = -4.

例：解 x² + 7x + 12 = 0。两个数是 3 和 4，因为 3 × 4 = 12，3 + 4 = 7。所以 (x + 3)(x + 4) = 0，因此 x = -3 或 x = -4。

When a ≠ 1, you may need to factorise by grouping or use trial and error. Always check by expanding your brackets.

当 a ≠ 1 时，可能需要分组分解或试错。务必通过展开括号来验证。

If AB = 0, then either A = 0 or B = 0.

The zero-product property is the key step: once you have a product equal to zero, set each bracket to zero and solve the two linear equations.

零乘积性质是关键步骤：一旦得到乘积等于零，就让每个括号等于零，然后解两个一次方程。

4. Solving by Quadratic Formula | 二次公式法

The quadratic formula works for every quadratic equation, even when factorisation is difficult. It is given in the Edexcel IGCSE formula booklet:

二次求根公式适用于所有二次方程，即使因式分解困难也无妨。Edexcel IGCSE 公式书给出该公式：

x = (−b ± √(b² − 4ac)) / (2a)

To use the formula, identify a, b, c from the equation ax² + bx + c = 0, substitute them into the formula, and simplify. The symbol ± means there are two solutions, one with + and one with −.

使用公式时，先从方程 ax² + bx + c = 0 中确定 a、b、c，然后代入公式并化简。符号 ± 表示有两个解，一个取加号，一个取减号。

Example: Solve 3x² – 4x – 2 = 0. Here a=3, b=-4, c=-2. Substitute: x = (4 ± √(16 + 24)) / 6 = (4 ± √40) / 6. Since √40 = 2√10, the solutions are x = (2 ± √10) / 3.

例：解 3x² – 4x – 2 = 0。这里 a=3，b=-4，c=-2。代入得 x = (4 ± √(16 + 24)) / 6 = (4 ± √40) / 6。由于 √40 = 2√10，所以解为 x = (2 ± √10) / 3。

When giving answers, pay attention to the required form: exact value, decimal place, or significant figure.

给出答案时，注意题目要求的形式：精确值、保留小数位还是有效数字。

5. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + p)² + q. This method is especially useful for finding turning points and for solving equations when factorisation fails.

配方法将一个二次式改写为 a(x + p)² + q 的形式。此法特别适用于求顶点坐标，以及在因式分解失败时求解方程。

If a = 1, take half of b and square it. For x² + 6x + 2 = 0, half of 6 is 3, and 3² = 9. So (x + 3)² – 9 + 2 = 0, hence (x + 3)² – 7 = 0.

若 a=1，取 b 的一半然后平方。对于 x² + 6x + 2 = 0，6 的一半是 3，3²=9。因此 (x + 3)² – 9 + 2 = 0，即 (x + 3)² – 7 = 0。

Then solve: (x + 3)² = 7, so x + 3 = ±√7, hence x = -3 ± √7.

然后求解：(x + 3)² = 7，所以 x + 3 = ±√7，即 x = -3 ± √7。

When a ≠ 1, first factor out a from the x² and x terms, then complete the square inside the bracket.

当 a ≠ 1 时，先从 x² 和 x 项提出公因子 a，然后再在括号内配方。

The completed square form reveals the vertex of the parabola: (x + p)² + q has its minimum (or maximum) at (−p, q).

配平方形式揭示抛物线的顶点：(x + p)² + q 的极小值（或极大值）出现在 (−p, q)。

6. The Discriminant | 判别式

The discriminant is the part of the quadratic formula under the square root: b² − 4ac. It tells us how many real roots exist without solving the whole equation.

判别式是二次公式中根号内的部分：b² − 4ac。它告诉我们方程有多少个实数根，而无需完整求解。

Discriminant Δ = b² − 4ac	Nature of roots	根的性质
Δ > 0 and a perfect square	Two rational roots	两个有理根
Δ > 0 and not a perfect square	Two irrational roots	两个无理根
Δ = 0	One repeated root	一个重根
Δ < 0	No real roots	无实数根

Example: For 2x² + 3x + 5 = 0, Δ = 9 − 40 = −31 < 0, so there are no real roots.

例：对于 2x² + 3x + 5 = 0，Δ = 9 − 40 = −31 < 0，因此没有实数根。

Exam questions often ask you to “show that the equation has two distinct real roots”. Just calculate the discriminant and state its sign.

考试常要求“证明方程有两个不同的实数根”。只需计算判别式并说明其正负号即可。

7. Graphical Interpretation | 图像解释

The graph of a quadratic equation y = ax² + bx + c is a parabola. If a > 0, it opens upwards (a “U” shape); if a < 0, it opens downwards (an "n" shape).

二次方程 y = ax² + bx + c 的图像是一条抛物线。当 a > 0 时开口向上（“U”形）；当 a < 0 时开口向下（“n”形）。

The x-intercepts of the parabola are exactly the roots of the equation ax² + bx + c = 0. The y-intercept is simply c.

抛物线与 x 轴的交点正是方程 ax² + bx + c = 0 的根。与 y 轴的交点就是 c。

The vertex (turning point) can be found by completing the square or by using x = −b/(2a). This is a frequent exam topic.

顶点（转向点）可以通过配方或使用 x = −b/(2a) 求得。这是考试常考内容。

If the discriminant Δ < 0, the parabola does not cross the x-axis at all. The graph lies entirely above or below the x-axis depending on the sign of a.

如果判别式 Δ < 0，抛物线根本不与 x 轴相交。图像完全位于 x 轴上方或下方，取决于 a 的符号。

8. Sum and Product of Roots | 根的和与积

For a quadratic equation ax² + bx + c = 0, the two roots α and β satisfy a useful pair of relationships:

对于二次方程 ax² + bx + c = 0，两根 α 和 β 满足以下有用关系：

α + β = −b/a, αβ = c/a

These formulas allow you to find the sum or product of the roots without solving the equation. They are also helpful in constructing new quadratic equations with given roots.

这些公式允许你不解方程直接求出根的和或积。同样有助于构造具有给定根的新二次方程。

Example: If a quadratic has roots 2 and 5, then sum = 7 and product = 10, so the equation is x² − 7x + 10 = 0.

例：若一个二次方程的根为 2 和 5，则和为 7，积为 10，所以方程为 x² − 7x + 10 = 0。

This topic appears in several Edexcel IGCSE papers, often as a short question or in the extended paper’s non-calculator section.

这个主题在 Edexcel IGCSE 多份试卷中出现，通常作为简答题或在扩展卷的无计算器部分考查。

9. Quadratic Inequalities | 二次不等式

Sometimes you are asked to solve inequalities such as x² − 5x + 6 > 0. The first step is always to factorise: (x − 2)(x − 3) > 0.

有时需要解形如 x² − 5x + 6 > 0 的不等式。第一步总是因式分解：(x − 2)(x − 3) > 0。

Then sketch a graph. Since the coefficient of x² is positive, the parabola opens upwards and crosses the x-axis at 2 and 3. The expression is positive outside the roots, so x < 2 or x > 3.

然后画草图。由于 x² 的系数为正，抛物线开口向上，与 x 轴交于 2 和 3。表达式在两根之外为正，所以 x < 2 或 x > 3。

For an inequality with < 0, the solution lies between the roots. For example, (x − 2)(x − 3) < 0 gives 2 < x < 3.

对于 < 0 的不等式，解在两根之间。例如 (x − 2)(x − 3) < 0 给出 2 < x < 3。

Remember to reverse the inequality sign when multiplying or dividing by a negative number during rearrangement.

记住：在整理过程中乘或除以负数时要翻转不等号。

10. Worded Problems | 应用题

Quadratic equations are often embedded in word problems involving areas, profit, projectile motion, or consecutive numbers. You must translate the text into an equation.

二次方程常出现在面积、利润、抛体运动或连续整数等应用题中。你必须将文字转化为方程。

Common technique: choose a variable for the unknown, write down any known relationships, and form a quadratic equation. Then solve and reject any solution that is physically impossible (e.g. a negative length).

常用技巧：设未知变量，写出已知关系，然后建立二次方程。求解后排除不符合物理意义的解（如负长度）。

Example: The area of a rectangle is 24 cm². Its length is 2 cm more than its width. Find the width w. Equation: w(w + 2) = 24, so w² + 2w − 24 = 0.

例：一个矩形面积为 24 cm²。长比宽多 2 cm。求宽 w。方程：w(w + 2) = 24，即 w² + 2w − 24 = 0。

Factorising (w + 6)(w − 4) = 0 gives w = −6 or w = 4. Since a width cannot be negative, w = 4 cm.

因式分解 (w + 6)(w − 4) = 0，得 w = −6 或 w = 4。由于宽度不可能为负，所以 w = 4 cm。

Always define your variable and state the final answer in the context of the question.

始终设好变量，并在题目情境中写出最终答案。

11. Common Exam Pitfalls | 常见易错点

Forgetting to set the equation to 0 before factorising or using the quadratic formula.

在分解因式或使用求根公式前忘记将方程整理为等于 0 的形式。
Sign errors when substituting b into the formula. If b is negative, then −b becomes positive.

把 b 代入公式时出现符号错误。如果 b 是负数，那么 −b 就变为正数。
Miscalculating b² − 4ac when b is a fraction or a negative number. Use brackets carefully.

当 b 是分数或负数时，误算 b² − 4ac。请小心使用括号。
Stopping at the factorised form and not solving for x.

停在因式分解形式而不继续解出 x。
Writing ± incorrectly when taking square roots: x² = 9 implies x = ±3, not x = 3 only.

开平方时错误处理 ±：x² = 9 意味着 x = ±3，而不仅仅是 x = 3。
Not simplifying surds such as √48 to 4√3.

没有化简根式，如 √48 应化为 4√3。

Check each solution by substituting it back into the original equation. This catches many small slips.

将每个解代回原方程检验。这能发现许多小错误。

12. Practice Questions | 练习题

Here are three exam-style questions for you to try. The answers are shown after each question.

以下有三道考试风格练习题供你尝试。答案附于每题之后。

1. Solve 2x² − 5x − 3 = 0.

1. 解 2x² − 5x − 3 = 0。

Answer: Factorise to (2x + 1)(x − 3) = 0, so x = −1/2 or x = 3.

答案：因式分解得 (2x + 1)(x − 3) = 0，所以 x = −1/2 或 x = 3。

2. Find the discriminant of x² + 4x + 5 = 0 and state the number of real roots.

2. 求 x² + 4x + 5 = 0 的判别式，并说明实数根的个数。

Answer: Δ = 16 − 20 = −4 < 0, so no real roots.

答案：Δ = 16 − 20 = −4 < 0，所以无实数根。

3. A right-angled triangle has sides x, x + 1, and hypotenuse x + 2. Use Pythagoras to form a quadratic and find x.

3. 一个直角三角形三边分别为 x、x + 1 和斜边 x + 2。用勾股定理建立二次方程并求 x。

Answer: x² + (x + 1)² = (x + 2)² → x² − 2x − 3 = 0 → (x − 3)(x + 1) = 0 → x = 3 (since length is positive).

答案：x² + (x + 1)² = (x + 2)² → x² − 2x − 3 = 0 → (x − 3)(x + 1) = 0 → x = 3（因为边长取正）。

Practise these until you can work through them confidently without a calculator.

反复练习，直到你能在不使用计算器的情况下自信解答。

In summary, solving quadratic equations requires a solid understanding of factorisation, the quadratic formula, completing the square, and the discriminant. Each method has its own strengths, and the best choice depends on the question. Keep the standard form in mind, check your answers, and you will be well prepared for the IGCSE Edexcel Mathematics exam.

总之，解二次方程需要扎实掌握因式分解、求根公式、配方和判别式。每种方法各有优势，最佳选择取决于题目。牢记标准形式，检验答案，你将为 IGCSE Edexcel 数学考试做好充分准备。

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IGCSE Edexcel Mathematics #083: Solving Quadratic Equations | 解二次方程