Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations appear in nearly every IGCSE Mathematics paper, whether as a direct solve question, a graph-sketching task, or a word problem. Mastering the three main solving methods — factorisation, the quadratic formula, and completing the square — is essential if you are aiming for a Grade 7, 8 or 9.

一元二次方程几乎出现在每一份 IGCSE 数学试卷中，无论是直接求解、画图题，还是应用题。掌握三种主要解法——因式分解法、求根公式法和配方法——是冲击 A*（7、8、9 分）的关键。

1. What Is a Quadratic Equation | 什么是一元二次方程

A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the unknown x is 2, which is why it is called ‘quadratic’ — from the Latin ‘quadratus’, meaning square.

一元二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。未知数 x 的最高次数是 2，因此被称为“二次”(quadratic)——源自拉丁语 “quadratus”，意为“平方”。

For example, 2x² + 5x − 3 = 0 is a quadratic equation, while 3x + 4 = 0 is linear because its highest power is 1.

例如，2x² + 5x − 3 = 0 是一元二次方程，而 3x + 4 = 0 是一次方程，因为它的最高次数是 1。

Key idea: a quadratic equation can have at most two real solutions, and these solutions are also called the ‘roots’ of the equation.

关键概念：一元二次方程最多有两个实数解，这些解也称为方程的“根”。

2. Method 1: Factorisation | 方法一：因式分解法

Factorisation is the fastest method, but it only works when the quadratic can be written as a product of two linear brackets. For a simple quadratic x² + bx + c = 0, look for two numbers that multiply to give c and add to give b.

因式分解法是最快的方法，但它只适用于能将二次式写成两个一次括号乘积的情况。对于简单的二次方程 x² + bx + c = 0，找到两个数，它们的乘积等于 c，和等于 b。

For example, solve x² − 5x + 6 = 0. We need two numbers whose product is 6 and whose sum is −5. The numbers are −2 and −3, so the equation becomes (x − 2)(x − 3) = 0.

例如，解方程 x² − 5x + 6 = 0。我们需要找到两个数，其乘积为 6，和为 −5。这两个数是 −2 和 −3，因此方程可写成 (x − 2)(x − 3) = 0。

(x − 2)(x − 3) = 0, so x = 2 or x = 3

(x − 2)(x − 3) = 0，所以 x = 2 或 x = 3

Remember the zero product property: if A × B = 0, then either A = 0 or B = 0. This is the entire logic behind solving by factorisation.

记住零乘积性质：如果 A × B = 0，那么 A = 0 或 B = 0。这正是因式分解法求解的全部逻辑。

For a coefficient of x² greater than 1, such as 2x² + 7x + 3 = 0, you may need to use the ‘ame’thod of grouping’: multiply a and c (2 × 3 = 6), find two numbers that multiply to 6 and add to 7 (1 and 6), then split the middle term and factor by grouping.

当 x² 的系数大于 1 时，例如 2x² + 7x + 3 = 0，可以使用“分组法”：将 a 与 c 相乘（2 × 3 = 6），找到两个数使其乘积为 6、和为 7（即 1 和 6），然后拆开中间项并分组因式分解。

2x² + 7x + 3 = 2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1)

So x = −3 or x = −½. Always check: does (x + 3)(2x + 1) expand back to 2x² + 7x + 3? Yes.

所以 x = −3 或 x = −½。务必检查：(x + 3)(2x + 1) 展开后是否还原为 2x² + 7x + 3？是的。

3. Method 2: The Quadratic Formula | 方法二：求根公式法

When factorisation is difficult or impossible, the quadratic formula works for every quadratic equation. Given ax² + bx + c = 0, the roots are given by the following formula.

当因式分解困难或无法进行时，求根公式适用于所有一元二次方程。对于 ax² + bx + c = 0，其根由以下公式给出。

x = (−b ± √(b² − 4ac)) / 2a

Let us solve 3x² + 5x − 2 = 0 using the formula. Here a = 3, b = 5, c = −2.

我们用公式法解 3x² + 5x − 2 = 0。这里 a = 3，b = 5，c = −2。

First compute the discriminant b² − 4ac = 5² − 4 × 3 × (−2) = 25 + 24 = 49. Since 49 is a perfect square, we could actually factorise this, but let us continue with the formula.

首先计算判别式 b² − 4ac = 5² − 4 × 3 × (−2) = 25 + 24 = 49。由于 49 是完全平方数，此题其实可以因式分解，但我们继续用公式法。

x = (−5 ± √49) / (2 × 3) = (−5 ± 7) / 6

So x = (−5 + 7)/6 = 2/6 = ⅓, or x = (−5 − 7)/6 = −12/6 = −2. The two roots are x = ⅓ and x = −2.

因此 x = (−5 + 7)/6 = 2/6 = ⅓，或 x = (−5 − 7)/6 = −12/6 = −2。两个根为 x = ⅓ 和 x = −2。

Always write down the values of a, b and c before substituting. This prevents sign errors, which are the most common mistake with the quadratic formula.

代入前务必先写出 a、b、c 的值。这可以避免符号错误——这是使用求根公式时最常见的失误。

4. The Discriminant | 判别式

The expression b² − 4ac under the square root is called the discriminant, often denoted by Δ. It tells you the nature of the roots without solving the full equation.

根号内的表达式 b² − 4ac 称为判别式，常用 Δ 表示。无需完整求解方程，它就能告诉我们根的性质。

If Δ > 0, the equation has two distinct real roots. If Δ = 0, the equation has exactly one real root (a repeated root). If Δ < 0, the equation has no real roots.

若 Δ > 0，方程有两个不等的实数根；若 Δ = 0，方程恰好有一个实数根（重根）；若 Δ < 0，方程没有实数根。

判别式 Δ	根的个数	图像与 x 轴
Δ > 0	两个不同实根	相交于两点
Δ = 0	一个重根	相切于一点
Δ < 0	无实根	不与 x 轴相交

For example, the equation x² + 2x + 5 = 0 has Δ = 4 − 20 = −16 < 0, so it has no real solutions. This matches the fact that the graph y = x² + 2x + 5 stays entirely above the x-axis.

例如，方程 x² + 2x + 5 = 0 的判别式为 Δ = 4 − 20 = −16 < 0，因此没有实数解。这与抛物线 y = x² + 2x + 5 完全位于 x 轴上方的事实相符。

5. Method 3: Completing the Square | 方法三：配方法

Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. This method is particularly useful for finding turning points and solving equations where factorisation fails.

配方法将 ax² + bx + c 改写为 a(x + p)² + q 的形式。此方法在求抛物线顶点以及解决因式分解无法处理的方程时尤为有用。

For x² + 6x + 2 = 0, take half of 6 to get 3, then square it to get 9. Write x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7.

对于 x² + 6x + 2 = 0，取 6 的一半得 3，再平方得 9。写成 x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7。

(x + 3)² − 7 = 0, so (x + 3)² = 7, and x + 3 = ±√7

(x + 3)² − 7 = 0，所以 (x + 3)² = 7，因此 x + 3 = ±√7

Hence x = −3 + √7 or x = −3 − √7. Notice that these are a pair of conjugate surd roots, which often occur when Δ is not a perfect square.

因此 x = −3 + √7 或 x = −3 − √7。注意这两个根是共轭根式对，当 Δ 不是完全平方数时经常出现这种情况。

When the coefficient of x² is not 1, first factor it out from the first two terms. For example, 2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3.

当 x² 的系数不为 1 时，先从前面两项中提取该系数。例如，2x² + 8x + 5 = 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3。

6. Which Method Should You Choose | 如何选择解法

In the exam, time is limited, so choosing the right method matters. The table below summarises a good decision strategy.

考试时间有限，因此选择合适的解法至关重要。下表总结了一个有效的决策策略。

情况	推荐方法
a = 1 且能快速找到因数	因式分解
a > 1 且 Δ 为完全平方数	因式分解（分组法）
Δ 不是完全平方数	求根公式
题目要求顶点或最值	配方法

A quick tip: if the question says ‘solve’, any valid method earns full marks. But if the question says ‘by factorisation’ or ‘using completing the square’, you must use the specified method or you will lose method marks.

小提示：如果题目只说“求解”，任何正确方法都能得满分。但如果题目明确说“用因式分解法”或“用配方法”，你必须使用指定方法，否则会扣方法分。

7. Solving Word Problems | 解应用题

Quadratic word problems follow a standard four-step process: define the variable, form the equation, solve it, and interpret the answers in context.

二次方程应用题遵循标准的四步流程：设未知数、列方程、解方程、结合题意解释答案。

Example: The length of a rectangle is 4 cm longer than its width. The area is 45 cm². Find the dimensions.

例题：一个长方形的长比宽长 4 厘米，面积为 45 平方厘米。求长方形的尺寸。

Let the width be x cm. Then the length is (x + 4) cm, and the area is x(x + 4) = 45, which expands to x² + 4x − 45 = 0.

设宽为 x 厘米，则长为 (x + 4) 厘米，面积方程为 x(x + 4) = 45，展开得 x² + 4x − 45 = 0。

x² + 4x − 45 = 0, so (x + 9)(x − 5) = 0, giving x = −9 or x = 5

x² + 4x − 45 = 0，所以 (x + 9)(x − 5) = 0，得 x = −9 或 x = 5

The width cannot be negative, so x = 5. Therefore the width is 5 cm and the length is 9 cm. Always reject negative answers if the context demands a positive quantity.

宽度不能为负数，所以 x = 5。因此宽为 5 厘米，长为 9 厘米。当题意要求正数时，一定要舍去负数解。

8. Sketching the Graph | 画函数图像

The graph of y = ax² + bx + c is a parabola. When a > 0, it opens upwards (U-shaped); when a < 0, it opens downwards (n-shaped).

函数 y = ax² + bx + c 的图像是抛物线。当 a > 0 时开口向上（U 形）；当 a < 0 时开口向下（∩ 形）。

The roots you compute are exactly the x-coordinates where the graph crosses the x-axis. The turning point (vertex) is found at x = −b/(2a).

你求得的根正是抛物线与 x 轴交点的横坐标。顶点（转折点）的横坐标为 x = −b/(2a)。

For y = x² − 4x + 3, the roots are x = 1 and x = 3 (since (x − 1)(x − 3) = 0). The vertex lies midway between the roots at x = 2, and y = 4 − 8 + 3 = −1, so the vertex is (2, −1).

对于 y = x² − 4x + 3，根为 x = 1 和 x = 3（因为 (x − 1)(x − 3) = 0）。顶点位于两根中点 x = 2 处，y = 4 − 8 + 3 = −1，所以顶点为 (2, −1)。

If a question asks for the minimum or maximum value, completing the square is often the most direct route: y = (x − 2)² − 1 immediately tells you the minimum y is −1 at x = 2.

如果题目要求最大值或最小值，配方法通常是最直接的途径：y = (x − 2)² − 1 立刻告诉我们最小值为 −1，此时 x = 2。

9. Common Mistakes and How to Avoid Them | 常见错误与避坑指南

Mistake 1: forgetting to rearrange the equation into the standard form ax² + bx + c = 0 before solving. Always move all terms to one side first.

错误一：求解前忘记将方程整理成标准形式 ax² + bx + c = 0。务必先将所有项移到等号同侧。

Mistake 2: sign errors when removing brackets. In (x + 3)(x − 2) = 0, many students write x = 2 or x = −3 but then forget the final answers. The correct answers are x = −3 and x = 2.

错误二：去括号时出现符号错误。在 (x + 3)(x − 2) = 0 中，许多学生得出 x = 2 或 x = −3，但最终答案应书写为 x = −3 和 x = 2。

Mistake 3: incorrectly simplifying fractions under the square root. Remember that √(a/b) = √a/√b only when both a and b are positive, and always simplify surds fully.

错误三：根号内分数化简错误。记住 √(a/b) = √a/√b 仅当 a、b 均为正数时成立，并且根式一定要化简到最简形式。

Mistake 4: losing solutions by dividing both sides by x when x could equal zero. If you have x² = 3x, do not divide by x; instead rearrange to x² − 3x = 0 and factorise x(x − 3) = 0.

错误四：当 x 可能等于 0 时，直接两边同除以 x 导致丢根。若遇到 x² = 3x，切勿除以 x；应移项得 x² − 3x = 0，再因式分解为 x(x − 3) = 0。

10. Exam-Style Practice Questions | 真题风格练习

Now it is time to put these skills to the test. Try these four questions before checking the solutions.

现在让我们用这四个题目检验所学技能。先独立尝试，再对照答案。

Question 1: Solve x² − 7x + 12 = 0. Answer: (x − 3)(x − 4) = 0, so x = 3 or x = 4.

题目 1：解 x² − 7x + 12 = 0。答案：(x − 3)(x − 4) = 0，所以 x = 3 或 x = 4。

Question 2: Solve 2x² + 3x − 5 = 0 by factorisation. Answer: (2x + 5)(x − 1) = 0, so x = −5/2 or x = 1.

题目 2：用因式分解法解 2x² + 3x − 5 = 0。答案：(2x + 5)(x − 1) = 0，所以 x = −5/2 或 x = 1。

Question 3: Solve x² + 4x + 1 = 0 by completing the square, giving your answers in surd form. Answer: (x + 2)² − 3 = 0, so x = −2 ± √3.

题目 3：用配方法解 x² + 4x + 1 = 0，答案以根式表示。答案：(x + 2)² − 3 = 0，所以 x = −2 ± √3。

Question 4: A ball is thrown upwards, and its height h metres after t seconds is given by h = 20t − 5t². When does the ball return to the ground? Answer: set h = 0, so 20t − 5t² = 0, giving 5t(4 − t) = 0, hence t = 0 or t = 4. It returns at t = 4 seconds.

题目 4：一个球被向上抛出，t 秒后距离地面的高度 h 米由 h = 20t − 5t² 给出。球何时落回地面？答案：令 h = 0，得 20t − 5t² = 0，即 5t(4 − t) = 0，故 t = 0 或 t = 4。球在 t = 4 秒时落回地面。

Quadratic equations reward careful, methodical work. Master factorisation, the quadratic formula, and completing the square, and you will find that every quadratic question on the IGCSE paper — from pure algebra to geometry to kinematics — becomes routine. Practice each method until it is automatic, and always check your roots by substituting them back into the original equation.

一元二次方程考验的是细心与条理。掌握因式分解、求根公式和配方法后，IGCSE 试卷上每一道二次方程题——无论是纯代数、几何还是运动学中的应用——都将变得得心应手。请将每种方法练习到自动化，并始终将根代回原方程进行检验。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Solving Quadratic Equations | 解一元二次方程