Mastering Quadratic Equations | 掌握一元二次方程

📚 Mastering Quadratic Equations | 掌握一元二次方程

Quadratic equations form the backbone of IGCSE Mathematics Algebra. Every year, examiners test students on factorisation, the quadratic formula, completing the square, and graph sketching. Mastering this single topic can secure 10–15% of your total marks, so it is essential to build a rock-solid understanding.

一元二次方程是 IGCSE 数学代数部分的核心内容。每年考官都会考查因式分解、求根公式、配方法以及图像绘制。熟练掌握这一专题可为你的总分锁定 10%–15% 的分数，因此建立牢固的理解至关重要。

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

A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is exactly 2. Its general form is written as ax² + bx + c = 0, where a, b and c are real constants and a ≠ 0. If a were zero, the equation would become linear, not quadratic.

一元二次方程是次数为 2 的多项式方程，即变量的最高次数恰好为 2。其一般形式写作 ax² + bx + c = 0，其中 a、b、c 为实数常数，且 a ≠ 0。若 a 为零，方程就变成了一次方程，而不是二次方程。

For example, 3x² − 5x + 2 = 0 is quadratic, while 2x + 1 = 0 is linear and x³ − 1 = 0 is cubic. In the IGCSE syllabus, you will encounter quadratic equations that can be solved by four main methods: factorisation, the quadratic formula, completing the square, and (for graphical problems) reading roots off a curve.

例如，3x² − 5x + 2 = 0 是二次方程，而 2x + 1 = 0 是一次方程，x³ − 1 = 0 是三次方程。在 IGCSE 考纲中，你会遇到可用四种主要方法求解的二次方程：因式分解、求根公式、配方法，以及（针对图像问题）从曲线上读取根。

2. Standard Form and Key Features | 标准形式与关键特征

The standard form ax² + bx + c = 0 gives us three pieces of information. The coefficient a controls the width and direction of the parabola; b affects the position of the axis of symmetry; c is the y-intercept of the graph. Understanding these roles helps you predict the shape of a quadratic graph before you plot it.

标准形式 ax² + bx + c = 0 提供了三条关键信息。系数 a 控制抛物线的宽度与开口方向；b 影响对称轴的位置；c 是图像与 y 轴的交点。理解这些作用有助于你在绘图之前预测二次函数图像的大致形状。

a > 0: the parabola opens upwards (U shape); a < 0: it opens downwards (n shape).

a > 0：抛物线开口向上（U 形）；a < 0：抛物线开口向下（n 形）。
c is the y-intercept, so the curve always passes through (0, c).

c 是 y 轴截距，所以曲线始终经过点 (0, c)。
The equation is usually written with all terms on one side, so it equals zero.

方程通常把所有项移到一边，使其等于零。

3. Solving by Factorisation | 因式分解法求解

Factorisation is the fastest method when the quadratic has simple integer roots. The principle is to rewrite ax² + bx + c as a product of two brackets, then use the zero product property: if p × q = 0, then p = 0 or q = 0.

当二次方程具有简单的整数根时，因式分解是最快的方法。其原理是将 ax² + bx + c 改写为两个括号相乘的形式，然后利用零乘积性质：若 p × q = 0，则 p = 0 或 q = 0。

Worked example: solve x² − 7x + 12 = 0. We look for two numbers that multiply to 12 and add to −7. Those numbers are −3 and −4. Therefore (x − 3)(x − 4) = 0, giving x = 3 or x = 4.

例题：解 x² − 7x + 12 = 0。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 (x − 3)(x − 4) = 0，得到 x = 3 或 x = 4。

(x − 3)(x − 4) = 0 → x = 3 或 x = 4

When the coefficient of x² is not 1, you must consider factor pairs of both a and c. For example, 2x² + 7x + 3 = 0 factorises as (2x + 1)(x + 3) = 0.

当 x² 的系数不为 1 时，你必须同时考虑 a 和 c 的因数对。例如，2x² + 7x + 3 = 0 可分解为 (2x + 1)(x + 3) = 0。

4. Solving by the Quadratic Formula | 公式法求解

The quadratic formula solves any quadratic equation, including those that cannot be factorised easily. The formula is derived from completing the square and is given on the IGCSE formula sheet, but you must know how to substitute correctly.

求根公式可以求解任何一元二次方程，包括那些不容易因式分解的方程。该公式由配方法推导而来，在 IGCSE 公式表中给出，但你必须掌握正确的代入方法。

x = (−b ± √(b² − 4ac)) / (2a)

Worked example: solve 2x² + 3x − 5 = 0. Here a = 2, b = 3, c = −5. Substitute into the formula:

例题：解 2x² + 3x − 5 = 0。这里 a = 2，b = 3，c = −5。代入公式：

x = (−3 ± √(3² − 4 × 2 × (−5))) / (2 × 2) = (−3 ± √(9 + 40)) / 4 = (−3 ± √49) / 4

Since √49 = 7, we get x = (−3 + 7) / 4 = 1, or x = (−3 − 7) / 4 = −2.5. Always show your substitution step clearly to earn method marks.

因为 √49 = 7，我们得到 x = (−3 + 7) / 4 = 1，或 x = (−3 − 7) / 4 = −2.5。一定要清楚写出代入步骤，以获得方法分。

5. Solving by Completing the Square | 配方法求解

Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. This is especially useful for finding turning points and solving equations where b is even. The key identity is x² + bx = (x + b/2)² − (b/2)².

配方法将 ax² + bx + c 改写为 a(x + p)² + q 的形式。这在求顶点坐标和解 b 为偶数的方程时尤其有用。关键恒等式是 x² + bx = (x + b/2)² − (b/2)²。

Worked example: solve x² − 6x + 5 = 0 by completing the square. First halve −6 to get −3, so (x − 3)². Then adjust the constant:

例题：用配方法解 x² − 6x + 5 = 0。先将 −6 取半得到 −3，于是有 (x − 3)²。然后调整常数项：

x² − 6x + 5 = (x − 3)² − 9 + 5 = (x − 3)² − 4

Set this equal to zero: (x − 3)² − 4 = 0, so (x − 3)² = 4. Taking square roots gives x − 3 = ±2, hence x = 5 or x = 1.

令其等于零：(x − 3)² − 4 = 0，所以 (x − 3)² = 4。开平方得 x − 3 = ±2，因此 x = 5 或 x = 1。

6. The Discriminant | 判别式

The expression b² − 4ac is called the discriminant, often written as Δ. It tells us how many real roots a quadratic equation has without actually solving it. This is a very common exam question that only requires substitution and a conclusion.

表达式 b² − 4ac 称为判别式，通常记作 Δ。它可以在不解方程的情况下告诉我们一元二次方程有多少个实数根。这是非常常见的考题，只需代入并写出结论。

If Δ > 0, the equation has two distinct real roots.

若 Δ > 0，方程有两个不相等的实数根。
If Δ = 0, the equation has exactly one repeated real root.

若 Δ = 0，方程有两个相等的实数根（一个重根）。
If Δ < 0, the equation has no real roots (two complex roots).

若 Δ < 0，方程没有实数根（有两个复数根）。

For example, for x² − 4x + 4 = 0, Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0, so there is one repeated root. This also means the graph touches the x-axis at exactly one point.

例如，对于 x² − 4x + 4 = 0，Δ = (−4)² − 4 × 1 × 4 = 16 − 16 = 0，因此有一个重根。这也意味着图像与 x 轴恰好相切于一点。

7. Quadratic Graphs: Parabola Basics | 二次函数图像：抛物线基础

The graph of y = ax² + bx + c is always a parabola. The sign of a determines the direction, and the magnitude of a determines how wide or narrow the curve is. A large |a| produces a steep, narrow parabola, while a small |a| produces a wide, flat one.

y = ax² + bx + c 的图像始终是抛物线。a 的正负决定开口方向，a 的绝对值大小决定曲线的宽窄。|a| 越大，抛物线越陡峭狭窄；|a| 越小，抛物线越宽扁。

Every parabola has a vertical axis of symmetry given by x = −b / (2a). The turning point (vertex) lies on this axis. The y-intercept is always at (0, c), and the x-intercepts, if they exist, are the roots of ax² + bx + c = 0.

每条抛物线都有一条竖直对称轴，其方程为 x = −b / (2a)。顶点（转向点）位于这条对称轴上。与 y 轴的交点始终为 (0, c)，而与 x 轴的交点（如果存在）就是方程 ax² + bx + c = 0 的根。

8. Finding Roots and Turning Points | 求根与顶点

Roots are the x-values where the curve crosses the x-axis. You can find them using any of the solving methods from Sections 3–5. The turning point, however, is best found by completing the square: if y = a(x − h)² + k, then the vertex is exactly at (h, k).

根是曲线与 x 轴交点的 x 值。你可以使用第 3–5 节中的任何求解方法来求根。然而，顶点最好通过配方法求得：若 y = a(x − h)² + k，则顶点恰好位于 (h, k)。

Example: for y = x² − 6x + 5, completing the square gives y = (x − 3)² − 4. Hence the vertex is at (3, −4) and the axis of symmetry is x = 3.

例如：对于 y = x² − 6x + 5，配方得到 y = (x − 3)² − 4。因此顶点位于 (3, −4)，对称轴为 x = 3。

y = (x − 3)² − 4 → 顶点 (3, −4)，对称轴 x = 3

Notice that the x-coordinate of the vertex is the midpoint of the two roots. In the example above, the roots are 1 and 5, whose midpoint is 3 — exactly the vertex’s x-coordinate.

注意，顶点的 x 坐标是两个根的中点。在上例中，根为 1 和 5，其中点为 3——恰好是顶点的 x 坐标。

9. Sketching Quartics | 绘制二次函数草图

Sketching a quadratic requires four key features: the direction of the parabola, the y-intercept, the x-intercepts (roots), and the vertex. Follow this checklist every time to avoid losing easy marks.

绘制二次函数草图需要四个关键特征：抛物线开口方向、y 轴截距、x 轴截距（根）以及顶点。每次绘图都按这个清单操作，避免丢失容易得到的分数。

Step 1: Determine a > 0 (smile) or a < 0 (frown).

第一步：判断 a > 0（开口向上）还是 a < 0（开口向下）。
Step 2: Mark the y-intercept at (0, c).

第二步：标出 y 轴截距 (0, c)。
Step 3: Solve ax² + bx + c = 0 and mark the roots on the x-axis.

第三步：解 ax² + bx + c = 0，并将根标在 x 轴上。
Step 4: Find the vertex and plot it. If roots are hard to find, complete the square.

第四步：求出顶点并描点。如果根较难求，可用配方法。
Step 5: Draw a smooth U-shaped or n-shaped curve through all points.

第五步：用平滑的 U 形或 n 形曲线穿过所有标记点。

10. Word Problems with Quadratics | 二次方程应用题

Examiners love placing quadratic equations in real-world contexts, especially area problems and projectile motion. The key is to translate the words into an algebraic equation, solve it, and clearly state which answer is sensible.

考官喜欢将二次方程置于实际情境中，尤其是面积问题和抛体运动。关键在于将文字转化为代数方程，求解，并明确说明哪个答案合理。

Example: a rectangle has length 4 cm longer than its width, and its area is 96 cm². Let the width be x. Then length = x + 4, so x(x + 4) = 96, giving x² + 4x − 96 = 0.

例题：一个长方形的长比宽长 4 cm，面积为 96 cm²。设宽为 x，则长 = x + 4，于是 x(x + 4) = 96，即 x² + 4x − 96 = 0。

x(x + 4) = 96 → x² + 4x − 96 = 0 → (x + 12)(x − 8) = 0

The solutions are x = −12 and x = 8. Since width cannot be negative, we reject −12 and conclude the width is 8 cm and the length is 12 cm. Always reject negative answers in measurement contexts.

解为 x = −12 和 x = 8。由于宽度不能为负数，我们舍去 −12，得出宽为 8 cm，长为 12 cm。在涉及测量的情境中，务必舍去负数解。

11. Common Exam Pitfalls | 常见考试误区

Even strong students lose marks on quadratics due to a few recurring mistakes. Make a personal checklist from the list below before every test.

即使是优秀的学生也会因为一些反复出现的错误而在二次方程题目上丢分。每次考试前，请根据下面的列表做一个个人检查清单。

Forgetting to rearrange the equation to zero before factorising or using the formula.

在因式分解或使用公式前，忘记将方程整理为等于零的形式。
Misidentifying a, b and c, especially when b or c is negative. For x² − 5x = 0, note b = −5 and c = 0.

错误识别 a、b、c，尤其是 b 或 c 为负数时。对于 x² − 5x = 0，注意 b = −5，c = 0。
Losing the ± sign when taking square roots: if x² = 9, then x = ±3.

开平方时漏掉 ± 号：若 x² = 9，则 x = ±3。
Stopping after factorisation and forgetting the second root: (x − 3)(x − 4) = 0 gives two roots.

因式分解后忘了第二个根：(x − 3)(x − 4) = 0 给出两个根。
Sketching a straight line instead of a smooth curve when drawing the parabola.

画抛物线时画成直线而不是平滑曲线。

12. Practice Questions | 练习题

Apply everything you have learned to the following exam-style questions. Attempt each one fully before checking the answers below.

将你所学的内容应用到以下仿考题中。请先完整作答，再核对下方的答案。

Question 1: Solve x² − 2x − 15 = 0 by factorisation.

第 1 题：用因式分解法解 x² − 2x − 15 = 0。

Question 2: Solve 3x² + 5x − 2 = 0 using the quadratic formula. Leave your answers as exact fractions.

第 2 题：用求根公式解 3x² + 5x − 2 = 0，答案以精确分数表示。

Question 3: Find the discriminant of 2x² − 4x + 1 = 0 and state the number of real roots.

第 3 题：求 2x² − 4x + 1 = 0 的判别式，并说明实数根的个数。

Question 4: Complete the square for y = x² + 6x + 2, then state the coordinates of the vertex.

第 4 题：用配方法改写 y = x² + 6x + 2，并写出顶点坐标。

Answers: 1) x = 5 or x = −3. 2) x = 1/3 or x = −2. 3) Δ = 8, two distinct real roots. 4) y = (x + 3)² − 7, vertex at (−3, −7).

答案：1) x = 5 或 x = −3。2) x = 1/3 或 x = −2。3) Δ = 8，两个不相等的实数根。4) y = (x + 3)² − 7，顶点为 (−3, −7)。

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Mastering Quadratic Equations | 掌握一元二次方程