Quadratic Equations | 二次方程

📚 Quadratic Equations | 二次方程

Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in almost every exam paper, either as direct solving questions or as part of word problems, graphs, and coordinate geometry. Mastering quadratics is essential for achieving a high grade.

二次方程是 IGCSE 数学中最重要的考点之一。几乎每份试卷都会出现二次方程，要么直接考查解法，要么出现在应用题、函数图像或坐标几何中。掌握二次方程是取得高分的关键。

1. Standard Form of a Quadratic Equation | 二次方程的标准形式

A quadratic equation is an equation in which the highest power of the unknown variable is 2. The standard form is written as:

二次方程是未知数最高次数为 2 的方程，其标准形式如下：

ax² + bx + c = 0

where a, b and c are constants, and a ≠ 0. If a = 0, the equation becomes linear, not quadratic. The values of a, b and c determine the shape and position of the parabola when the equation is graphed.

其中 a、b、c 为常数，且 a ≠ 0。若 a = 0，方程就变成一次方程而非二次方程。a、b、c 的值决定了抛物线图像的形状与位置。

For example, in the equation 2x² – 3x + 5 = 0, we have a = 2, b = -3, c = 5. Note that a quadratic equation may also appear in a rearranged form, such as x² = 4x – 3, which can be rewritten as x² – 4x + 3 = 0.

例如，在方程 2x² – 3x + 5 = 0 中，a = 2，b = -3，c = 5。注意二次方程也可能以其他形式出现，如 x² = 4x – 3，可改写为 x² – 4x + 3 = 0。

2. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic has simple integer roots. The method involves writing the quadratic as a product of two linear factors, then setting each factor equal to zero.

因式分解法是当二次方程有简单整数根时最快捷的解法。该方法将二次式写成两个一次因式的乘积，然后令每个因式等于零。

The general principle is based on the zero product property: if p × q = 0, then either p = 0 or q = 0. For a quadratic of the form x² + bx + c, we look for two numbers that multiply to give c and add to give b.

其基本原理是零乘积性质：若 p × q = 0，则 p = 0 或 q = 0。对于形如 x² + bx + c 的二次式，我们需要找两个数，它们的乘积为 c，和为 b。

Solve x² + 5x + 6 = 0

We need two numbers that multiply to 6 and add to 5. These are 2 and 3. Therefore:

我们需要找到两个数，乘积为 6 且和为 5。这两个数是 2 和 3。因此：

(x + 2)(x + 3) = 0

x + 2 = 0 or x + 3 = 0

x = -2 or x = -3

When the coefficient of x² is not 1, the factorisation is slightly harder. For ax² + bx + c, we multiply a by c and find two numbers that multiply to ac and add to b. These numbers are then used to split the middle term.

当 x² 的系数不为 1 时，因式分解稍微复杂一些。对于 ax² + bx + c，我们将 a 乘以 c，找到两个数的乘积为 ac、和为 b，然后用这两个数拆分中间项。

Solve 2x² + 7x + 3 = 0

Here, a × c = 2 × 3 = 6. Two numbers that multiply to 6 and add to 7 are 1 and 6. Split the middle term:

这里 a × c = 2 × 3 = 6。乘积为 6、和为 7 的两个数是 1 和 6。拆分中间项：

2x² + x + 6x + 3 = 0

x(2x + 1) + 3(2x + 1) = 0

(2x + 1)(x + 3) = 0

x = -½ or x = -3

Some quadratics are special cases, such as the difference of two squares: x² – 9 = (x – 3)(x + 3), or perfect squares: x² – 6x + 9 = (x – 3)². Recognising these patterns saves time in the exam.

有些二次式是特殊情况，如平方差公式：x² – 9 = (x – 3)(x + 3)，或完全平方：x² – 6x + 9 = (x – 3)²。在考试中识别这些模式可以节省时间。

3. Solving by Completing the Square | 配方法

Completing the square is a powerful technique that works for any quadratic, including those that cannot be factorised. The goal is to rewrite the quadratic in the form (x + p)² + q, from which the solutions can be found directly.

配方法是一种强大的解题技巧，适用于任何二次方程，包括无法因式分解的情况。其目标是将二次式改写为 (x + p)² + q 的形式，从而直接求出解。

For a quadratic x² + bx + c, we take half of b, square it, and adjust the constant term:

对于二次式 x² + bx + c，我们取 b 的一半并平方，然后调整常数项：

x² + bx + c = (x + b/2)² – (b/2)² + c

Solve x² + 6x + 2 = 0

Half of 6 is 3, and 3² = 9. Therefore:

6 的一半是 3，3² = 9。因此：

x² + 6x + 2 = (x + 3)² – 9 + 2

(x + 3)² – 7 = 0

(x + 3)² = 7

x + 3 = ±√7

x = -3 ± √7

When the coefficient of x² is not 1, first factor out that coefficient from the x² and x terms. For example, to complete the square for 2x² – 8x + 5, write 2(x² – 4x) + 5, then complete the square inside the bracket.

当 x² 的系数不为 1 时，先将该系数从 x² 项和 x 项中提取出来。例如，对 2x² – 8x + 5 配方，先写成 2(x² – 4x) + 5，再在括号内配方。

2(x² – 4x) + 5 = 2[(x – 2)² – 4] + 5

= 2(x – 2)² – 8 + 5 = 2(x – 2)² – 3

Completing the square also helps identify the turning point (vertex) of the parabola: for y = (x + p)² + q, the vertex is at (-p, q). Exams frequently ask for the coordinates of the minimum or maximum point.

配方还能帮助确定抛物线的顶点：对于 y = (x + p)² + q，顶点坐标为 (-p, q)。考试经常要求写出最大值或最小值点的坐标。

4. The Quadratic Formula | 求根公式

The quadratic formula is the most general method and works for every quadratic equation, whether the roots are rational, irrational, or complex (though in IGCSE we generally focus on real roots).

求根公式是最通用的解法，适用于所有二次方程，无论根是有理数、无理数还是复数（不过 IGCSE 通常只关注实数根）。

x = (-b ± √(b² – 4ac)) / 2a

This formula is derived from completing the square on the general form ax² + bx + c = 0. You are expected to memorise it, as it is not provided in the formula booklet for most IGCSE papers.

该公式通过对一般形式 ax² + bx + c = 0 配方推导而来。你需要牢记此公式，因为大多数 IGCSE 试卷的公式册中不会提供。

Solve 3x² – 5x – 2 = 0

Here a = 3, b = -5, c = -2. Substitute into the formula:

这里 a = 3，b = -5，c = -2。代入公式：

x = (5 ± √((-5)² – 4 × 3 × (-2))) / (2 × 3)

x = (5 ± √(25 + 24)) / 6

x = (5 ± √49) / 6 = (5 ± 7) / 6

x = 2 or x = -1/3

Always write down the substituted expression first before simplifying. This shows your working clearly and earns method marks even if you make a calculation error.

务必先写出代入后的完整表达式再化简。这样可以清晰展示解题过程，即使出现计算错误也能获得方法分。

5. The Discriminant | 判别式

The expression b² – 4ac inside the square root is called the discriminant. It tells us the nature of the roots without actually solving the equation.

根号内的表达式 b² – 4ac 称为判别式。它无需解方程就能告诉我们根的性质。

Value of b² – 4ac	Nature of Roots	判别式取值	根的性质
b² – 4ac > 0	Two distinct real roots	b² – 4ac > 0	两个不相等的实数根
b² – 4ac = 0	One repeated real root (equal roots)	b² – 4ac = 0	两个相等的实数根（重根）
b² – 4ac < 0	No real roots	b² – 4ac < 0	没有实数根

If b² – 4ac is a perfect square (and a, b, c are rational), the roots are rational and the equation can likely be factorised. If not, the roots are irrational or surds.

若 b² – 4ac 是一个完全平方数（且 a、b、c 为有理数），则根为有理数，方程通常可以因式分解。若不是，则根为无理数或根式。

Example: Find the range of k such that x² + 4x + k = 0 has two distinct real roots.

For two distinct real roots, the discriminant must be positive:

要使方程有两个不相等的实数根，判别式必须大于零：

b² – 4ac > 0

4² – 4 × 1 × k > 0

16 – 4k > 0

k < 4

Discriminant problems are extremely common in IGCSE exams. They often appear as “find the value of k” questions or in the context of tangents to curves.

判别式问题在 IGCSE 考试中极为常见，通常以”求 k 的值”的形式出现，或与曲线切线相关。

6. Solving Quadratics Graphically | 图像法解二次方程

The graph of a quadratic function y = ax² + bx + c is a parabola. The roots of the equation ax² + bx + c = 0 correspond to the x-coordinates where the parabola crosses the x-axis.

二次函数 y = ax² + bx + c 的图像是一条抛物线。方程 ax² + bx + c = 0 的根对应抛物线与 x 轴交点的横坐标。

If the discriminant > 0, the parabola crosses the x-axis at two points.
如果判别式 > 0，抛物线与 x 轴有两个交点。
If the discriminant = 0, the parabola touches the x-axis at exactly one point (the vertex lies on the axis).
如果判别式 = 0，抛物线与 x 轴只有一个切点（顶点在 x 轴上）。
If the discriminant < 0, the parabola does not intersect the x-axis at all.
如果判别式 < 0，抛物线与 x 轴没有交点。

For y = ax² + bx + c, the graph opens upwards when a > 0 (a U-shape with a minimum point) and opens downwards when a < 0 (an inverted U-shape with a maximum point).

对于 y = ax² + bx + c，当 a > 0 时图像开口向上（U 形，有最小值）；当 a < 0 时图像开口向下（倒 U 形，有最大值）。

The y-intercept is simply c, since setting x = 0 gives y = c. The line of symmetry is x = -b/(2a), which passes through the vertex.

抛物线的 y 截距就是 c，因为令 x = 0 时 y = c。对称轴为 x = -b/(2a)，它经过顶点。

In the exam, you may be given a graph and asked to solve a related equation, such as finding x when y = k. This involves drawing a horizontal line at y = k and reading off the x-coordinates of the intersections.

在考试中，你可能会拿到一张图，要求解相关方程，例如求 y = k 时的 x 值。这需要在 y = k 处画一条水平线，读取交点横坐标。

7. Applications: Word Problems | 应用：文字应用题

Quadratic equations frequently arise in real-world problems involving area, projectile motion, number theory, and geometry. The key skill is translating the words into an equation.

二次方程经常出现在涉及面积、抛体运动、数论和几何的实际问题中。关键技能是将文字转化为方程。

Example: A rectangle has length 4 cm longer than its width. Its area is 96 cm². Find its dimensions.

例题：一个矩形的长比宽长 4 cm，面积为 96 cm²。求其尺寸。

Let the width be x cm. Then the length is (x + 4) cm. The area is:

设宽为 x cm，则长为 (x + 4) cm。面积为：

x(x + 4) = 96

x² + 4x – 96 = 0

(x + 12)(x – 8) = 0

x = -12 or x = 8

Since a length cannot be negative, we reject x = -12. The width is 8 cm and the length is 12 cm.

由于长度不能为负数，我们舍去 x = -12。宽为 8 cm，长为 12 cm。

When solving word problems, always state what your variable represents, form the equation clearly, solve it, and then check whether each solution makes sense in the context. Negative values are often discarded, but sometimes both solutions are valid.

解应用题时，始终要说明变量代表什么，清楚地列出方程，解方程，然后检查每个解在情境中是否合理。负值通常舍去，但有时两个解都有效。

8. Simultaneous Equations with a Quadratic | 含二次方程的联立方程组

IGCSE exams often ask you to solve a system where one equation is linear and the other is quadratic, such as y = x + 1 and y = x² – 2x + 3. The method is substitution: replace y in the quadratic with the linear expression.

IGCSE 考试经常要求解一个线性方程与一个二次方程组成的方程组，如 y = x + 1 和 y = x² – 2x + 3。解法是代入法：将线性表达式代入二次方程中的 y。

Example: Solve y = 2x + 1 and y = x² + x – 3.

2x + 1 = x² + x – 3

0 = x² – x – 4

x = (1 ± √(1 + 16)) / 2 = (1 ± √17) / 2

Then substitute each x value back into the linear equation to find the corresponding y values. The solutions represent the intersection points of a straight line and a parabola.

然后将每个 x 值代回线性方程求出对应的 y 值。这些解代表直线与抛物线的交点。

Graphically, the number of solutions equals the number of intersection points: two solutions if the line cuts the parabola, one if the line is tangent, and zero if they do not meet. The discriminant of the resulting quadratic tells us which case we have.

从图像上看，解的个数等于交点个数：直线与抛物线相交时有两个解，相切时有一个解，没有交点时无解。所得二次方程的判别式可以告诉我们属于哪种情况。

9. Choosing the Right Method | 如何选择合适的方法

With three main methods available, choosing the fastest one saves valuable time in the exam. Here is a practical guide:

掌握三种主要解法后，选择最快的方法能为考试节省宝贵时间。以下是一个实用的选择指南：

Situation	Recommended Method
a = 1 and easy factors; question includes the word “factorise”	Factorisation
Question asks for turning point or coordinates of vertex	Completing the square
Equation cannot be factorised easily; irrational roots expected	Quadratic formula
Checking nature of roots without solving	Discriminant only

对于 a = 1 且容易分解的情况，或因式分解提示词：选因式分解法；当题目要求顶点坐标：选配方法；当方程无法简单分解或根为无理数：选求根公式；只需判断根的性质：只用判别式。

A useful tip: if the discriminant is a perfect square, factorisation will always work. If you spot this quickly, you can skip the formula.

实用技巧：若判别式为完全平方数，则因式分解必定可行。若能快速看出这一点，就可以跳过求根公式。

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

Many students lose marks on quadratics due to avoidable errors. Here are the most frequent pitfalls and how to avoid them:

许多学生在二次方程上失分是因为一些可以避免的错误。以下是最常见的陷阱及避免方法：

Forgetting to rearrange to standard form: Always write ax² + bx + c = 0 before solving. If the equation is given as 2x² = 5x – 3, rearrange first.
忘记整理成标准形式：解题前务必写成 ax² + bx + c = 0。如果方程是 2x² = 5x – 3，要先移项。
Sign errors when substituting into the formula: Pay careful attention to negative values of b and c. Use brackets when substituting.
代入公式时符号错误：特别注意 b 和 c 为负数的情况。代入时使用括号。
Dividing both sides by x: If you divide by x, you lose the root x = 0. Factorise instead: x² – 5x = 0 becomes x(x – 5) = 0.
两边同除以 x：除以 x 会丢失根 x = 0。应改用因式分解：x² – 5x = 0 应化为 x(x – 5) = 0。
Forgetting ± before the square root: Both positive and negative square roots must be considered.
忘记平方根前的 ± 号：正负平方根都必须考虑。
Rounding too early: Keep values in surd form or as fractions until the final step.
过早四舍五入：在最后一步之前，保持根式或分数形式。

Always check your answers by substituting them back into the original equation. This takes only seconds and catches most careless errors.

始终将答案代回原方程验算。这只需几秒钟，却能发现大多数粗心错误。

11. Practice Questions | 练习题

Try these IGCSE-style questions to test your understanding. Answers are provided below.

请尝试以下 IGCSE 风格练习题以检验你的理解。答案附在下方。

1. Solve x² – 7x + 12 = 0.

2. Solve 2x² + 5x – 3 = 0 using the quadratic formula.

3. Express x² + 8x + 15 in the form (x + p)² + q and state the minimum value of the expression.

4. Find the values of k for which the equation kx² + 3x + 2 = 0 has no real roots.

1. 解方程 x² – 7x + 12 = 0。

2. 用求根公式解 2x² + 5x – 3 = 0。

3. 将 x² + 8x + 15 表示为 (x + p)² + q 的形式，并指出该表达式的最小值。

4. 求方程 kx² + 3x + 2 = 0 无实数根时 k 的取值范围。

Solutions 解答

1. (x – 3)(x – 4) = 0, so x = 3 or x = 4. 因此 x = 3 或 x = 4。

2. x = (-5 ± √(25 + 24)) / 4 = (-5 ± 7) / 4, so x = ½ or x = -3. 因此 x = ½ 或 x = -3。

3. x² + 8x + 15 = (x + 4)² – 16 + 15 = (x + 4)² – 1. The minimum value is -1. 最小值为 -1。

4. For no real roots, the discriminant must be negative: 9 – 8k < 0, so k > 9/8.

要使方程无实数根，判别式必须小于零：9 – 8k < 0，因此 k > 9/8。

12. Summary | 总结

Quadratic equations are a core topic that connects to many areas of IGCSE Mathematics, including graphs, geometry, and functions. To master them, you must be fluent in all three solving methods: factorisation, completing the square, and the quadratic formula.

二次方程是 IGCSE 数学的核心内容，与图像、几何和函数等多个领域相关联。要掌握它，你必须熟练三种解法：因式分解、配方和求根公式。

Remember the key ideas: the discriminant reveals the nature of the roots; the completed square form gives the vertex; the graph gives a visual representation. Practise a variety of problems, especially word problems and simultaneous equations, and always check your answers.

记住关键要点：判别式揭示根的性质；配方法给出顶点坐标；图像提供直观理解。多练习不同类型的题目，尤其是应用题和联立方程组，并始终验算答案。

With consistent practice, quadratic equations will become one of the most reliable topic areas for scoring marks in your IGCSE exam.

通过持续练习，二次方程将成为你在 IGCSE 考试中最稳定的得分板块之一。

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Quadratic Equations | 二次方程