Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most important topics in IGCSE Mathematics. They appear in many different contexts, from pure algebra to real-world problem solving. In this revision guide, we will explore all the methods you need to solve them confidently.

二次方程是 IGCSE 数学中最重要的主题之一。它们出现在许多不同的情境中，从纯代数到现实世界的问题解决。在本复习指南中，我们将探索所有你需要自信解决二次方程的方法。

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

A quadratic equation is a polynomial equation of degree 2. This means the highest power of the variable (usually \(x\)) is 2. The general form is:

二次方程是次数为 2 的多项式方程。这意味着变量（通常是 \(x\)）的最高次幂是 2。其一般形式为：

ax² + bx + c = 0, (a ≠ 0)

Here, \(a\), \(b\), and \(c\) are constants, with \(a\) being the coefficient of \(x²\), \(b\) the coefficient of \(x\), and \(c\) the constant term. Because \(a \neq 0\), the equation is truly quadratic; if \(a = 0\), it becomes linear.

这里，\(a\)、\(b\) 和 \(c\) 是常数，其中 \(a\) 是 \(x²\) 的系数，\(b\) 是 \(x\) 的系数，\(c\) 是常数项。因为 \(a \neq 0\)，方程才真正是二次的；如果 \(a = 0\)，它就变成一次方程。

For example, \(3x² – 5x + 2 = 0\) is a quadratic equation, while \(2x + 7 = 0\) is not.

例如，\(3x² – 5x + 2 = 0\) 是二次方程，而 \(2x + 7 = 0\) 不是。

2. Standard Form | 标准形式

To solve a quadratic equation, it is often convenient to rewrite it in standard form: \(ax² + bx + c = 0\). This means all terms are on one side, and the right-hand side is zero.

为了解二次方程，通常方便将其重写为标准形式：\(ax² + bx + c = 0\)。这意味着所有项都在一边，右边为零。

For example, the equation \(2x² = 3x – 4\) can be rearranged as:

例如，方程 \(2x² = 3x – 4\) 可以重新排列为：

2x² – 3x + 4 = 0

Always check that the equation is in this form before applying any solving technique. If there are fractions or brackets, expand and simplify first.

在应用任何求解技巧之前，务必检查方程是否为此形式。如果有分数或括号，先展开并化简。

Expand brackets: Remove parentheses by multiplying out.
Combine like terms: Add or subtract terms with the same power of \(x\).
Move all terms to one side: Ensure the right side equals zero.

展开括号：通过乘法去除括号。
合并同类项：将具有相同 \(x\) 次幂的项相加或相减。
将所有项移到一边：确保右边等于零。

3. Factorisation Method | 因式分解法

If the quadratic expression can be factorised into two linear factors, we can solve it using the zero product property. The method works when \(ax² + bx + c\) can be written as \((px + q)(rx + s)\).

如果二次表达式可以因式分解为两个一次因式，我们可以利用零乘积性质来求解。当 \(ax² + bx + c\) 可以写成 \((px + q)(rx + s)\) 时，这个方法有效。

For the simple case where \(a = 1\), look for two numbers that multiply to give \(c\) and add to give \(b\). For example, solve \(x² – 5x + 6 = 0\):

对于 \(a = 1\) 的简单情况，找两个数，它们相乘等于 \(c\)，相加等于 \(b\)。例如，解 \(x² – 5x + 6 = 0\)：

x² – 5x + 6 = (x – 2)(x – 3) = 0

Then set each factor to zero:

然后将每个因式设为零：

x – 2 = 0 or x – 3 = 0

So \(x = 2\) or \(x = 3\). Always verify by substituting back into the original equation.

所以 \(x = 2\) 或 \(x = 3\)。务必通过代回原方程进行验证。

When \(a \neq 1\), use the method of grouping or trial and error. For example, solve \(2x² + 7x + 3 = 0\):

当 \(a \neq 1\) 时，使用分组法或试错法。例如，解 \(2x² + 7x + 3 = 0\)：

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

Then \(2x + 1 = 0\) gives \(x = -½\), and \(x + 3 = 0\) gives \(x = -3\).

然后 \(2x + 1 = 0\) 得 \(x = -½\)，\(x + 3 = 0\) 得 \(x = -3\)。

4. Quadratic Formula | 求根公式

When factorisation is difficult or impossible, the quadratic formula solves any quadratic equation:

当因式分解困难或不可能时，求根公式可以解任何二次方程：

x = (−b ± √(b² − 4ac)) / (2a)

This formula is derived from completing the square and works for all \(a\), \(b\), \(c\) as long as \(a \neq 0\). Let’s use it on \(2x² – 4x – 3 = 0\):

该公式由配方法推导而来，适用于所有 \(a\)、\(b\)、\(c\)，只要 \(a \neq 0\)。我们用它在 \(2x² – 4x – 3 = 0\) 上：

Here \(a = 2\), \(b = -4\), \(c = -3\). Substitute into the formula:

这里 \(a = 2\)，\(b = -4\)，\(c = -3\)。代入公式：

x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4 = (4 ± 2√10) / 4

Thus the two solutions are \(x = (2 + √10)/2\) and \(x = (2 – √10)/2\). This method always gives exact solutions, even when they are irrational.

因此两个解是 \(x = (2 + √10)/2\) 和 \(x = (2 – √10)/2\)。这种方法总是给出精确的解，即使它们是无理数。

5. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form \(a(x + h)² + k\). This technique is useful for finding turning points and solving equations.

配方法将二次式改写为 \(a(x + h)² + k\) 的形式。这种技巧有助于找到转折点和解方程。

For a quadratic \(x² + bx + c\), add and subtract \((b/2)²\). Example: solve \(x² + 6x + 4 = 0\) by completing the square.

对于二次式 \(x² + bx + c\)，加上并减去 \((b/2)²\)。示例：通过配方法解 \(x² + 6x + 4 = 0\)。

First halve the coefficient of \(x\): \(6/2 = 3\). Square it: \(3² = 9\). Rewrite the equation as:

首先将 \(x\) 的系数减半：\(6/2 = 3\)。平方：\(3² = 9\)。将方程改写为：

x² + 6x + 9 − 9 + 4 = 0

This gives \((x + 3)² − 5 = 0\). Hence \((x + 3)² = 5\). Taking square roots:

这得到 \((x + 3)² − 5 = 0\)。因此 \((x + 3)² = 5\)。开平方：

x + 3 = ±√5, so x = −3 ± √5

This method is especially valuable when the quadratic cannot be factorised easily.

当二次式不易因式分解时，这种方法尤为有价值。

6. The Discriminant | 判别式

The expression \(b² − 4ac\) inside the quadratic formula is called the discriminant, often denoted by \(Δ\). It tells us the nature of the roots without solving the equation.

求根公式中的表达式 \(b² − 4ac\) 称为判别式，通常用 \(Δ\) 表示。它不需要解方程就能告诉我们根的性质。

Discriminant	Nature of roots
Δ > 0	Two distinct real roots
Δ = 0	One repeated real root (equal roots)
Δ < 0	No real roots (complex roots)

判别式

判别式	根的性质
Δ > 0	两个不相等的实数根
Δ = 0	一个重复的实数根（相等根）
Δ < 0	没有实数根（复数根）

For example, consider \(x² – 4x + 4 = 0\). Here \(Δ = 16 – 16 = 0\), so there is exactly one root: \(x = 2\).

例如，考虑 \(x² – 4x + 4 = 0\)。这里 \(Δ = 16 – 16 = 0\)，所以恰好有一个根：\(x = 2\)。

If a question asks you to “show that the equation has no real roots”, simply calculate \(Δ\) and show it is negative.

如果题目要求你“证明方程没有实数根”，只需计算 \(Δ\) 并证明它为负。

7. Graphs of Quadratic Functions | 二次函数图像

The graph of \(y = ax² + bx + c\) is a parabola. The sign of \(a\) determines its orientation:

函数 \(y = ax² + bx + c\) 的图像是抛物线。\(a\) 的符号决定其开口方向：

If \(a > 0\), the parabola opens upwards and has a minimum point.
If \(a < 0\), it opens downwards and has a maximum point.

如果 \(a > 0\)，抛物线开口向上，且有最小值点。
如果 \(a < 0\)，抛物线开口向下，且有最大值点。

The roots of the equation \(ax² + bx + c = 0\) correspond to the x-intercepts of the graph. The y-intercept is simply \(c\). The vertex is the point where the curve changes direction; its x-coordinate is given by:

方程 \(ax² + bx + c = 0\) 的根对应图像与 x 轴的交点。y 截距就是 \(c\)。顶点是曲线改变方向的点；其 x 坐标为：

x = −b/(2a)

Then the y-coordinate can be found by substituting this x-value back into the original equation.

然后将这个 x 值代回原方程即可求出 y 坐标。

For example, for \(y = x² – 2x – 3\), the vertex x-coordinate is \(2/2 = 1\). Substituting \(x = 1\) gives \(y = 1 – 2 – 3 = -4\). So the vertex is \((1, -4)\).

例如，对于 \(y = x² – 2x – 3\)，顶点的 x 坐标为 \(2/2 = 1\)。代入 \(x = 1\) 得 \(y = 1 – 2 – 3 = -4\)。因此顶点是 \((1, -4)\)。

8. Sum and Product of Roots | 根与系数的关系

For a quadratic equation \(ax² + bx + c = 0\) with roots \(\alpha\) and \(\beta\), the following relationships hold:

对于具有根 \(\alpha\) 和 \(\beta\) 的二次方程 \(ax² + bx + c = 0\)，以下关系成立：

Sum of roots: α + β = −b/a

Product of roots: αβ = c/a

These are extremely useful for checking answers or constructing equations with given roots.

这些关系对于检查答案或构造具有给定根的方程非常有用。

For example, if the roots of \(2x² + 8x + 6 = 0\) are \(\alpha\) and \(\beta\), then \(\alpha + \beta = -4\) and \(\alpha \beta = 3\). Indeed, factorising \(2(x+1)(x+3)=0\) gives roots \(-1\) and \(-3\), and their sum is \(-4\), product is \(3\).

例如，如果 \(2x² + 8x + 6 = 0\) 的根是 \(\alpha\) 和 \(\beta\)，则 \(\alpha + \beta = -4\)，\(\alpha \beta = 3\)。事实上，因式分解 \(2(x+1)(x+3)=0\) 得到根 \(-1\) 和 \(-3\)，其和为 \(-4\)，积为 \(3\)。

To form a new quadratic equation whose roots are \(\alpha\) and \(\beta\), use \(x² – (\alpha + \beta)x + \alpha \beta = 0\).

若要构造一个根为 \(\alpha\) 和 \(\beta\) 的新二次方程，使用 \(x² – (\alpha + \beta)x + \alpha \beta = 0\)。

9. Word Problems | 应用题

Quadratic equations often model real-life situations involving areas, projectile motion, or numbers. The key steps are:

二次方程常用于模拟涉及面积、抛体运动或数字的现实情境。关键步骤如下：

Define the unknown variable.
Translate the problem into an equation.
Solve the equation using any suitable method.
Interpret the solutions in the context of the problem.

定义未知变量。
将问题转化为方程。
使用任何合适的方法解方程。
在问题语境中解释解。

Example: A rectangular garden has length 4 metres more than its width. If its area is 45 m², find its dimensions.

示例：一个矩形花园的长度比宽度多 4 米。如果其面积为 45 平方米，求其尺寸。

Let the width be \(x\) metres. Then the length is \(x + 4\). The area is \(x(x + 4) = 45\), so:

设宽度为 \(x\) 米。则长度为 \(x + 4\)。面积为 \(x(x + 4) = 45\)，所以：

x² + 4x − 45 = 0

Factorise: \((x + 9)(x − 5) = 0\). Hence \(x = -9\) or \(x = 5\). Since width cannot be negative, \(x = 5\). The length is \(9\) m.

因式分解：\((x + 9)(x − 5) = 0\)。因此 \(x = -9\) 或 \(x = 5\)。由于宽度不能为负，\(x = 5\)。长度为 9 米。

Always discard solutions that do not make sense in the context.

始终舍弃在上下文中不合理的解。

10. Practice Questions | 练习

To master quadratic equations, practice is essential. Try these questions by yourself:

要掌握二次方程，练习必不可少。请自行尝试以下问题：

Solve \(x² – 7x + 10 = 0\).
Solve \(3x² + 5x – 2 = 0\) using the quadratic formula.
Find the roots of \(x² + 4x + 7 = 0\). What does the discriminant tell you?
Write \(x² + 6x – 1\) in completed square form.
The sum of the squares of two consecutive integers is 365. Find the integers.

解 \(x² – 7x + 10 = 0\)。
使用求根公式解 \(3x² + 5x – 2 = 0\)。
求 \(x² + 4x + 7 = 0\) 的根。判别式告诉你什么？
将 \(x² + 6x – 1\) 写成配方法形式。
两个连续整数的平方和为 365。求这两个整数。

Check your answers using the sum and product of roots. If you get stuck, go back to the relevant section.

利用根与系数的关系检查答案。如果卡住了，返回相关章节。

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Solving Quadratic Equations | 解二次方程