Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations appear frequently in the Edexcel IGCSE Mathematics exam. Mastering the standard form, factorisation, the quadratic formula, completing the square, and the discriminant is essential for achieving a high grade.

二次方程在 Edexcel IGCSE 数学考试中频繁出现。掌握标准形式、因式分解法、二次公式、配方法以及判别式，是获得高分的关键。

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

A quadratic equation is an equation that can be written in the form:

ax² + bx + c = 0

where a, b and c are constants, and a ≠ 0. The term ax² is the quadratic term, bx is the linear term, and c is the constant term.

二次方程是可以写成以下形式的方程：

ax² + bx + c = 0

其中 a、b、c 是常数，且 a ≠ 0。ax² 是二次项，bx 是一次项，c 是常数项。

2. Solving by Factorisation | 因式分解法

If the quadratic expression can be factorised into two linear factors, then the solution is found by setting each factor equal to zero. This is often the fastest method when the coefficient of x² is 1 or a simple number.

如果二次式可以分解为两个一次因式，那么令每个因式等于零即可得到解。当 x² 的系数为 1 或简单数字时，这通常是最快的方法。

Example: Solve x² – 5x + 6 = 0.

例子：解方程 x² – 5x + 6 = 0。

x² – 5x + 6 = (x – 2)(x – 3) = 0

So x – 2 = 0 or x – 3 = 0, giving x = 2 or x = 3.

因此 x – 2 = 0 或 x – 3 = 0，得到 x = 2 或 x = 3。

For equations like 2x² + 5x – 3 = 0, look for a pair of brackets (px + q)(rx + s) such that pr = 2, qs = -3, and ps + qr = 5.

对于形如 2x² + 5x – 3 = 0 的方程，要寻找一对括号 (px + q)(rx + s)，使得 pr = 2，qs = -3，且 ps + qr = 5。

2x² + 5x – 3 = (2x – 1)(x + 3) = 0

Hence x = 1/2 or x = -3.

因此 x = 1/2 或 x = -3。

3. The Quadratic Formula | 二次求根公式

The quadratic formula solves any quadratic equation, even when factorisation is difficult or impossible. For ax² + bx + c = 0, the solutions are:

二次公式可以求解任何二次方程，即使因式分解困难或无法分解。对于 ax² + bx + c = 0，解为：

x = (-b ± √(b² – 4ac)) / (2a)

Be careful with signs. Always substitute the values of a, b and c correctly, including negative signs.

注意正负号。务必正确代入 a、b、c 的值，包括负号。

Example: Solve 3x² + 4x – 5 = 0. Here a = 3, b = 4, c = -5.

例子：解方程 3x² + 4x – 5 = 0。这里 a = 3，b = 4，c = -5。

x = (-4 ± √(4² – 4 × 3 × (-5))) / (2 × 3) = (-4 ± √76) / 6

So x ≈ 0.79 or x ≈ -2.12. Leave the answer in exact surd form if requested: x = (-2 ± √19) / 3.

因此 x ≈ 0.79 或 x ≈ -2.12。如果题目要求精确根式形式，可写成 x = (-2 ± √19) / 3。

4. Completing the Square | 配方法

Completing the square rewrites a quadratic expression in the form:

配方法将二次式改写为以下形式：

(x + p)² + q = 0

For x² + bx + c = 0, take half of b, square it, and adjust the constant term.

对于 x² + bx + c = 0，取 b 的一半，平方，然后调整常数项。

Example: Solve x² + 6x + 4 = 0 by completing the square.

例子：用配方法解方程 x² + 6x + 4 = 0。

x² + 6x = (x + 3)² – 9

So the equation becomes (x + 3)² – 9 + 4 = 0, i.e. (x + 3)² = 5.

于是方程变为 (x + 3)² – 9 + 4 = 0，即 (x + 3)² = 5。

x + 3 = ±√5

Therefore x = -3 ± √5.

因此 x = -3 ± √5。

If the coefficient of x² is not 1, first divide the whole equation by that coefficient.

如果 x² 的系数不是 1，先将整个方程除以该系数。

5. The Discriminant and the Nature of Roots | 判别式与根的性质

The expression inside the square root in the quadratic formula is called the discriminant, usually denoted by Δ:

二次公式中根号内的表达式称为判别式，通常用 Δ 表示：

Δ = b² – 4ac

The sign of Δ tells us the number of real roots without solving the equation:

Δ 的符号告诉我们方程有多少个实数根，无需实际求解：

If Δ > 0: two distinct real roots.
If Δ = 0: one repeated real root (a double root).
If Δ < 0: no real roots (the graph never crosses the x-axis).

如果 Δ > 0：方程有两个不相等的实数根。
如果 Δ = 0：方程有一个重根（二重根）。
如果 Δ < 0：方程没有实数根（图像不与 x 轴相交）。

Example: For 2x² + 3x – 1 = 0, Δ = 3² – 4 × 2 × (-1) = 9 + 8 = 17 > 0, so there are two distinct real roots.

例子：对于 2x² + 3x – 1 = 0，Δ = 3² – 4 × 2 × (-1) = 9 + 8 = 17 > 0，因此有两个不相等的实数根。

6. Graphical Interpretation | 图像的几何意义

The equation y = ax² + bx + c represents a parabola. The real roots of ax² + bx + c = 0 are the x-coordinates of the points where the parabola crosses the x-axis.

方程 y = ax² + bx + c 表示一条抛物线。ax² + bx + c = 0 的实数根就是抛物线与 x 轴交点的横坐标。

If the parabola touches the x-axis at exactly one point, the equation has a repeated root. If it does not touch the x-axis at all, there are no real roots.

如果抛物线恰好与 x 轴相切于一点，则方程有重根。如果抛物线完全不与 x 轴相交，则没有实数根。

You may also be asked to write a quadratic in vertex form:

你可能会被要求将二次式写成顶点式：

y = a(x – h)² + k

where (h, k) is the vertex of the parabola. Completing the square is the algebraic tool used to obtain this form.

其中 (h, k) 是抛物线的顶点。配方法就是得到该形式的代数工具。

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

Read the problem carefully, define a variable, and translate the information into a quadratic equation. After solving, check whether each answer makes sense in the context of the problem (for example, lengths and times cannot be negative).

仔细阅读题目，设未知量，并将信息转化为二次方程。求出解后，检查每个答案在问题情境中是否合理（例如长度和时间不能为负）。

Example: The length of a rectangle is 3 cm longer than its width. Its area is 40 cm². Find its width.

例子：一个长方形的长比宽多 3 cm，面积为 40 cm²。求它的宽。

Let the width be x cm. Then the length is (x + 3) cm. The area equation is:

设宽为 x cm，则长为 (x + 3) cm。面积方程为：

x(x + 3) = 40

So x² + 3x – 40 = 0. Factorising gives (x + 8)(x – 5) = 0, so x = -8 or x = 5. Since width cannot be negative, x = 5. The width is 5 cm.

因此 x² + 3x – 40 = 0。因式分解得 (x + 8)(x – 5) = 0，所以 x = -8 或 x = 5。因为宽不能为负，x = 5。宽为 5 cm。

8. Common Pitfalls and Tips | 常见陷阱与技巧

Many marks are lost through avoidable errors. Keep these points in mind:

很多分数因为可以避免的错误而丢失。请记住以下要点：

Always rearrange the equation into the form ax² + bx + c = 0 before factorising or using the formula.
When using the quadratic formula, write down the values of a, b and c first.
Remember that √(b² – 4ac) is applied to the whole discriminant, not to b² and 4ac separately.
Do not forget the ± sign in front of the square root.
Check your answers by substituting them back into the original equation.
In word problems, discard any solution that is not possible in the real-life context.

在因式分解或使用公式前，务必先将方程整理成 ax² + bx + c = 0 的形式。
使用二次公式时，先写出 a、b、c 的值。
记住 √(b² – 4ac) 是对整个判别式开方，而不是分别对 b² 和 4ac 开方。
不要忘记根号前的 ± 号。
将答案代回原方程检查是否正确。
在应用题中，舍去在实际情境中不可能的解。

9. Exam-Style Practice Questions | 考试风格练习

Try these questions without looking at the solutions until you have attempted them fully.

先尝试独立完成以下题目，再核对答案。

1. Solve x² – 7x + 12 = 0.

1. 解方程 x² – 7x + 12 = 0。

2. Solve 2x² + 5x – 12 = 0 using the quadratic formula.

2. 使用二次公式解方程 2x² + 5x – 12 = 0。

3. Find the value of k such that x² + kx + 9 = 0 has exactly one real root.

3. 求 k 的值，使得方程 x² + kx + 9 = 0 恰好有一个实数根。

4. A right-angled triangle has legs of lengths x and x + 2, and a hypotenuse of length 4. Find x.

4. 一个直角三角形，两条直角边长为 x 和 x + 2，斜边长为 4。求 x。

Answers:

答案：

1. x = 3 or x = 4.

1. x = 3 或 x = 4。

2. x = (-5 ± √(25 + 96)) / 4 = (-5 ± 11) / 4, so x = 1.5 or x = -4.

2. x = (-5 ± √(25 + 96)) / 4 = (-5 ± 11) / 4，因此 x = 1.5 或 x = -4。

3. For one real root, Δ = 0: k² – 36 = 0, so k = ±6.

3. 有一个实数根时 Δ = 0：k² – 36 = 0，所以 k = ±6。

4. Using Pythagoras: x² + (x + 2)² = 16. This simplifies to 2x² + 4x – 12 = 0, or x² + 2x – 6 = 0. Using the quadratic formula gives x = -1 ± √7. Since x must be positive, x = -1 + √7.

4. 由勾股定理得 x² + (x + 2)² = 16。化简得 2x² + 4x – 12 = 0，即 x² + 2x – 6 = 0。用二次公式得 x = -1 ± √7。由于 x 必须为正，x = -1 + √7。

Quadratic equations are a cornerstone of IGCSE Mathematics. With consistent practice in factorisation, the quadratic formula, completing the square, and understanding the discriminant, you will approach any quadratic problem with confidence.

二次方程是 IGCSE 数学的基石。通过持续练习因式分解、二次公式、配方法并理解判别式，你将自信地面对任何二次方程问题。

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Solving Quadratic Equations | 解二次方程