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. In this revision guide, we will explore their standard forms, solving methods, the discriminant, and real-world applications—step by step with clear worked examples.

二次方程是 IGCSE 数学中最核心的考点之一。本复习指南将系统地讲解它的标准形式、求解方法、判别式以及实际应用，并通过分步例题帮助你完全掌握。

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

A quadratic equation in one variable is any equation that can be written in the form ax² + bx + c = 0, where a, b, and c are constants, a ≠ 0. The highest power of the variable is always 2.

一元二次方程是能写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 为常数，且 a ≠ 0。变量的最高次数始终为 2。

The coefficient a is called the leading coefficient, b is the linear coefficient, and c is the constant term.
If a = 0, the equation becomes linear, not quadratic.
Some equations require rearrangement into this standard form before solving.

a 称为二次项系数，b 称为一次项系数，c 称为常数项。
若 a = 0，方程退化为一次方程，不再是二次方程。
有些方程需要先移项、整理成标准形式后才能求解。

ax² + bx + c = 0 (a ≠ 0)

2. Solving by Factorisation | 因式分解法

If the quadratic expression can be factorised into two linear brackets, then we can use the zero product property: if p × q = 0, then p = 0 or q = 0.

如果二次式能因式分解为两个一次括号相乘，我们就可以利用零积性质：若 p × q = 0，则 p = 0 或 q = 0。

Example: Solve x² – 5x + 6 = 0.

例题：解方程 x² – 5x + 6 = 0。

We look for two numbers that multiply to 6 and add to –5. These are –2 and –3. So x² – 5x + 6 = (x – 2)(x – 3) = 0. Hence x = 2 or x = 3.

我们寻找两个数，乘积为 6，和为 –5。这两个数是 –2 和 –3。因此 x² – 5x + 6 = (x – 2)(x – 3) = 0。所以 x = 2 或 x = 3。

Always factorise out any common factor first. If the coefficient of x² is not 1, use the method of splitting the middle term.

切记先提取公因式。如果 x² 的系数不为 1，可使用拆中项法。

3. Solving by Completing the Square | 配方法

Completing the square rewrites the quadratic in the form a(x + p)² + q. This is especially useful when factorisation is difficult or when we want to find the vertex of a parabola.

配方法将二次式改写为 a(x + p)² + q 的形式。当因式分解困难或需要求抛物线顶点时，这种方法尤为实用。

Example: Solve x² + 6x – 7 = 0 by completing the square.

例题：用配方法解 x² + 6x – 7 = 0。

Take half of 6, which is 3, and square it to get 9. Then x² + 6x = (x + 3)² – 9. The equation becomes (x + 3)² – 9 – 7 = 0, i.e. (x + 3)² = 16. Taking square roots gives x + 3 = ±4, so x = 1 or x = –7.

取 6 的一半得 3，平方得 9。于是 x² + 6x = (x + 3)² – 9。原方程变为 (x + 3)² – 9 – 7 = 0，即 (x + 3)² = 16。两边开平方得 x + 3 = ±4，所以 x = 1 或 x = –7。

x² + bx = (x + b/2)² – (b/2)²

4. The Quadratic Formula | 求根公式

The quadratic formula solves any quadratic equation directly. For ax² + bx + c = 0, the solutions are given by:

求根公式可以直接解任何二次方程。对于 ax² + bx + c = 0，解为：

x = (−b ± √(b² − 4ac)) / 2a

The expression under the square root, b² – 4ac, is called the discriminant. It determines the nature of the roots.

根号下的表达式 b² – 4ac 称为判别式。它决定根的性质。

Example: Solve 2x² – 3x – 2 = 0 using the formula.

例题：用公式法解 2x² – 3x – 2 = 0。

Here a = 2, b = –3, c = –2. The discriminant is (–3)² – 4 × 2 × (–2) = 9 + 16 = 25. Therefore x = (3 ± √25) / 4 = (3 ± 5) / 4, giving x = 2 or x = –½.

这里 a = 2，b = –3，c = –2。判别式为 (–3)² – 4 × 2 × (–2) = 9 + 16 = 25。因此 x = (3 ± √25) / 4 = (3 ± 5) / 4，得 x = 2 或 x = –½。

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

For real coefficients, the discriminant Δ = b² – 4ac has three cases:

对于实系数，判别式 Δ = b² – 4ac 有三种情况：

Δ = b² – 4ac	Nature of roots \| 根的性质
Δ > 0	Two distinct real roots \| 两个不相等的实数根
Δ = 0	Two equal real roots (one repeated root) \| 两个相等的实数根（一个重根）
Δ < 0	No real roots (two complex roots) \| 没有实数根（两个虚根）

If the question only asks for the number of real roots, you do not need to solve the full equation—just evaluate the discriminant.

如果题目只问实数根的个数，无需完整解方程，只需计算判别式即可。

6. Solving Quadratic Equations with Fractions or Surds | 含分数或根式的二次方程

Sometimes a quadratic equation appears inside a fraction, a rational expression, or after squaring a surd. The key is to rewrite it as a standard quadratic equation by multiplying through by a common denominator or by isolating terms.

有时二次方程隐藏在分式、有理式或根式平方之后。关键是通过乘以公分母或移项，将其整理为标准二次方程。

Example: Solve 3/x + x = 5.

例题：解方程 3/x + x = 5。

Multiply both sides by x (assuming x ≠ 0): 3 + x² = 5x, so x² – 5x + 3 = 0. Then use the formula to get x = (5 ± √13)/2.

两边同乘 x（假设 x ≠ 0）：3 + x² = 5x，即 x² – 5x + 3 = 0。然后用公式得 x = (5 ± √13)/2。

When the solution involves a surd, leave it in simplified surd form unless the question asks for a decimal.

当解含根式时，除非题目要求小数，否则保留最简根式形式。

7. Quadratic Equations from Word Problems | 二次方程应用题

Many IGCSE problems require translating a real-world situation into a quadratic equation. Common models include area, product of consecutive integers, projectile motion, and number relationships.

许多 IGCSE 题目需要将实际问题转化为二次方程。常见模型包括面积、连续整数乘积、抛体运动和数字关系。

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

例题：一个长方形的长比宽长 3 cm，面积为 40 cm²，求宽。

Let the width be x. Then the length is x + 3. Area: x(x + 3) = 40, so x² + 3x – 40 = 0. Factorise: (x + 8)(x – 5) = 0. Since width cannot be negative, x = 5. The width is 5 cm.

设宽为 x，则长为 x + 3。面积：x(x + 3) = 40，即 x² + 3x – 40 = 0。因式分解：(x + 8)(x – 5) = 0。因宽不能为负，所以 x = 5。宽为 5 cm。

Always check whether both solutions are valid in the context of the problem; discard any solution that violates physical or geometric constraints.

务必检查两个解在题目背景下是否都成立，舍弃违背实际或几何约束的解。

8. The Graph of a Quadratic Function | 二次函数图象

A quadratic function y = ax² + bx + c has a parabola-shaped graph. The sign of a tells us whether it opens upwards (a > 0) or downwards (a < 0).

二次函数 y = ax² + bx + c 的图象是抛物线。a 的符号决定开口方向：a > 0 开口向上，a < 0 开口向下。

The y-intercept is at (0, c).
The roots of the equation ax² + bx + c = 0 are the x-intercepts of the graph.
The vertex (turning point) lies on the axis of symmetry x = −b/(2a).

y 轴截距为 (0, c)。
方程 ax² + bx + c = 0 的根就是图象与 x 轴的交点。
顶点（转向点）位于对称轴 x = −b/(2a) 上。

If the quadratic has one root, the graph touches the x-axis at the vertex. If it has no real roots, the graph never crosses the x-axis.

若方程有一个重根，图象顶点恰好与 x 轴相切。若无实数根，图象与 x 轴无交点。

9. Solving Quadratic Inequalities | 二次不等式

Once you can solve ax² + bx + c = 0, you can solve quadratic inequalities like ax² + bx + c > 0 or ≤ 0 by sketching the graph.

一旦能解 ax² + bx + c = 0，就可以通过画草图来解类似 ax² + bx + c > 0 或 ≤ 0 的二次不等式。

Example: Solve x² – x – 6 > 0.

例题：解不等式 x² – x – 6 > 0。

Factorise: (x – 3)(x + 2) > 0. The roots are x = –2 and x = 3. Since the parabola opens upward, the expression is positive when x < –2 or x > 3. So the solution is x < –2 or x > 3.

因式分解：(x – 3)(x + 2) > 0。根为 x = –2 和 x = 3。因抛物线开口向上，表达式的值为正时满足 x < –2 或 x > 3。所以解为 x < –2 或 x > 3。

Remember: for a > 0, the “positive” region lies outside the roots; for a < 0, it lies between the roots.

记住：当 a > 0 时，正值区域位于两根之外；当 a < 0 时，正值区域位于两根之间。

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

Students often lose marks in this topic because of small sign errors, forgetting to set the equation to zero, or discarding valid roots incorrectly.

学生在二次方程题目中常因符号错误、忘记将方程化为零，或错误舍去有效根而失分。

Before solving, always rearrange to the form ax² + bx + c = 0.
If a common factor exists, divide by it first.
When using the quadratic formula, write down a, b, c explicitly first.
Check your answers by substituting back into the original equation.
In word problems, include units and state which solution is valid.

求解前，务必整理成 ax² + bx + c = 0 的形式。
若有公因式，先提取。
使用求根公式前，先明确写出 a、b、c 的值。
将答案代回原方程检验。
应用题要写单位，并说明哪个解有效。

Practising past paper questions from your exam board is the best way to avoid these pitfalls.

练习你所考考试局的最新真题是避免这些陷阱的最佳方法。

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Solving Quadratic Equations | 解二次方程