Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most frequently tested topics in IGCSE Mathematics. This revision guide covers everything you need: standard form, factorisation, the quadratic formula, completing the square, the discriminant, graphs, and common pitfalls — all with clear worked examples.

二次方程是 IGCSE 数学中考频最高的知识点之一。本复习指南涵盖你所需的一切：标准形式、因式分解法、求根公式、配方法、判别式、图像以及常见易错点，并配有清晰的例题讲解。

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

A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is 2. Its general form is:

二次方程是次数为 2 的多项式方程，即变量的最高次幂为 2。它的一般形式为：

ax² + bx + c = 0, 其中 a ≠ 0

Here, a, b and c are constants, and x is the unknown variable. If a = 0, the equation becomes linear, not quadratic.

其中 a、b、c 为常数，x 为未知数。如果 a = 0，方程退化为一次方程，不再是二次方程。

Examples: x² − 5x + 6 = 0; 2x² + 3x − 1 = 0; x² = 9.
Non-examples: x + 2 = 0 (linear); x³ − 1 = 0 (cubic).

示例：x² − 5x + 6 = 0；2x² + 3x − 1 = 0；x² = 9。
反例：x + 2 = 0（一次方程）；x³ − 1 = 0（三次方程）。

2. Standard Form and Rearranging | 标准形式与整理方程

Before solving, always rearrange the equation into the standard form ax² + bx + c = 0. All terms must be on one side, with zero on the other.

在求解之前，始终将方程整理成标准形式 ax² + bx + c = 0。所有项移到一边，另一边为零。

Worked Example: Solve x² = 3x + 4.

例题：解方程 x² = 3x + 4。

Step 1: Bring all terms to the left: x² − 3x − 4 = 0.

第一步：将所有项移到左边：x² − 3x − 4 = 0。

Step 2: Now the equation is in standard form with a = 1, b = −3, c = −4.

第二步：现在方程成为标准形式，其中 a = 1，b = −3，c = −4。

Always check that the signs are correct after rearranging. A common error is forgetting to change the sign when moving terms across the equals sign.

整理后务必检查符号是否正确。一个常见错误是在移项时忘记变号。

3. Solving by Factorisation | 因式分解法

Factorisation is the fastest method when the quadratic has simple integer factors. The principle is: if the product of two expressions is zero, then at least one of them must be zero.

当二次式具有简单的整数因式时，因式分解是最快捷的方法。其原理是：若两个表达式的乘积为零，则其中至少有一个必为零。

若 p × q = 0，则 p = 0 或 q = 0

Worked Example: Solve x² − 5x + 6 = 0.

例题：解方程 x² − 5x + 6 = 0。

Step 1: Find two numbers whose product is +6 and whose sum is −5. Those numbers are −2 and −3.

第一步：找两个数，其乘积为 +6，其和为 −5。这两个数是 −2 和 −3。

Step 2: Write in factored form: (x − 2)(x − 3) = 0.

第二步：写成因式形式：(x − 2)(x − 3) = 0。

Step 3: Set each factor to zero: x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.

第三步：令每个因式为零：x − 2 = 0 或 x − 3 = 0，解得 x = 2 或 x = 3。

Type	Factored Form	Example
Difference of squares	(x − p)(x + p)	x² − 9 = (x − 3)(x + 3)
Perfect square	(x ± p)²	x² + 6x + 9 = (x + 3)²
General trinomial	(mx + p)(nx + q)	2x² + 5x + 2 = (2x + 1)(x + 2)

If the coefficient of x² is not 1, use trial and error or the “cross method” to find the factors.

如果 x² 的系数不为 1，可用试错法或十字相乘法来寻找因式。

4. The Quadratic Formula | 求根公式法

The quadratic formula works for every quadratic equation, even when factorisation is difficult or impossible. For ax² + bx + c = 0:

求根公式适用于所有二次方程，即使无法因式分解也能求解。对于 ax² + bx + c = 0：

x = (−b ± √(b² − 4ac)) / 2a

Worked Example: Solve 2x² + 3x − 5 = 0.

例题：解方程 2x² + 3x − 5 = 0。

Step 1: Identify a = 2, b = 3, c = −5.

第一步：确定 a = 2，b = 3，c = −5。

Step 2: Substitute into the formula:

第二步：代入公式：

x = (−3 ± √(3² − 4 × 2 × (−5))) / (2 × 2)

Step 3: Simplify: x = (−3 ± √(9 + 40)) / 4 = (−3 ± √49) / 4 = (−3 ± 7) / 4.

第三步：化简：x = (−3 ± √(9 + 40)) / 4 = (−3 ± √49) / 4 = (−3 ± 7) / 4。

Step 4: So x = 4/4 = 1, or x = −10/4 = −2.5.

第四步：因此 x = 4/4 = 1，或 x = −10/4 = −2.5。

Always write the formula down first, then substitute carefully. A common mistake is mis-stating the sign of c when substituting.

务必先写出公式，再仔细代入。常见错误是代入时弄错 c 的符号。

5. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. This is especially useful for finding turning points and solving equations without using the formula.

配方法将二次式改写为 (x + p)² + q 的形式。它在求顶点坐标和不解方程的情况下尤为有用。

Worked Example: Solve x² + 6x + 2 = 0 by completing the square.

例题：用配方法解方程 x² + 6x + 2 = 0。

Step 1: Halve the coefficient of x: 6 ÷ 2 = 3. Write (x + 3)².

第一步：将 x 的系数除以 2：6 ÷ 2 = 3。写出 (x + 3)²。

Step 2: Expand (x + 3)² = x² + 6x + 9. This is 7 more than the original quadratic, so subtract 7:

第二步：展开 (x + 3)² = x² + 6x + 9。这比原式多了 7，因此减去 7：

(x + 3)² − 7 = 0

Step 3: Solve: (x + 3)² = 7, so x + 3 = ±√7, giving x = −3 ± √7.

第三步：求解：(x + 3)² = 7，所以 x + 3 = ±√7，得 x = −3 ± √7。

For quadratics where a ≠ 1, first divide the whole equation by a before completing the square.

当 a ≠ 1 时，先将整个方程除以 a，再进行配方。

6. The Discriminant | 判别式

The discriminant is the expression b² − 4ac inside the square root of the quadratic formula. It tells us the nature of the roots without solving the equation.

判别式是求根公式根号内的表达式 b² − 4ac。它无需解方程即可判断根的性质。

Value of b² − 4ac	Nature of Roots	Graph
b² − 4ac > 0	Two distinct real roots	Crosses the x-axis twice
b² − 4ac = 0	One repeated real root	Touches the x-axis once
b² − 4ac < 0	No real roots	Does not intersect the x-axis

b² − 4ac 的值	根的性质	图像
b² − 4ac > 0	两个不等实根	与 x 轴交于两点
b² − 4ac = 0	一个重根	与 x 轴相切于一点
b² − 4ac < 0	无实根	与 x 轴无交点

Worked Example: Determine the nature of the roots of x² + 4x + 5 = 0.

例题：判断方程 x² + 4x + 5 = 0 的根的性质。

Here a = 1, b = 4, c = 5. The discriminant is 4² − 4 × 1 × 5 = 16 − 20 = −4. Since −4 < 0, the equation has no real roots.

这里 a = 1，b = 4，c = 5。判别式为 4² − 4 × 1 × 5 = 16 − 20 = −4。因为 −4 < 0，所以方程无实根。

7. Sum and Product of Roots | 根的和与积

For a quadratic equation ax² + bx + c = 0 with roots α and β, the following relationships hold:

对于根为 α 和 β 的二次方程 ax² + bx + c = 0，有以下关系：

α + β = −b/a

α × β = c/a

Worked Example: Find the sum and product of the roots of 2x² + 8x + 3 = 0.

例题：求方程 2x² + 8x + 3 = 0 的根的和与积。

Sum = −8/2 = −4. Product = 3/2.

和 = −8/2 = −4。积 = 3/2。

This relationship is useful when checking answers or when constructing a quadratic equation from given roots.

这个关系可用于检验答案，也可用于根据已知根构造二次方程。

8. Quadratic Graphs | 二次函数图像

The graph of y = ax² + bx + c is a parabola. Its shape and position depend on the signs of a and the discriminant.

y = ax² + bx + c 的图像是一条抛物线。其形状和位置取决于 a 的符号和判别式。

If a > 0: the parabola opens upwards (U shape) and has a minimum point.
If a < 0: the parabola opens downwards (∩ shape) and has a maximum point.

若 a > 0：抛物线开口向上（U 形），存在最小值点。
若 a < 0：抛物线开口向下（∩ 形），存在最大值点。

The turning point can be found by completing the square. For y = (x + p)² + q, the turning point is (−p, q).

顶点坐标可通过配方法求得。对于 y = (x + p)² + q，顶点坐标为 (−p, q)。

Worked Example: Find the turning point of y = x² − 6x + 5.

例题：求 y = x² − 6x + 5 的顶点坐标。

Complete the square: y = (x − 3)² − 9 + 5 = (x − 3)² − 4. The turning point is (3, −4). Since a = 1 > 0, this is a minimum point.

配方：y = (x − 3)² − 9 + 5 = (x − 3)² − 4。顶点坐标为 (3, −4)。因为 a = 1 > 0，这是最小值点。

9. Word Problems | 应用题

Quadratic equations often appear in geometry and number problems. The key steps are: define the variable, form the equation, solve it, and check the validity of answers.

二次方程常出现在几何和数字问题中。关键步骤为：设未知数、列方程、解方程、检验答案的合理性。

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

例题：一个长方形的长比宽长 4 cm，面积为 96 cm²。求其尺寸。

Step 1: Let the width be x cm. Then the length is (x + 4) cm.

第一步：设宽为 x cm，则长为 (x + 4) cm。

Step 2: Area = x(x + 4) = 96, so x² + 4x − 96 = 0.

第二步：面积 = x(x + 4) = 96，即 x² + 4x − 96 = 0。

Step 3: Factorise: (x + 12)(x − 8) = 0, giving x = −12 or x = 8.

第三步：因式分解：(x + 12)(x − 8) = 0，得 x = −12 或 x = 8。

Step 4: Reject x = −12 because length cannot be negative. So width = 8 cm and length = 12 cm.

第四步：舍去 x = −12，因为长度不能为负。所以宽 = 8 cm，长 = 12 cm。

Always check whether each solution makes sense in the context of the problem.

务必检查每个解是否符合实际问题的情境。

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

Here are the most frequent errors students make in the exam, and how to avoid them.

以下是学生在考试中最常见的错误以及如何避免它们。

Mistake 1: Forgetting to write the equation in standard form before factorising. Fix: Always rearrange to ax² + bx + c = 0 first.
Mistake 2: Sign errors when substituting into the quadratic formula. Fix: Use brackets around negative numbers.
Mistake 3: Dividing by a variable that may be zero. Fix: Never divide both sides by x unless x ≠ 0 is guaranteed.
Mistake 4: Losing one solution. Fix: Every quadratic has up to two solutions; always write both.

错误一：因式分解前忘记整理成标准形式。对策：先移项整理为 ax² + bx + c = 0。
错误二：代入求根公式时符号出错。对策：负数代入时加括号。
错误三：除以可能为零的变量。对策：除非确定 x ≠ 0，否则不要两边同时除以 x。
错误四：漏掉一个解。对策：每个二次方程最多有两个解，务必都写出来。

In the exam, if factorisation does not work within a few seconds, switch to the quadratic formula. For multiple-choice questions, check answers by substituting back into the original equation.

考试中，如果因式分解几秒内无头绪，立即改用求根公式。对于选择题，可将答案代回原方程进行验证。

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Solving Quadratic Equations | 解二次方程