Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

Quadratic equations are one of the most important topics in the IGCSE Mathematics syllabus. Mastering them unlocks access to harder topics such as functions, graphs, and coordinate geometry. In this revision guide, we will explore every method you need to solve quadratics confidently, step by step.

一元二次方程是 IGCSE 数学课程中最重要的内容之一。掌握它，你就能顺利进阶到函数、图象、坐标几何等更难的专题。在本复习指南中，我们将逐一讲解解二次方程所需的每种方法，带你稳步提升。

1. The Standard Form | 标准形式

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

凡能写成标准形式的方程都称为一元二次方程：

ax² + bx + c = 0, where a ≠ 0

Here, a is the coefficient of x², b is the coefficient of x, and c is the constant term. The condition a ≠ 0 is essential; if a = 0, the equation becomes linear, not quadratic.

其中 a 是 x² 的系数，b 是 x 的系数，c 是常数项。条件 a ≠ 0 至关重要；若 a = 0，方程就变成一次方程，而不是二次方程。

For example, 3x² − 2x + 5 = 0 is a quadratic equation with a = 3, b = −2, c = 5. However, 2x + 1 = 0 is linear because it contains no x² term.

例如，3x² − 2x + 5 = 0 是一个二次方程，其中 a = 3，b = −2，c = 5。而 2x + 1 = 0 是一次方程，因为其中不含 x² 项。

2. Solving by Factorisation | 因式分解法

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

当二次方程具有简单的整数根时，因式分解是最快的方法。核心思路是零积性质：若两个表达式的乘积为零，则至少其中一个必须为零。

Consider the equation x² − 5x + 6 = 0. We look for two numbers that multiply to give 6 and add to give −5. These numbers are −2 and −3, so we factor as:

考虑方程 x² − 5x + 6 = 0。我们需要找到两个数，相乘得 6，相加得 −5。这两个数是 −2 和 −3，因此因式分解为：

(x − 2)(x − 3) = 0

Applying the zero product property gives x − 2 = 0 or x − 3 = 0, hence x = 2 or x = 3. Always check your answers by substituting them back into the original equation.

应用零积性质得 x − 2 = 0 或 x − 3 = 0，因此 x = 2 或 x = 3。一定要将答案代回原方程检验。

If the coefficient of x² is not 1, for example 2x² + 7x + 3 = 0, look for two factors of 2 × 3 = 6 that add to 7. Those are 1 and 6. Then rewrite the middle term: 2x² + x + 6x + 3 = 0, and factor by grouping: x(2x + 1) + 3(2x + 1) = 0, which gives (2x + 1)(x + 3) = 0, so x = −½ or x = −3.

如果 x² 的系数不为 1，例如 2x² + 7x + 3 = 0，则先找 2 × 3 = 6 的两个因数使其和为 7。它们为 1 和 6。然后重写中间项：2x² + x + 6x + 3 = 0，再分组提取公因式：x(2x + 1) + 3(2x + 1) = 0，得 (2x + 1)(x + 3) = 0，故 x = −½ 或 x = −3。

3. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. This method works for any quadratic, even when factorisation is not obvious.

配方法将二次方程改写为 (x + p)² + q 的形式。该方法适用于任何二次方程，即使因式分解不明显也能使用。

To complete the square for x² + bx + c, take half of b, square it, then add and subtract that value. For example, solve x² + 6x + 2 = 0:

对 x² + bx + c 配方时，取 b 的一半、平方，然后加上再减去该值。例如，解 x² + 6x + 2 = 0：

x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7

Set this equal to zero: (x + 3)² − 7 = 0, so (x + 3)² = 7. Taking square roots gives x + 3 = ±√7, therefore x = −3 ± √7.

令其等于零：(x + 3)² − 7 = 0，所以 (x + 3)² = 7。两边开平方得 x + 3 = ±√7，因此 x = −3 ± √7。

When the coefficient of x² is not 1, first divide the entire equation by that coefficient. For instance, 2x² − 4x − 1 = 0 becomes x² − 2x − ½ = 0, then complete the square on x² − 2x.

当 x² 的系数不为 1 时，先将整个方程除以该系数。例如，2x² − 4x − 1 = 0 化为 x² − 2x − ½ = 0，再对 x² − 2x 配方。

4. The Quadratic Formula | 二次求根公式

The quadratic formula is a universal tool. It solves any quadratic equation ax² + bx + c = 0 directly:

二次求根公式是万能工具。它可直接解任意二次方程 ax² + bx + c = 0：

x = (−b ± √(b² − 4ac)) ÷ (2a)

For example, solve 2x² − 3x − 2 = 0. Here a = 2, b = −3, c = −2. Substitute into the formula:

例如，解 2x² − 3x − 2 = 0。其中 a = 2，b = −3，c = −2。代入公式：

x = (3 ± √((−3)² − 4 × 2 × (−2))) ÷ (2 × 2) = (3 ± √25) ÷ 4

Since √25 = 5, we have x = (3 + 5) ÷ 4 = 2 or x = (3 − 5) ÷ 4 = −½. The formula is especially useful when the roots are irrational or the quadratic cannot be easily factorised.

因为 √25 = 5，所以 x = (3 + 5) ÷ 4 = 2 或 x = (3 − 5) ÷ 4 = −½。当根为无理数或二次式不易因式分解时，该公式尤其有用。

Be careful with signs: when b is negative, −b becomes positive, and b² is always positive. Use brackets when substituting to avoid sign errors.

注意符号：当 b 为负数时，−b 变为正数，且 b² 永远为正。代入时务必加上括号，以避免符号错误。

5. The Discriminant | 判别式

The expression b² − 4ac under the square root in the quadratic formula is called the discriminant, denoted by Δ. It tells us the number and type of roots without solving the equation.

二次公式根号下的表达式 b² − 4ac 称为判别式，记作 Δ。它无需解方程即可告诉我们根的个数和类型。

If Δ > 0: the equation has two distinct real roots.
如果 Δ > 0：方程有两个不同的实数根。
If Δ = 0: the equation has exactly one repeated real root.
如果 Δ = 0：方程有且仅有一个重根（两个相等的实数根）。
If Δ < 0: the equation has no real roots; the roots are complex.
如果 Δ < 0：方程没有实数根；根为复数。

For example, for 3x² − 2x + 4 = 0, Δ = (−2)² − 4 × 3 × 4 = 4 − 48 = −44 < 0, so the equation has no real solutions. This result is often stated as "the curve does not intersect the x-axis".

例如，对于 3x² − 2x + 4 = 0，Δ = (−2)² − 4 × 3 × 4 = 4 − 48 = −44 < 0，所以方程无实数解。这个结果常表述为"曲线与 x 轴不相交"。

6. Sum and Product of Roots | 根与系数的关系

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

对于二次方程 ax² + bx + c = 0，若其两根为 α 和 β，则以下关系恒成立：

α + β = −b ÷ a, αβ = c ÷ a

These relationships allow you to form a quadratic equation when given its roots. For instance, if the roots are 3 and −4, then the sum is −1 and the product is −12. The equation is x² − (sum)x + (product) = 0, which gives x² + x − 12 = 0.

这些关系使你在已知根的情况下也能构造出对应的二次方程。例如，若两根为 3 和 −4，则和为 −1，积为 −12。方程为 x² − (和)x + (积) = 0，即 x² + x − 12 = 0。

These formula are particularly useful in IGCSE questions that ask you to find the value of expressions such as α² + β². Since (α + β)² = α² + β² + 2αβ, we rearrange to get α² + β² = (α + β)² − 2αβ.

这些公式在 IGCSE 中特别常用于求解 α² + β² 之类的表达式。因为 (α + β)² = α² + β² + 2αβ，所以移项得 α² + β² = (α + β)² − 2αβ。

7. Solving Equations Reducible to Quadratics | 可化为二次方程的方程

Many non-quadratic equations can be transformed into quadratic equations using a substitution. A common type is the quartic equation of the form ax⁴ + bx² + c = 0, which is quadratic in x².

许多非二次方程可以通过换元转化为二次方程。常见类型是 ax⁴ + bx² + c = 0 形式四次方程，它关于 x² 是二次的。

For example, solve x⁴ − 5x² + 4 = 0. Let u = x². Then the equation becomes u² − 5u + 4 = 0, which factorises as (u − 1)(u − 4) = 0. Therefore u = 1 or u = 4. Since u = x², we have x² = 1 or x² = 4, so x = ±1 or x = ±2.

例如，解 x⁴ − 5x² + 4 = 0。令 u = x²，则方程化为 u² − 5u + 4 = 0，因式分解得 (u − 1)(u − 4) = 0。因此 u = 1 或 u = 4。由于 u = x²，有 x² = 1 或 x² = 4，故 x = ±1 或 x = ±2。

Another common case is equations involving √x. Let u = √x, then the equation containing x = u² becomes a quadratic in u. Remember to check that each solution for u leads to a valid non-negative value of x.

另一个常见类型是含 √x 的方程。令 u = √x，则含 x = u² 的方程化为关于 u 的二次方程。记住检查每个 u 的解是否能推出 x 的非负有效值。

8. Solving Word Problems | 应用题

Many IGCSE word problems lead naturally to quadratic equations. The key steps are: define the unknown, translate the conditions into an equation, solve it, and interpret the answers in the context of the problem.

许多 IGCSE 应用题天然会导出二次方程。关键步骤是：设定未知数、将条件转化为方程、求解、再结合题意解释答案。

Example: The length of a rectangle is 3 cm longer than its width, and its area is 40 cm². Find its dimensions. Let the width be x cm; then the length is (x + 3) cm. The area equation is x(x + 3) = 40, which gives x² + 3x − 40 = 0. Factoring: (x + 8)(x − 5) = 0, so x = −8 or x = 5. Since a width cannot be negative, x = 5. The rectangle is 5 cm by 8 cm.

例题：长方形的长比宽长 3 cm，面积为 40 cm²。求其尺寸。设宽为 x cm，则长为 (x + 3) cm。面积方程为 x(x + 3) = 40，即 x² + 3x − 40 = 0。因式分解得 (x + 8)(x − 5) = 0，所以 x = −8 或 x = 5。由于宽不能为负，取 x = 5。故长方形为 5 cm × 8 cm。

Always reject negative solutions when the variable represents a physical quantity such as length, time, or number of items. However, keep both solutions when the problem allows negative values, such as coordinates on a number line.

当变量代表长度、时间或物品数量等物理量时，务必舍去负数解。但若题目允许负值（如数轴上的坐标），则应保留两个解。

9. Graphs of Quadratic Functions | 二次函数图象

The graph of y = ax² + bx + c is a parabola. When a > 0, the parabola opens upward; when a < 0, it opens downward. The solutions of ax² + bx + c = 0 correspond to the x-coordinates where the parabola crosses the x-axis.

y = ax² + bx + c 的图象是抛物线。当 a > 0 时，抛物线开口向上；当 a < 0 时，开口向下。方程 ax² + bx + c = 0 的解就是抛物线与 x 轴交点的横坐标。

The vertex (turning point) of the parabola is located at x = −b ÷ (2a). Substituting this value back into the equation gives the y-coordinate of the vertex. The axis of symmetry is the vertical line x = −b ÷ (2a).

抛物线的顶点位于 x = −b ÷ (2a) 处。将该值代回方程即可得到顶点的纵坐标。对称轴是竖直线 x = −b ÷ (2a)。

For example, for y = x² − 4x + 3, the axis of symmetry is x = 2. Substituting x = 2 gives y = 4 − 8 + 3 = −1, so the vertex is at (2, −1). The roots are x = 1 and x = 3, found by factorising x² − 4x + 3 = (x − 1)(x − 3).

例如，对 y = x² − 4x + 3，对称轴为 x = 2。代入 x = 2 得 y = 4 − 8 + 3 = −1，故顶点为 (2, −1)。由因式分解 x² − 4x + 3 = (x − 1)(x − 3) 得两根为 x = 1 和 x = 3。

The discriminant also determines the number of x-intercepts of the parabola: Δ > 0 gives two intersections, Δ = 0 gives one (the vertex touches the axis), and Δ < 0 gives zero intersections with the x-axis.

判别式还决定抛物线与 x 轴的交点个数：Δ > 0 有两个交点，Δ = 0 有一个交点（顶点与轴相切），Δ < 0 则没有交点。

10. Common Pitfalls and Tips | 常见误区与技巧

Students frequently make the following errors when solving quadratic equations. Avoiding these will earn you valuable marks in the exam.

学生在解二次方程时经常犯以下错误。避免这些错误能帮你在考试中拿到宝贵的分数。

Forgetting to rearrange to standard form: Always bring all terms to one side so that the equation equals zero before factorising or using the formula.
忘记化为标准形式：在使用因式分解或公式之前，务必把所有项移到一边，使方程等于零。
Dividing by x: If x is a common factor, do not simply divide by x; you will lose the root x = 0. Factor by x instead.
除以 x：若 x 是公因式，切勿直接除以 x，否则会丢失根 x = 0。而应提取 x 作为因式。
Missing ± when taking square roots: If x² = 9, then x = ±3, not just x = 3.
开平方时遗漏 ±：若 x² = 9，则 x = ±3，而不仅仅是 x = 3。
Sign errors in the quadratic formula: When b is negative, substituting −b carefully with brackets is essential.
二次公式中的符号错误：当 b 为负时，务必用括号仔细代入 −b。
Skipping the check: Always verify your roots by substitution, especially in word problems.
跳过检验：始终通过代入来验证你的根，尤其是在应用题中。

A useful tip: if a quadratic factorises quickly by inspection, use factorisation; otherwise, use the quadratic formula directly to save time. Completing the square is best reserved for questions on turning points or graph sketching.

实用技巧：如果二次式能通过观察快速分解，就用因式分解；否则直接用求根公式以节省时间。配方法最好用于涉及顶点或画图的问题。

11. Practice Questions | 练习

Try these IGCSE-style questions. They cover all the methods discussed above.

试试以下 IGCSE 风格练习题。它们涵盖了上面讨论的所有方法。

Question 1: Solve x² − 7x + 12 = 0 by factorisation.

第 1 题：用因式分解法解 x² − 7x + 12 = 0。

Question 2: Solve 3x² + 5x − 2 = 0 using the quadratic formula. Give your answers correct to 2 decimal places.

第 2 题：用二次求根公式解 3x² + 5x − 2 = 0。答案精确到小数点后两位。

Question 3: The equation 2x² + kx + 3 = 0 has exactly one real root. Find the possible values of k.

第 3 题：方程 2x² + kx + 3 = 0 只有一个实数根。求 k 的可能值。

Question 4: A right-angled triangle has legs of length x and x + 2, and hypotenuse √34. Form a quadratic equation and solve it to find x.

第 4 题：一个直角三角形两条直角边长分别为 x 和 x + 2，斜边长为 √34。建立二次方程并求解 x。

Question 5: If the roots of x² + 4x − 6 = 0 are α and β, find the value of αβ + α + β.

第 5 题：若 x² + 4x − 6 = 0 的两根为 α 和 β，求 αβ + α + β 的值。

Answers: Q1: x = 3 or x = 4. Q2: x = 0.33 or x = −2.00. Q3: k = ±2√6. Q4: x = 3. Q5: α + β = −4, αβ = −6, hence αβ + α + β = −10.

答案：第 1 题：x = 3 或 x = 4。第 2 题：x = 0.33 或 x = −2.00。第 3 题：k = ±2√6。第 4 题：x = 3。第 5 题：α + β = −4，αβ = −6，故 αβ + α + β = −10。

12. Key Formula Summary | 核心公式总结

Memorise the following table for a quick revision before the exam.

考前请熟记下表，快速回顾核心公式。

Concept \| 概念	Formula \| 公式
Standard form \| 标准形式	ax² + bx + c = 0, a ≠ 0
Quadratic formula \| 求根公式	x = (−b ± √(b² − 4ac)) ÷ (2a)
Discriminant \| 判别式	Δ = b² − 4ac
Sum of roots \| 两根之和	α + β = −b ÷ a
Product of roots \| 两根之积	αβ = c ÷ a
Axis of symmetry \| 对称轴	x = −b ÷ (2a)

Remember: the more questions you practise, the faster you will recognise which method to use. Good luck with your revision!

记住：练习越多，你就能越快地识别该用哪种方法。祝你复习顺利！

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Solving Quadratic Equations | 解一元二次方程