Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most tested topics in IGCSE Mathematics. A quadratic equation is any equation of the form ax² + bx + c = 0, where a ≠ 0. Mastery of this topic requires fluency in factorisation, the quadratic formula, completing the square, and interpreting graphs.

二次方程是 IGCSE 数学中考查最多的考点之一。二次方程是形如 ax² + bx + c = 0 的方程，其中 a ≠ 0。掌握这一考点需要熟练运用因式分解、求根公式、配方法以及解读图像。

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

A quadratic equation is a polynomial equation of degree 2. Its standard form is ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The term ax² makes it quadratic; if a = 0, the equation becomes linear.

二次方程是次数为 2 的多项式方程，其标准形式为 ax² + bx + c = 0，其中 a、b、c 为常数且 a ≠ 0。ax² 这一项使其成为二次；若 a = 0，方程退化为一次方程。

Example / 例：2x² − 5x + 3 = 0 has a = 2, b = −5, c = 3. / 中 a = 2，b = −5，c = 3。
Example / 例：x² = 9 can be rewritten as x² − 9 = 0. / 可改写为 x² − 9 = 0。

Not every quadratic equation appears in standard form. Often you must rearrange terms first: move all terms to one side and simplify.

并非所有二次方程都以标准形式出现，通常需要先移项整理：将所有项移到一边并化简。

2. Expanding and Simplifying Quadratics | 展开与化简二次式

Before solving, you must be able to expand brackets. The key rule is the distributive law: (x + p)(x + q) = x² + (p + q)x + pq.

在求解之前，必须掌握展开括号的方法。核心法则是分配律：(x + p)(x + q) = x² + (p + q)x + pq。

(x + 3)(x − 5) = x² − 5x + 3x − 15 = x² − 2x − 15

When the coefficient of x² is not 1, use FOIL carefully: multiply First, Outer, Inner, Last, then collect like terms.

当 x² 的系数不为 1 时，用 FOIL 方法仔细运算：先乘首项、外项、内项、末项，然后合并同类项。

Example / 例：(2x + 1)(x − 4) = 2x² − 8x + x − 4 = 2x² − 7x − 4
Special case / 特例：(x + 3)² = x² + 6x + 9; (x − 2)² = x² − 4x + 4
Difference of squares / 平方差：x² − 16 = (x + 4)(x − 4)

Mastering expansion is essential because factorisation is simply the reverse process.

熟练掌握展开是必要的，因为因式分解正是这一过程的逆运算。

3. Solving by Factorisation | 因式分解法求解

If a quadratic can be factorised into two brackets, set each bracket equal to zero. This uses the zero product property: if AB = 0, then A = 0 or B = 0.

若二次方程可分解为两个括号相乘，则令每个括号等于零。这基于零积性质：若 AB = 0，则 A = 0 或 B = 0。

x² − 7x + 12 = 0 → (x − 3)(x − 4) = 0 → x = 3 or x = 4

Step-by-step method:

分步方法：

Step 1: Rearrange into standard form ax² + bx + c = 0. / 第一步：化为标准形式 ax² + bx + c = 0。
Step 2: Find two numbers that multiply to give ac and add to give b. / 第二步：找出两个数，乘积为 ac，和为 b。
Step 3: Split the middle term and factor by grouping. / 第三步：拆分中间项并分组因式分解。
Step 4: Set each bracket to zero and solve. / 第四步：令每个括号为零并求解。

For equations like x² = 6x, do not divide both sides by x — this loses the root x = 0. Instead, rearrange: x² − 6x = 0, then x(x − 6) = 0, giving x = 0 or x = 6.

对于 x² = 6x 这类方程，切勿两边同除以 x——这会丢失根 x = 0。正确做法是移项：x² − 6x = 0，然后 x(x − 6) = 0，得 x = 0 或 x = 6。

4. Solving by the Quadratic Formula | 公式法求解

When factorisation is difficult or impossible, use the quadratic formula. For ax² + bx + c = 0:

当因式分解困难或不可行时，使用求根公式。对于 ax² + bx + c = 0：

x = (−b ± √(b² − 4ac)) / 2a

This formula works for all quadratic equations, including those with irrational or complex roots. Ensure the equation is in standard form before substituting.

该公式适用于所有二次方程，包括无理根或复数根的方程。代入前务必确保方程为标准形式。

Example / 例：Solve 2x² + 3x − 5 = 0. / 求解 2x² + 3x − 5 = 0。
Here a = 2, b = 3, c = −5. / 其中 a = 2，b = 3，c = −5。
x = (−3 ± √(9 + 40)) / 4 = (−3 ± 7) / 4 → x = 1 or x = −5/2

x = (−3 ± √(3² − 4 × 2 × (−5))) / (2 × 2) = (−3 ± √49) / 4 = (−3 ± 7) / 4

Always show substitution clearly in the exam — this earns method marks even if your final arithmetic is wrong.

考试中务必清晰写出代入过程——即使最终计算有误，也能获得方法分数。

5. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form a(x + h)² + k. This is especially useful for finding turning points and for solving equations without factorisation.

配方法将二次式改写为 a(x + h)² + k 的形式，特别适用于求顶点坐标以及无需因式分解即可求解方程。

For x² + bx, add and subtract (b/2)² to form a perfect square.

对于 x² + bx，加减 (b/2)² 以构成完全平方。

x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4

To solve x² + 6x + 5 = 0 in this form:

以此形式求解 x² + 6x + 5 = 0：

(x + 3)² − 4 = 0 → (x + 3)² = 4 → x + 3 = ±2 → x = −1 or x = −5

When a ≠ 1, factor a out of the first two terms first. For example, 2x² + 8x + 3 = 2(x² + 4x) + 3 = 2(x + 2)² − 8 + 3 = 2(x + 2)² − 5.

当 a ≠ 1 时，先将 a 从前两项中提出。例如，2x² + 8x + 3 = 2(x² + 4x) + 3 = 2(x + 2)² − 8 + 3 = 2(x + 2)² − 5。

6. The Discriminant Δ = b² − 4ac | 判别式 Δ = b² − 4ac

The discriminant determines the nature of the roots without solving the full equation.

判别式可判断根的性质，而无需完整求解方程。

Discriminant / 判别式值	Nature of Roots / 根的性质
Δ > 0 and a perfect square	Two distinct rational roots / 两个不同的有理根
Δ > 0 but not a perfect square	Two distinct irrational roots / 两个不同的无理根
Δ = 0	One repeated real root / 一个重根（两个相等实根）
Δ < 0	No real roots / 无实数根

For example, the equation x² + 2x + 5 = 0 has Δ = 4 − 20 = −16 < 0, so it has no real roots. This means its graph never crosses the x-axis.

例如，方程 x² + 2x + 5 = 0 的 Δ = 4 − 20 = −16 < 0，因此没有实数根，这意味着其图像与 x 轴不相交。

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

The graph of y = ax² + bx + c is a parabola. If a > 0, it opens upward (a “smile”); if a < 0, it opens downward (a "frown").

y = ax² + bx + c 的图像是抛物线。若 a > 0，开口向上；若 a < 0，开口向下。

The y-intercept is c. / y 截距为 c。
The x-intercepts are the real roots of ax² + bx + c = 0. / x 截距是方程 ax² + bx + c = 0 的实数根。
The axis of symmetry is x = −b/(2a). / 对称轴为 x = −b/(2a)。
The vertex (turning point) is (−b/(2a), c − b²/(4a)). / 顶点坐标为 (−b/(2a), c − b²/(4a))。

For y = x² − 4x + 3: axis of symmetry x = 2, vertex (2, −1), roots x = 1 and x = 3

对于 y = x² − 4x + 3：对称轴 x = 2，顶点 (2, −1)，根为 x = 1 和 x = 3。

When sketching, always label the y-intercept, x-intercepts (if any), and the turning point.

画草图时，务必标出 y 截距、x 截距（若有）以及顶点。

8. Word Problems | 应用题

Quadratic equations often arise from geometric or numeric problems. Set up the equation carefully, then solve.

二次方程常出现在几何或数字类应用题中。先仔细建立方程，再求解。

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

例：一个长方形的长比宽多 3 cm，面积为 40 cm²，求宽。

Let width = x → x(x + 3) = 40 → x² + 3x − 40 = 0 → (x + 8)(x − 5) = 0 → x = 5

设宽为 x → x(x + 3) = 40 → x² + 3x − 40 = 0 → (x + 8)(x − 5) = 0 → x = 5

Reject the negative solution x = −8 because a length cannot be negative. Always check whether your solutions make sense in the context of the question.

舍去负解 x = −8，因为长度不能为负。务必检验解在题目情境中是否合理。

9. Exam Tips and Common Mistakes | 考试技巧与常见错误

Below are the most frequent errors students make in quadratic equations — and how to avoid them.

以下是学生在二次方程中最高频的错误及其避免方法。

Mistake / 错误	Correction / 正确做法
Dividing both sides by x / 两边同除以 x	Always factorise; never lose the root x = 0 / 务必因式分解，切勿丢失 x = 0 这个根
Forgetting to rearrange to standard form / 忘记化为标准形式	Set all terms to one side first / 先将所有项移到一边
Sign errors in b and c / 代入公式时 b、c 符号错误	Write brackets: x = (−b ± √(b² − 4ac)) / 2a / 加括号代入
Not rejecting negative roots in word problems / 应用题中未舍去负根	Check answers against the context / 根据题目情境检验答案
Rounding too early / 过早四舍五入	Keep exact values until the final step / 保留精确值直到最后一步

Always spend a few seconds checking your solutions by substituting them back into the original equation.

每次求解后花几秒钟将解代回原方程验证。

Quadratic equations connect algebra, graphs and problem-solving. Practise all four solving methods — factorisation, formula, completing the square, and graphical interpretation — and you will be well prepared for any IGCSE question.

二次方程将代数、图像与问题解决紧密相连。熟练掌握四种求解方法——因式分解、求根公式、配方法以及图像解读——你就能从容应对任何 IGCSE 题目。

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Solving Quadratic Equations | 解二次方程