Solving Quadratic Equations: A Complete IGCSE Guide | IGCSE二次方程求解全攻略

📚 Solving Quadratic Equations: A Complete IGCSE Guide | IGCSE二次方程求解全攻略

Quadratic equations appear in almost every IGCSE Mathematics paper. Mastering the three core solving methods — factorisation, completing the square, and the quadratic formula — is essential if you are aiming for a top grade. In this guide, we break down each method step by step, explain the discriminant, and show how to apply these skills to inequalities and word problems.

二次方程几乎出现在每一份 IGCSE 数学试卷中。掌握三种核心解法——因式分解、配方和二次公式——是冲击高分的关键。本指南将逐步拆解每种方法，解释判别式，并展示如何将这些技能应用到不等式和应用题中。

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

A quadratic equation is any equation that can be written in the standard form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the variable is 2, which is why it is called “quadratic”. When a = 0, the equation becomes linear, so the term ax² must be present for the equation to be genuinely quadratic.

二次方程是任何可以写成标准形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。变量的最高次数为 2，因此称为“二次”。当 a = 0 时，方程退化为一次方程，所以必须有 ax² 项，方程才是真正的二次方程。

Common examples include:

x² − 7x + 12 = 0
x² − 7x + 12 = 0
2x² + 3x − 5 = 0
2x² + 3x − 5 = 0
x² = 16, which can be rearranged to x² − 16 = 0
x² = 16，可改写为 x² − 16 = 0

2. Solving by Factorisation | 因式分解法

Factorisation is the simplest method when the quadratic factorises neatly. Write the quadratic as a product of two brackets, set each bracket equal to zero, and solve the resulting linear equations. This method is quick but only works when the expression can be factorised into rational factors.

当二次式可以整齐地分解时，因式分解是最简单的方法。将二次式写成两个括号的乘积，令每个括号等于零，然后解所得的线性方程。此方法快捷，但仅当表达式可以分解为有理因式时才能使用。

Follow these steps:

Step 1: Rearrange the equation into standard form ax² + bx + c = 0.
第一步：将方程整理为标准形式 ax² + bx + c = 0。
Step 2: Factorise the left-hand side into two linear brackets.
第二步：将左边分解为两个一次括号的乘积。
Step 3: Use the zero product property: if AB = 0, then A = 0 or B = 0.
第三步：运用零积性质：若 AB = 0，则 A = 0 或 B = 0。
Step 4: Solve each linear equation to find the two roots.
第四步：解每个一次方程，得到两个根。

Solve x² − 5x + 6 = 0. We look for two numbers that multiply to 6 and add to −5: they are −2 and −3.

解 x² − 5x + 6 = 0。我们要找两个数，它们相乘得 6、相加得 −5：这两个数是 −2 和 −3。

x² − 5x + 6 = (x − 2)(x − 3) = 0

So x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3. Always check by substituting both values back into the original equation.

因此 x − 2 = 0 或 x − 3 = 0，解得 x = 2 或 x = 3。务必把两个值代回原方程进行验算。

3. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic in the form (x + p)² + q. For x² + bx + c = 0, we add and subtract (b/2)² to create a perfect square. This method is especially useful when the coefficient of x is even, and it directly reveals the turning point of the parabola.

配方法将二次式改写为 (x + p)² + q 的形式。对于 x² + bx + c = 0，我们通过加减 (b/2)² 来构造完全平方。此方法在 x 的系数为偶数时尤其方便，并能直接揭示抛物线的顶点。

Worked example: Solve x² + 6x + 5 = 0 by completing the square.

示例：用配方法解 x² + 6x + 5 = 0。

Half of 6 is 3, so we write (x + 3)² − 9 + 5 = 0.

6 的一半是 3，因此写成 (x + 3)² − 9 + 5 = 0。

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

Therefore x = −3 + 2 = −1 or x = −3 − 2 = −5. The turning point of y = x² + 6x + 5 is (−3, −4), read directly from the completed-square form.

因此 x = −3 + 2 = −1 或 x = −3 − 2 = −5。函数 y = x² + 6x + 5 的顶点为 (−3, −4)，可直接从配方式中读出。

General formula for the vertex: when the quadratic is (x + p)² + q, the turning point is (−p, q).

顶点的一般公式：当二次式为 (x + p)² + q 时，顶点为 (−p, q)。

4. The Quadratic Formula | 二次公式法

The quadratic formula is a universal method that solves any quadratic equation, even when factorisation is difficult or impossible. For ax² + bx + c = 0, the roots are given by:

二次公式是一种通用方法，可以解任何二次方程，即使因式分解困难或不可行。对于 ax² + bx + c = 0，根由下式给出：

x = (−b ± √(b² − 4ac)) / (2a)

Worked example: Solve 2x² − 4x − 3 = 0 using the formula.

示例：用公式法解 2x² − 4x − 3 = 0。

Here a = 2, b = −4 and c = −3. Substitute into the formula:

这里 a = 2，b = −4，c = −3。代入公式：

x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4 = 1 ± √10 / 2

So x ≈ 2.58 or x ≈ −0.58 (to 2 decimal places). The formula is always reliable, but you must be careful with signs when b is negative.

因此 x ≈ 2.58 或 x ≈ −0.58（精确到两位小数）。公式法始终可靠，但当 b 为负数时，要特别注意符号。

5. The Discriminant | 判别式

The discriminant is the expression under the square root in the quadratic formula: Δ = b² − 4ac. It tells us how many real roots a quadratic equation has, without solving the equation fully.

判别式是二次公式中根号内的表达式：Δ = b² − 4ac。它无需完整解方程，即可告诉我们二次方程有多少个实数根。

Value of Δ 判别式值	Nature of roots 根的性质
Δ > 0 Δ > 0	Two distinct real roots 两个不相等的实数根
Δ = 0 Δ = 0	One repeated root (two equal roots) 一个重根（两个相等的根）
Δ < 0 Δ < 0	No real roots 没有实数根

For example, the equation x² + 2x + 5 = 0 has Δ = 4 − 20 = −16 < 0, so it has no real roots. In addition, if Δ is a perfect square and a, b, c are rational, the quadratic can be factorised.

例如，方程 x² + 2x + 5 = 0 的判别式 Δ = 4 − 20 = −16 < 0，因此没有实数根。此外，若 Δ 是完全平方数且 a、b、c 为有理数，则二次式可以因式分解。

6. Solving Quadratic Inequalities | 二次不等式

To solve a quadratic inequality, first find the roots of the corresponding equation, then use a sign diagram or sketch the parabola to determine which intervals satisfy the inequality. The key is to remember that the parabola changes sign only at its roots.

解二次不等式时，先求出对应方程的根，然后利用符号图或抛物线草图判断哪些区间满足不等式。关键在于记住抛物线只在其根处改变符号。

Worked example: Solve x² − 3x − 10 < 0.

示例：解 x² − 3x − 10 < 0。

First factorise: (x − 5)(x + 2) < 0. The critical values are x = −2 and x = 5. Since the coefficient of x² is positive, the parabola opens upward, so the expression is negative between the roots.

首先因式分解：(x − 5)(x + 2) < 0。临界值为 x = −2 和 x = 5。由于 x² 的系数为正，抛物线开口向上，因此表达式在两个根之间为负。

−2 < x < 5

For the inequality x² − 3x − 10 ≥ 0, the solution would be x ≤ −2 or x ≥ 5, because the parabola is positive or zero outside the roots.

对于不等式 x² − 3x − 10 ≥ 0，解为 x ≤ −2 或 x ≥ 5，因为抛物线在两根之外为正或为零。

7. Graphical Interpretation | 图像解释

The graph of a quadratic function y = ax² + bx + c is a parabola. The roots of the equation are the x-intercepts of the graph. If a > 0, the parabola opens upward (U-shaped); if a < 0, it opens downward (N-shaped). This shape determines the sign of the quadratic in different intervals, which is why sketching is so useful for inequalities.

二次函数 y = ax² + bx + c 的图像是一条抛物线。方程的根就是抛物线与 x 轴的交点。当 a > 0 时，抛物线开口向上（U 形）；当 a < 0 时，开口向下（N 形）。图像形状决定了二次式在不同区间的正负号，这正是画草图在解不等式时非常有用的原因。

The turning point (vertex) can be found by completing the square: y = (x + p)² + q gives vertex (−p, q). The axis of symmetry is the vertical line x = −p. The y-intercept is always c, since setting x = 0 gives y = c.

顶点可以通过配方法求得：y = (x + p)² + q 的顶点为 (−p, q)。对称轴是竖直线 x = −p。y 截距始终为 c，因为令 x = 0 时 y = c。

Remember the connection: when a quadratic has no real roots, its graph does not cross the x-axis at all; the entire parabola lies above or below the x-axis.

记住对应关系：当二次方程没有实数根时，其图像完全不与 x 轴相交；整个抛物线位于 x 轴上方或下方。

8. Word Problems | 应用题

Many IGCSE questions require forming a quadratic equation from a real-life context. Read the problem carefully, define a variable, translate the conditions into an equation, then solve it using the most appropriate method. In geometry problems, remember that lengths and areas must be positive, so always reject negative solutions that do not make sense.

许多 IGCSE 题目要求从实际情境中建立二次方程。仔细阅读题目，定义变量，将条件转化为方程，然后用最合适的方法求解。在几何问题中，长度和面积必须为正，因此要舍去没有实际意义的负数解。

Worked example: The area of a rectangle is 24 cm² and its length is 5 cm longer than its width. Find the width.

示例：一个长方形的面积为 24 cm²，长比宽长 5 cm。求宽。

Let the width be x. Then the length is x + 5, so x(x + 5) = 24. Expanding gives x² + 5x − 24 = 0. Factorise: (x + 8)(x − 3) = 0, so x = −8 or x = 3. Since width cannot be negative, the width is 3 cm.

设宽为 x，则长为 x + 5，所以 x(x + 5) = 24。展开得 x² + 5x − 24 = 0。因式分解：(x + 8)(x − 3) = 0，故 x = −8 或 x = 3。由于宽不能为负，所以宽为 3 cm。

9. Common Mistakes and Exam Tips | 常见错误与考试提示

In the exam, candidates often lose marks for small but avoidable errors. Here are the most common mistakes and how to avoid them.

考试中，考生常因细小但可避免的错误而失分。以下是最常见的错误及规避方法。

Mistake: Forgetting to rearrange the equation to zero before factorising.
错误：因式分解前忘记将方程整理为零。
Fix: Always write the equation in the form ax² + bx + c = 0 first.
对策：始终先将方程写成 ax² + bx + c = 0 的形式。
Mistake: Dropping the ± sign when taking square roots.
错误：开平方时遗漏 ± 号。
Fix: Remember that x² = k gives x = +√k or x = −√k.
对策：记住 x² = k 给出 x = +√k 或 x = −√k。
Mistake: Sign errors when substituting b into the quadratic formula.
错误：代入二次公式时 b 的符号出错。
Fix: Use brackets when substituting negative values, e.g. (−4)².
对策：代入负值时加括号，例如 (−4)²。
Mistake: Confusing the axis of symmetry with the roots.
错误：混淆对称轴与方程的根。
Fix: Sketch the parabola to check whether your answers are reasonable.
对策：画抛物线草图，检查答案是否合理。

Top tip: if factorisation is not obvious within ten seconds, switch to the quadratic formula. Also, always test your solutions by substituting them into the original equation.

重要提示：如果在十秒内看不出因式分解，就改用二次公式。另外，务必把解代回原方程进行检验。

10. Choosing the Right Method | 如何选择合适的方法

Different situations call for different methods. The table below summarises when each method is most effective.

不同情况适用不同方法。下表总结了每种方法最有效的使用时机。

Method 方法	Best used when 最佳适用场景
Factorisation 因式分解	Coefficients are small and factors are easy to spot 系数较小且因式容易看出
Completing the square 配方法	You need the vertex, or the coefficient of x is even 需要求顶点，或 x 的系数为偶数
Quadratic formula 二次公式	Factorisation is not obvious; works for any quadratic 因式分解不明显；适用于任何二次方程

Whichever method you choose, you should always check whether the discriminant is non-negative before expecting real solutions. A quick sketch of the parabola can also prevent many careless errors.

无论选择哪种方法，在期望实数解之前，都应先检查判别式是否非负。快速画出抛物线草图也能避免许多粗心错误。

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Solving Quadratic Equations: A Complete IGCSE Guide | IGCSE二次方程求解全攻略