Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations appear throughout the IGCSE Mathematics syllabus, from algebra and graphs to problem solving in real-life contexts. Mastering the three main methods — factorisation, the quadratic formula and completing the square — is essential for achieving high marks in both Paper 1 (non-calculator) and Paper 2 (calculator).

二次方程贯穿 IGCSE 数学整个大纲，从代数、图像到实际情境中的问题解决都会涉及。熟练掌握三种主要方法——因式分解、求根公式和配方法——是在 Paper 1（不可使用计算器）和 Paper 2（可使用计算器）中取得高分的关键。

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

A quadratic equation is a polynomial equation in which the highest power of the variable x is 2. Its general form is:

二次方程是变量 x 的最高次数为 2 的多项式方程，其一般形式为：

ax² + bx + c = 0, where a ≠ 0

Here a, b and c are constants, and a cannot be zero; if a = 0 the equation becomes linear. The value of a also determines the shape of the graph — if a > 0 the parabola opens upwards, and if a < 0 it opens downwards.

其中 a、b、c 为常数，且 a 不能为零；若 a = 0，方程退化为一次方程。a 的值还决定图像形状——若 a > 0，抛物线开口向上；若 a < 0，抛物线开口向下。

For example, 2x² − 3x + 1 = 0 is quadratic, while 3x + 5 = 0 is not. An equation like x² − 4 = 0 is a special case where b = 0 and c = −4.

例如，2x² − 3x + 1 = 0 是二次方程，而 3x + 5 = 0 不是。形如 x² − 4 = 0 的方程是 b = 0、c = −4 的特殊情况。

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2. Solving by Factorisation | 因式分解法

Factorisation relies on the zero-product property: if the product of two expressions is zero, then at least one of them must be zero.

因式分解法依赖于零积性质：若两个表达式的乘积为零，则其中至少有一个必须为零。

If AB = 0, then A = 0 or B = 0

To solve x² − 5x + 6 = 0, first factorise the left-hand side. We look for two numbers that multiply to +6 and add to −5: these are −2 and −3.

解 x² − 5x + 6 = 0 时，先将左边因式分解。我们需要找两个数，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

Applying the zero-product property gives x − 2 = 0 or x − 3 = 0, so the solution set is x = 2 or x = 3. Always check your answer by substituting back into the original equation.

应用零积性质得 x − 2 = 0 或 x − 3 = 0，所以解为 x = 2 或 x = 3。务必把答案代回原方程进行检验。

Some quadratics factorise using the difference of two squares. For x² − 9 = 0, write (x − 3)(x + 3) = 0, giving x = 3 or x = −3. Notice that b = 0 here, so the two roots are opposites.

部分二次式可用平方差公式分解。对于 x² − 9 = 0，写成 (x − 3)(x + 3) = 0，得 x = 3 或 x = −3。注意此处 b = 0，因此两根互为相反数。

For quadratics with a leading coefficient not equal to 1, such as 2x² + 5x − 3 = 0, factorise by grouping or by trial: (2x − 1)(x + 3) = 0, giving x = ½ or x = −3.

对于二次项系数不为 1 的方程，如 2x² + 5x − 3 = 0，可用分组分解或试根法：(2x − 1)(x + 3) = 0，解得 x = ½ 或 x = −3。

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3. Solving by the Quadratic Formula | 求根公式法

When factorisation is difficult or impossible, use the quadratic formula. It works for every quadratic equation in the form ax² + bx + c = 0.

当因式分解困难或无法进行时，使用求根公式。它适用于所有 ax² + bx + c = 0 形式的二次方程。

x = (−b ± √(b² − 4ac)) / 2a

Consider 3x² − 4x − 2 = 0. Here a = 3, b = −4, c = −2. Substitute carefully, being alert to negative signs.

以 3x² − 4x − 2 = 0 为例，其中 a = 3，b = −4，c = −2。代入时要格外细心，注意负号。

x = (4 ± √(16 + 24)) / 6 = (4 ± √40) / 6

This gives x = (4 + √40)/6 ≈ 1.72 and x = (4 − √40)/6 ≈ −0.39 (to 2 d.p.). On Paper 2 you may give decimal answers, but on Paper 1 leave the answer in exact surd form.

从而 x = (4 + √40)/6 ≈ 1.72，x = (4 − √40)/6 ≈ −0.39（精确到 2 位小数）。在 Paper 2 中可以给出小数答案，但在 Paper 1 中应保留根式的精确形式。

A helpful memory aid is to write the formula down first, then label a, b and c. This simple step prevents sign errors and mis-substitution, which are the most common causes of lost marks.

一个有用的技巧是先把公式写下来，再标出 a、b、c 的值。这一简单步骤能防止符号错误和代错值，这两个问题是失分最常见的原因。

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4. Solving by Completing the Square | 配方法

Completing the square rewrites a quadratic in the form p(x + q)² + r. This method is particularly useful for finding turning points and proving inequalities.

配方法将二次式改写为 p(x + q)² + r 的形式。该方法在求顶点坐标和证明不等式时特别有用。

Solve x² + 6x + 2 = 0. Take half of 6, which is 3, and write (x + 3)² − 9 + 2 = 0, so (x + 3)² = 7.

解 x² + 6x + 2 = 0。取 6 的一半即 3，写成 (x + 3)² − 9 + 2 = 0，即 (x + 3)² = 7。

x + 3 = ±√7, so x = −3 ± √7

For a coefficient of x² greater than 1, factor it out first. For 2x² + 8x + 5 = 0, write 2(x² + 4x) + 5 = 2[(x + 2)² − 4] + 5

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