Solving Quadratic Equations | 二次方程解析

📚 Solving Quadratic Equations | 二次方程解析

A quadratic equation is one of the most important algebraic topics in the IGCSE Mathematics syllabus. Whether you are solving by factorising, using the quadratic formula or completing the square, a solid understanding of quadratics will support your performance in many exam questions.

二次方程是 IGCSE 数学课程中最重要的代数主题之一。无论是通过因式分解、使用二次公式还是配方法来求解，扎实掌握二次方程都能帮助你在许多考试题目中取得更好表现。

1. Standard Form | 标准形式

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

二次方程是可以用标准形式书写的方程：

ax² + bx + c = 0

where a, b and c are constants, and a is not equal to 0. The coefficient a determines whether the parabola opens upwards or downwards, and the values of b and c affect the position of the curve.

其中 a、b 和 c 是常数，且 a 不等于 0。系数 a 决定抛物线的开口方向，而 b 和 c 的值影响曲线的位置。

For example, 3x² – 2x + 1 = 0 has a = 3, b = -2 and c = 1. If a were 0, the equation would become linear, not quadratic.

例如，3x² – 2x + 1 = 0 中 a = 3，b = -2，c = 1。如果 a 为 0，方程将变为一次方程，而不是二次方程。

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2. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when the quadratic expression can be written as a product of two linear factors. The key principle is the zero-product property: if AB = 0, then A = 0 or B = 0.

当二次表达式可以写成两个一次因式的乘积时，因式分解通常是最快的方法。关键原理是零积性质：如果 AB = 0，那么 A = 0 或 B = 0。

For example, solve x² – 5x + 6 = 0. We look for two numbers that multiply to 6 and add to -5. These numbers are -2 and -3, so the factors are (x – 2)(x – 3). Setting each factor to zero gives x = 2 or x = 3.

例如，解 x² – 5x + 6 = 0。我们寻找两个相乘等于 6 且相加等于 -5 的数。这两个数是 -2 和 -3，因此因式为 (x – 2)(x – 3)。令每个因式等于零，得到 x = 2 或 x = 3。

When the coefficient of x² is not 1, factorisation requires extra care. For example, 2x² + 5x – 3 = 0 can be factorised as (2x – 1)(x + 3) = 0, giving x = ½ or x = -3.

当 x² 的系数不为 1 时，因式分解需要格外小心。例如，2x² + 5x – 3 = 0 可以分解为 (2x – 1)(x + 3) = 0，得到 x = ½ 或 x = -3。

• Always expand your factors to check they produce the original expression.
• 务必展开因式，检查是否得到原表达式。
• If the expression cannot be factorised, use another method.
• 如果表达式无法因式分解，请使用其他方法。

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3. The Quadratic Formula | 二次公式

The quadratic formula works for every quadratic equation, including those that cannot be factorised easily. For ax² + bx + c = 0, the solutions are given by

二次公式适用于所有二次方程，包括那些不容易因式分解的方程。对于 ax² + bx + c = 0，解为：

x = (-b ± √(b² – 4ac)) / (2a)

In this formula, the symbol ± means that we calculate two values: one with a plus sign and one with a minus sign.

在这个公式中，符号 ± 表示我们需要计算两个值：一个用加号，一个用减号。

For example, solve 2x² – 4x – 3 = 0. Here a = 2, b = -4 and c = -3. Substituting into the formula gives x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4, which simplifies to x ≈ 2.58 or x ≈ -0.58.

例如，解 2x² – 4x – 3 = 0。这里 a = 2，b = -4，c = -3。代入公式得到 x = (4 ± √(16 + 24)) / 4 = (4 ± √40) / 4，化简后 x ≈ 2.58 或 x ≈ -0.58。

Always simplify the square root if possible, and round only at the final step to avoid rounding errors.

如果可以，务必化简根号；只在最后一步四舍五入，以避免舍入误差。

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4. Completing the Square | 配方法

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

配方法将二次式改写为 a(x + p)² + q 的形式。这对于求抛物线的顶点以及在不使用公式的情况下解方程非常有用。

For a quadratic x² + bx, we add and subtract (b/2)². Since x² + bx + (b/2)² = (x + b/2)², we get x² + bx = (x + b/2)² – (b/2)².

对于 x² + bx，我们加上并减去 (b/2)²。因为 x² + bx + (b/2)² = (x + b/2)²，所以 x² + bx = (x + b/2)² – (b/2)²。

For example, x² + 6x – 1 = (x + 3)² – 9 – 1 = (x + 3)² – 10. Setting this to zero gives (x + 3)² = 10, so x = -3 ± √10.

例如，x² + 6x – 1 = (x + 3)² – 9 – 1 = (x + 3)² – 10。令其等于零，得到 (x + 3)² = 10，所以 x = -3 ± √10。

When a is not 1, factor out a first. For 2x² + 8x + 5 = 0, write 2(x² + 4x) + 5 = 2[(x + 2)² – 4] + 5 = 2(x + 2)² – 3.

当 a 不是 1 时，先提出 a。对于 2x² + 8x + 5 = 0，写成 2(x² + 4x) + 5 = 2[(x + 2)² – 4] + 5 = 2(x + 2)² – 3。

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5. The Discriminant | 判别式

Inside the quadratic formula, the expression b² – 4ac is called the discriminant,

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