A-Level Maths: Simplification and Operations with Radicals | A-Level 数学：根式的化简与运算方法

📚 A-Level Maths: Simplification and Operations with Radicals | A-Level 数学：根式的化简与运算方法

Radicals (also called surds) appear frequently in A-Level Mathematics, from quadratic equations to trigonometric identities and calculus. Mastering the simplification and arithmetic of radicals is not only essential for obtaining full marks on pure mathematics questions but also forms the foundation for more advanced topics such as rationalising denominators, solving equations, and working with complex numbers.

根式（也称为无理式）在 A-Level 数学中频繁出现，从二次方程到三角恒等式再到微积分，处处可见。掌握根式的化简与运算，不仅是在纯数学题目中拿到满分的关键，更是学习有理化分母、解方程以及复数等进阶内容的基础。

1. Core Definitions and Properties | 基本定义与性质

A radical is an expression of the form ⁿ√a, where n is the index (or order) and a is the radicand. When n = 2, we simply write √a and call it a square root. For A-Level purposes, we most often work with square roots, though cube roots and higher roots also appear.

根式是形如 ⁿ√a 的表达式，其中 n 为根指数（或次数），a 为被开方数。当 n = 2 时，我们简写为 √a，称为平方根。在 A-Level 考试中，最常处理的是平方根，但立方根及更高次根也会出现。

The three fundamental properties that govern all radical operations are:

支配所有根式运算的三条基本性质是：

√(ab) = √a × √b (for a, b ≥ 0) — the product rule | 乘法法则（a, b ≥ 0）
√(a/b) = √a / √b (for a ≥ 0, b > 0) — the quotient rule | 除法法则（a ≥ 0, b > 0）
(√a)² = a — the cancellation rule | 消去法则

These properties seem simple, but they are the engine behind nearly every technique in this article. For example, √50 can be rewritten as √(25 × 2) = √25 × √2 = 5√2. This process is called simplifying a radical.

这些性质看似简单，却是本文几乎所有技巧的核心驱动力。例如，√50 可以改写为 √(25 × 2) = √25 × √2 = 5√2。这个过程称为根式的化简。

2. Simplifying Square Roots | 平方根的化简

To simplify a square root, factor the radicand into a product of a perfect square and another factor. Extract the square root of the perfect square and leave the remaining factor under the radical sign. The goal is to make the radicand as small as possible while keeping it an integer.

化简平方根的方法是将被开方数分解为一个完全平方数与另一个因数的乘积，取出完全平方数的平方根，将其余部分留在根号内。目标是使被开方数尽可能小，同时保持其为整数。

Consider √72. We factor 72 = 36 × 2, and since √36 = 6, we obtain √72 = 6√2. Alternatively, using prime factorisation: 72 = 2³ × 3² = 2² × 2 × 3², so √72 = 2 × 3 × √2 = 6√2.

以 √72 为例。我们将 72 分解为 36 × 2，因为 √36 = 6，所以 √72 = 6√2。或者用质因数分解：72 = 2³ × 3² = 2² × 2 × 3²，因此 √72 = 2 × 3 × √2 = 6√2。

For higher-order radicals, the same principle applies: look for perfect nth powers inside the radicand. For example, ∛54 = ∛(27 × 2) = ∛27 × ∛2 = 3∛2, because 27 = 3³ is a perfect cube.

对于更高次的根式，同样的原理适用：在被开方数中寻找完全 n 次幂。例如，∛54 = ∛(27 × 2) = ∛27 × ∛2 = 3∛2，因为 27 = 3³ 是完全立方数。

When simplifying, always check that the radicand has no remaining square factors (for square roots). A common mistake is stopping early — for instance, saying √48 = √(4 × 12) = 2√12, but 12 still contains a factor of 4, so the fully simplified form is 2√(4 × 3) = 4√3.

化简时，务必检查被开方数是否还含有平方因子（对于平方根）。一个常见错误是过早停止——例如，写出 √48 = √(4 × 12) = 2√12，但 12 仍含有因子 4，因此完全化简的形式应为 2√(4 × 3) = 4√3。

Expression \| 表达式	Factorisation \| 分解	Simplified Form \| 化简结果
√18	9 × 2	3√2
√200	100 × 2	10√2
√128	64 × 2	8√2
√a³b⁵	a² · a · b⁴ · b	ab²√(ab)

3. Adding and Subtracting Radicals | 根式的加减运算

Radicals can be added or subtracted only when they are like radicals — that is, they have the same index and the same radicand. The coefficients are then combined in the same way as like terms in algebra.

根式只有在它们是同类根式时才能相加减——即根指数相同且被开方数相同。此时，系数像代数中的同类项一样合并。

For example, 3√5 + 2√5 = 5√5, and 7√2 − 4√2 = 3√2. However, √2 + √3 cannot be simplified further because the radicands differ.

例如，3√5 + 2√5 = 5√5，7√2 − 4√2 = 3√2。然而，√2 + √3 无法进一步化简，因为被开方数不同。

The key challenge in addition and subtraction is recognising that radicals must first be simplified before checking whether they are like terms. For instance, √8 + √18 may not look like an addition of like radicals, but after simplification we get 2√2 + 3√2 = 5√2.

加减运算的关键挑战在于：必须先化简根式，再判断是否为同类项。例如，√8 + √18 乍看不是同类根式的加法，但化简后得到 2√2 + 3√2 = 5√2。

√8 + √18 = 2√2 + 3√2 = 5√2

Also be careful with expressions like √a + √a. Since both are identical, we get 2√a. But √a + √b (a ≠ b) remains as it is. A common exam trap is presenting √12 + √27 and expecting students to simplify both to 2√3 and 3√3 respectively.

还要注意像 √a + √a 这样的表达式。由于两者完全相同，结果为 2√a。但 √a + √b（a ≠ b）保持原样。考试中常见的陷阱是给出 √12 + √27，期望学生分别化简为 2√3 和 3√3。

4. Multiplying Radicals | 根式的乘法

Multiplication of radicals follows directly from the product rule: multiply the coefficients together and multiply the radicands together. For square roots, √a × √b = √(ab). When coefficients are present, c√a × d√b = cd√(ab).

根式的乘法直接遵循乘法法则：系数相乘，被开方数相乘。对于平方根，√a × √b = √(ab)。当存在系数时，c√a × d√b = cd√(ab)。

Example: 2√3 × 5√6. Multiply the coefficients: 2 × 5 = 10. Multiply the radicands: 3 × 6 = 18. So we get 10√18. Now simplify √18 = 3√2, giving the final answer 30√2.

例：2√3 × 5√6。系数相乘：2 × 5 = 10。被开方数相乘：3 × 6 = 18。因此得到 10√18。再化简 √18 = 3√2，最终答案为 30√2。

When multiplying a radical by itself, we use the cancellation rule: (√a)² = a. This is particularly useful in rationalising denominators, as we shall see later. More generally, (m√a)² = m²a.

当根式自乘时，利用消去法则：(√a)² = a。这在后续有理化分母时特别有用。更一般地，(m√a)² = m²a。

A very common A-Level question involves expanding brackets with radicals, such as (√3 + 2)(√3 − 5). This is exactly like expanding (x + 2)(x − 5), except x is replaced by √3:

A-Level 中非常常见的题型涉及含根式的括号展开，例如 (√3 + 2)(√3 − 5)。这完全类似于展开 (x + 2)(x − 5)，只需将 x 替换为 √3：

(√3 + 2)(√3 − 5) = √3·√3 − 5√3 + 2√3 − 10 = 3 − 3√3 − 10 = −7 − 3√3

Notice that √3·√3 = 3, which is a rational number. This observation is the basis for the difference of two squares pattern: (√a + √b)(√a − √b) = a − b.

注意 √3·√3 = 3，这是一个有理数。这个观察结果是平方差公式的基础：(√a + √b)(√a − √b) = a − b。

5. Dividing Radicals and Rationalising the Denominator | 根式的除法与分母有理化

Division of radicals uses the quotient rule: √a ÷ √b = √(a/b). However, a fundamental convention in mathematics is that a simplified expression should not have a radical in the denominator. The process of removing a radical from the denominator is called rationalising the denominator.

根式的除法使用除法法则：√a ÷ √b = √(a/b)。然而，数学中的一个基本约定是：化简后的表达式不应在分母中含有根式。将分母中的根式移除的过程称为分母有理化。

Case 1: The denominator is a single radical, e.g., 3/√7. Multiply both numerator and denominator by √7:

情形 1：分母是单个根式，例如 3/√7。将分子分母同时乘以 √7：

3/√7 = (3 × √7)/(√7 × √7) = 3√7/7

Case 2: The denominator is a binomial involving radicals, e.g., 5/(√3 + 2). Here we multiply by the conjugate of the denominator. The conjugate of √a + b is √a − b (and vice versa). Since (√a + b)(√a − b) = a − b², the denominator becomes rational.

情形 2：分母是含根式的二项式，例如 5/(√3 + 2)。此时乘以分母的共轭式。√a + b 的共轭式为 √a − b（反之亦然）。因为 (√a + b)(√a − b) = a − b²，分母变为有理数。

5/(√3 + 2) = [5(√3 − 2)] / [(√3 + 2)(√3 − 2)] = (5√3 − 10)/(3 − 4) = 10 − 5√3

Note that 3 − 4 = −1, so the fraction simplifies to −(5√3 − 10) = 10 − 5√3. Always check the sign carefully — this is a frequent source of errors.

注意 3 − 4 = −1，因此分数化简为 −(5√3 − 10) = 10 − 5√3。务必仔细检查符号——这是常见的错误来源。

When the denominator contains two radicals, such as 1/(√5 − √2), multiply by the conjugate √5 + √2:

当分母含有两个根式时，例如 1/(√5 − √2)，乘以共轭式 √5 + √2：

1/(√5 − √2) = (√5 + √2)/[(√5 − √2)(√5 + √2)] = (√5 + √2)/(5 − 2) = (√5 + √2)/3

6. Nested Radicals and Special Patterns | 嵌套根式与特殊模式

Occasionally, A-Level questions involve nested radicals such as √(a + √b). While there is no universal simplification formula, some expressions can be simplified by recognising perfect square patterns.

偶尔，A-Level 题目会涉及嵌套根式，如 √(a + √b)。虽然没有通用的化简公式，但某些表达式可以通过识别完全平方模式来化简。

Consider √(7 + 4√3). We look for two numbers p and q such that p + q = 7 and 2√(pq) = 4√3, i.e., pq = 12. Solving, p = 3 and q = 4 work: (√3 + √4)² = 3 + 4 + 2√12 = 7 + 4√3. Therefore, √(7 + 4√3) = √3 + 2.

考虑 √(7 + 4√3)。我们寻找两个数 p 和 q，使得 p + q = 7 且 2√(pq) = 4√3，即 pq = 12。解得 p = 3，q = 4 满足：(√3 + √4)² = 3 + 4 + 2√12 = 7 + 4√3。因此，√(7 + 4√3) = √3 + 2。

Another important pattern is the difference of two squares applied backwards: a − b = (√a + √b)(√a − √b). This identity is extremely useful for rationalising and for simplifying expressions like (√5 + √2)(√5 − √2) = 5 − 2 = 3.

另一个重要模式是平方差公式的逆向运用：a − b = (√a + √b)(√a − √b)。这一恒等式在有理化和化简如 (√5 + √2)(√5 − √2) = 5 − 2 = 3 的表达式时极为有用。

Also be aware of the identity (a + √b)² = a² + b + 2a√b. When squared, a radical expression produces a term with a radical. This is why equations involving radicals often require squaring both sides, which we discuss in the next section.

还要注意恒等式 (a + √b)² = a² + b + 2a√b。根式表达式平方后会产生一个含根式的项。这就是为什么含根式的方程通常需要两边平方，我们将在下一节讨论。

7. Solving Equations Involving Radicals | 含根式方程的求解

To solve an equation involving radicals, the standard strategy is to isolate the radical on one side and then raise both sides of the equation to the power of the index. For square roots, square both sides; for cube roots, cube both sides.

求解含根式方程的标准策略是：先将根式单独隔离到等式一侧，然后对方程两边同时进行与根指数相同的幂运算。对于平方根，两边平方；对于立方根，两边立方。

Example: Solve √(2x + 3) = 5. Squaring both sides gives 2x + 3 = 25, so 2x = 22 and x = 11. Always verify by substituting back: √(2(11) + 3) = √25 = 5, which is correct.

例：解方程 √(2x + 3) = 5。两边平方得 2x + 3 = 25，因此 2x = 22，x = 11。务必回代验证：√(2(11) + 3) = √25 = 5，正确。

When the equation has two radicals, isolate one radical first. Solve √(x + 1) + √(x − 1) = 3:

当方程含有两个根式时，先隔离其中一个。解 √(x + 1) + √(x − 1) = 3：

√(x + 1) = 3 − √(x − 1)

Squaring both sides:

两边平方：

x + 1 = 9 − 6√(x − 1) + (x − 1)

Simplify: x + 1 = x + 8 − 6√(x − 1), so 6√(x − 1) = 7, hence √(x − 1) = 7/6. Squaring again: x − 1 = 49/36, so x = 85/36. Verify that this satisfies the original equation.

化简：x + 1 = x + 8 − 6√(x − 1)，所以 6√(x − 1) = 7，即 √(x − 1) = 7/6。再次平方：x − 1 = 49/36，因此 x = 85/36。验证该解满足原方程。

Critical warning: Squaring both sides can introduce extraneous roots. For example, solve √(2x − 1) = x − 2. Squaring gives 2x − 1 = x² − 4x + 4, or x² − 6x + 5 = 0, so x = 1 or x = 5. However, substituting x = 1 gives √1 = 1 − 2 = −1, which is false. Therefore the only valid solution is x = 5. Always check your answers in the original equation.

重要警告：两边平方可能引入增根。例如，解 √(2x − 1) = x − 2。平方得 2x − 1 = x² − 4x + 4，即 x² − 6x + 5 = 0，所以 x = 1 或 x = 5。然而，将 x = 1 代入得 √1 = 1 − 2 = −1，不成立。因此唯一有效解为 x = 5。务必在原始方程中检验答案。

8. Radicals in Algebraic Fractions | 根式在代数分式中的应用

When radicals appear in algebraic fractions, the same simplification rules apply, but extra care is needed when factoring numerators and denominators. The key is to simplify the radicands first, then factor and cancel.

当根式出现在代数分式中时，同样的化简规则适用，但在分解分子和分母的因式时需要格外小心。关键是先化简被开方数，然后分解因式并约分。

Consider (√12 + √27)/√3. Simplify the numerator: 2√3 + 3√3 = 5√3. Then divide by √3 to get 5.

考虑 (√12 + √27)/√3。化简分子：2√3 + 3√3 = 5√3。然后除以 √3 得到 5。

Another common type involves fractions with radicals in both numerator and denominator, such as (3 + √5)/(2 − √5). Rationalise the denominator by multiplying by 2 + √5:

另一种常见类型是分子分母都含根式的分式，例如 (3 + √5)/(2 − √5)。通过乘以 2 + √5 来有理化分母：

= [(3 + √5)(2 + √5)] / [(2 − √5)(2 + √5)] = (6 + 3√5 + 2√5 + 5)/(4 − 5) = (11 + 5√5)/(−1) = −11 − 5√5

When the denominator is a simple radical and the numerator is a sum, split the fraction into separate terms first. For example, (√8 + √18)/√2 = √8/√2 + √18/√2 = √4 + √9 = 2 + 3 = 5.

当分母是简单根式而分子是和式时，先将分数拆分为若干项。例如，(√8 + √18)/√2 = √8/√2 + √18/√2 = √4 + √9 = 2 + 3 = 5。

It is also worth remembering that √(a²) = |a|, not simply a. In A-Level, unless the variable is specified as positive, you should include the absolute value when simplifying expressions like √(x²). In many exam questions, variables are assumed positive, but in pure mathematics without restrictions, the absolute value is the correct general answer.

还要记住 √(a²) = |a|，而不仅仅是 a。在 A-Level 中，除非题目明确变量为正，化简像 √(x²) 这样的表达式时应包含绝对值。在许多考试题目中，变量默认取正值，但在无限制的纯数学中，绝对值才是正确的一般答案。

9. Common Mistakes and Exam Tips | 常见错误与考试技巧

Below are the most frequent mistakes students make with radicals, along with strategies to avoid them.

以下是学生在处理根式时最常见的错误以及避免这些错误的策略。

Mistake \| 错误	Correct \| 正确	Tip \| 提示
√(a + b) = √a + √b	√(a + b) cannot be simplified	The radical does not distribute over addition
√(a²) = a	√(a²) = \|a\|	Remember the absolute value
Leaving a radical in the denominator	Rationalise the denominator	Multiply by the conjugate if needed
Not simplifying fully	e.g., √50 = 5√2	Always check for square factors

Time management tip: when facing a complicated radical expression, simplify each radical separately before performing addition or multiplication. This reduces the chance of arithmetic errors and often reveals cancellations.

时间管理技巧：面对复杂的根式表达式时，先分别化简每个根式，再进行加减或乘法运算。这能减少算术错误的机会，且常常能揭示出可以约分的项。

Exam questions often test radicals in the context of exact values — for instance, finding the exact value of sin 15° or cos 75° using compound angle formulae, which yields answers involving √6 ± √2 over 4. Being fluent with radical arithmetic ensures you can present such answers in their simplest exact form.

考试题目常在”精确值”的背景下考查根式——例如，使用和角公式求 sin 15° 或 cos 75° 的精确值，答案会涉及 (√6 ± √2)/4。熟练掌握根式运算能确保你以最简精确形式呈现答案。

10. Worked Examples | 综合例题精讲

Let us work through a few complete examples that resemble typical A-Level exam questions.

让我们完整求解几个典型的 A-Level 考试风格例题。

Example 1: Simplify (√5 + √3)(√5 − √3) and hence evaluate (√5 + √3)(√5 − √3) + 2√15.

例 1：化简 (√5 + √3)(√5 − √3)，并据此计算 (√5 + √3)(√5 − √3) + 2√15 的值。

Using the difference of two squares: (√5 + √3)(√5 − √3) = 5 − 3 = 2. Therefore the expression equals 2 + 2√15.

利用平方差公式：(√5 + √3)(√5 − √3) = 5 − 3 = 2。因此原表达式等于 2 + 2√15。

Example 2: Rationalise the denominator of 4/(3 − √5) and express the answer in the form a + b√5, where a and b are integers.

例 2：将 4/(3 − √5) 的分母有理化，并将答案表示为 a + b√5 的形式，其中 a 和 b 为整数。

4/(3 − √5) = [4(3 + √5)] / [(3 − √5)(3 + √5)] = (12 + 4√5)/(9 − 5) = (12 + 4√5)/4 = 3 + √5

Thus a = 3 and b = 1.

因此 a = 3，b = 1。

Example 3: Solve the equation √(2x + 3) − √(x − 1) = 1.

例 3：解方程 √(2x + 3) − √(x − 1) = 1。

Isolate: √(2x + 3) = 1 + √(x − 1). Square both sides:

隔离：√(2x + 3) = 1 + √(x − 1)。两边平方：

2x + 3 = 1 + 2√(x − 1) + (x − 1)

Simplify: 2x + 3 = x + 2√(x − 1), so x + 3 = 2√(x − 1). Square again:

化简：2x + 3 = x + 2√(x − 1)，因此 x + 3 = 2√(x − 1)。再次平方：

(x + 3)² = 4(x − 1) ⇒ x² + 6x + 9 = 4x − 4 ⇒ x² + 2x + 13 = 0

The discriminant is 2² − 4(13) = 4 − 52 = −48 < 0, so there are no real solutions. Always check whether solutions exist — the equation may have no real roots.

判别式为 2² − 4(13) = 4 − 52 = −48 < 0，因此没有实数解。务必检查解是否存在——方程可能没有实数根。

11. Practice Problems | 练习题

Test your understanding with the following problems. Full solutions are provided after the list.

通过以下题目检验你的理解。完整解答列在题目之后。

1. Simplify √45 + √20 − √5.
2. Rationalise: 7/(2√3 − 1).
3. Expand and simplify: (2√3 + √2)².
4. Solve: √(3x + 4) = 8.
5. Solve: 5/(√2 + 1) − 3/(√2 − 1).
6. Express √(11 + 6√2) in the form m + n√2.

1. 化简 √45 + √20 − √5。
2. 有理化：7/(2√3 − 1)。
3. 展开并化简：(2√3 + √2)²。
4. 解方程：√(3x + 4) = 8。
5. 计算：5/(√2 + 1) − 3/(√2 − 1)。
6. 将 √(11 + 6√2) 表示为 m + n√2 的形式。

Solutions / 解答：

1. √45 = 3√5, √20 = 2√5, so 3√5 + 2√5 − √5 = 4√5.

1. √45 = 3√5，√20 = 2√5，所以 3√5 + 2√5 − √5 = 4√5。

2. Multiply by 2√3 + 1: 7(2√3 + 1)/(12 − 1) = (14√3 + 7)/11.

2. 乘以 2√3 + 1：7(2√3 + 1)/(12 − 1) = (14√3 + 7)/11。

3. (2√3)² + 2(2√3)(√2) + (√2)² = 12 + 4√6 + 2 = 14 + 4√6.

3. (2√3)² + 2(2√3)(√2) + (√2)² = 12 + 4√6 + 2 = 14 + 4√6。

4. Square both sides: 3x + 4 = 64, so 3x = 60, x = 20. Check: √(60 + 4) = √64 = 8, valid.

4. 两边平方：3x + 4 = 64，因此 3x = 60，x = 20。检验：√(60 + 4) = √64 = 8，成立。

5. Rationalise each: 5(√2 − 1)/(2 − 1) = 5√2 − 5; 3(√2 + 1)/(2 − 1) = 3√2 + 3. Difference: (5√2 − 5) − (3√2 + 3) = 2√2 − 8.

5. 分别有理化：5(√2 − 1)/(2 − 1) = 5√2 − 5；3(√

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A-Level Maths: Simplification and Operations with Radicals | A-Level 数学：根式的化简与运算方法