📚 Example 2.8.2: Solving a Quadratic Equation with Surd Roots | 示例2.8.2：二次方程的无理根求解
This article works through Example 2.8.2 from an AQA A-Level Mathematics textbook. The example focuses on solving a quadratic equation that does not factorise neatly, using the quadratic formula and presenting the answers as simplified surds. This is a core skill assessed throughout the pure mathematics papers.
本文详细讲解AQA A-Level数学教科书中的示例2.8.2。该示例重点在于运用二次公式求解一个无法简便因式分解的二次方程，并以最简根式（surd）形式呈现答案。这是纯数学试卷中反复考查的核心技能。
1. The Problem | 题目
The equation we are asked to solve is x² − 6x + 2 = 0. At first glance, this quadratic does not factorise into simple integer brackets, because no two integers multiply to give 2 and add to give −6. We therefore need the quadratic formula.
我们需要求解的方程是 x² − 6x + 2 = 0。乍一看，这个二次多项式无法分解为简单的整数括号因式，因为不存在两个整数相乘为2且相加为−6。因此，我们必须借助二次公式。
The requirement in this example is to give the roots in simplified surd form. A surd form keeps the exact values of the roots, whereas a decimal answer would be rounded and therefore only approximate.
该示例要求以最简根式形式给出根。根式形式保留了根的精确值，而小数答案经过四舍五入仅为近似值。
2. Identifying the Coefficients | 确定系数
For a quadratic equation written in the standard form ax² + bx + c = 0, we compare our equation with the standard form. We have a = 1, b = −6 and c = 2. Note carefully that b is negative.
对于标准形式 ax² + bx + c = 0 的二次方程，我们将原方程与标准形式对照。可得 a = 1，b = −6，c = 2。请特别注意b为负数。
The coefficient a is the multiplier of x², b is the multiplier of x, and c is the constant term. Getting these three values correct is essential before substituting into the formula.
系数a是x²的乘数，b是x的乘数，c是常数项。在代入公式之前，正确确定这三个数值至关重要。
3. The Quadratic Formula | 二次公式
The quadratic formula states that for ax² + bx + c = 0, the solutions are given by:
二次公式指出，对于 ax² + bx + c = 0，其解为：
x = (−b ± √(b² − 4ac)) ⁄ (2a)
The symbol ± means we take two separate roots: one with the plus sign and one with the minus sign. The expression under the square root, b² − 4ac, is called the discriminant.
符号±表示我们取两个不同的根：一个取加号，一个取减号。平方根下的表达式 b² − 4ac 称为判别式。
This formula works for every quadratic equation, provided a ≠ 0. It is especially useful when the equation does not factorise, as in this example.
该公式适用于一切二次方程，前提是a ≠ 0。当方程无法因式分解时（如本例），此公式尤为有用。
4. Substituting the Values | 代入数值
We substitute a = 1, b = −6 and c = 2 into the formula. It is helpful to write the formula with brackets around every substituted value to avoid sign errors.
我们将 a = 1，b = −6，c = 2 代入公式。为避免符号错误，建议在代入每个数值时都加上括号。
x = (−(−6) ± √((−6)² − 4 × 1 × 2)) ⁄ (2 × 1)
Now simplify the numerator step by step. First, −(−6) becomes +6. Second, (−6)² = 36. The product 4 × 1 × 2 = 8. So the expression becomes:
现在逐步化简分子。首先，−(−6) 变为 +6。其次，(−6)² = 36。乘积 4 × 1 × 2 = 8。因此表达式变为：
x = (6 ± √(36 − 8)) ⁄ 2 = (6 ± √28) ⁄ 2
Notice that the square root sign extends over the entire expression 36 − 8 = 28, not just over 36 or just over 8 individually.
注意，根号覆盖整个表达式 36 − 8 = 28，而不是分别只覆盖36或只覆盖8。
5. Simplifying the Surd | 化简根式
The next step is to simplify √28. We look for the largest square factor of 28. Since 28 = 4 × 7, and 4 is a perfect square, we write:
下一步是化简 √28。我们寻找28的最大平方因子。由于 28 = 4 × 7，而4是完全平方数，因此可写为：
√28 = √(4 × 7) = √4 × √7 = 2√7
Thus the expression for x becomes x = (6 ± 2√7) ⁄ 2. We are not finished: each term in the numerator must be divided by 2.
因此x的表达式变为 x = (6 ± 2√7) ⁄ 2。我们尚未完成：分子中的每一项都必须除以2。
Divide 6 by 2 to get 3, and divide 2√7 by 2 to get √7. This yields the simplified final answer.
6除以2得3，2√7除以2得√7。由此得到最简形式的最终答案。
6. Writing the Solution Set | 写出解集
The two exact roots are x = 3 + √7 and x = 3 − √7. We can present the solution set as:
两个精确根为 x = 3 + √7 和 x = 3 − √7。解集可表示为：
x = 3 ± √7
This compact notation is an acceptable answer in an AQA examination, provided the surd is fully simplified. The roots are irrational numbers because √7 cannot be written as a terminating or repeating decimal.
在AQA考试中，只要根式已完全化简，这种简洁写法就是可接受的答案。由于√7无法写成有限小数或循环小数，这两个根是无理数。
It is worth noting that there is no further simplification possible. The number 3 is rational and √7 is irrational, so the sum and difference cannot be combined into a single surd term.
值得指出的是，无法进一步化简。3是有理数而√7是无理数，因此和与差不能合并为单个根式项。
7. Verification by Substitution | 代入验证
We can check the correctness of our roots by substituting x = 3 + √7 back into the original equation. The verification process also provides useful algebraic practice.
我们可以通过将 x = 3 + √7 代回原方程来检验答案的正确性。验证过程本身也是有用的代数练习。
Compute x² = (3 + √7)² = 9 + 6√7 + 7 = 16 + 6√7. Then −6x = −6(3 + √7) = −18 − 6√7. Adding the constant term 2 gives:
计算 x² = (3 + √7)² = 9 + 6√7 + 7 = 16 + 6√7。然后 −6x = −6(3 + √7) = −18 − 6√7。加上常数项2得到：
x² − 6x + 2 = (16 + 6√7) + (−18 − 6√7) + 2 = 0
The terms 6√7 and −6√7 cancel, and 16 − 18 + 2 = 0. Therefore the root is verified. The same check works for x = 3 − √7.
6√7 与 −6√7 相互抵消，且 16 − 18 + 2 = 0。因此该根验证成立。对 x = 3 − √7 的检验同理。
8. The Discriminant and the Nature of Roots | 判别式与根的性质
The discriminant Δ = b² − 4ac = (−6)² − 4 × 1 × 2 = 36 − 8 = 28. Because Δ > 0 and 28 is not a perfect square, the equation has two distinct real irrational roots.
判别式 Δ = b² − 4ac = (−6)² − 4 × 1 × 2 = 36 − 8 = 28。由于Δ > 0且28不是完全平方数，该方程有两个不同的实数无理根。
If the discriminant were zero, there would be exactly one repeated root. If it were negative, there would be no real roots. This example therefore illustrates the intermediate case where both roots are real and different.
若判别式为零，则方程只有一个重根。若判别式为负，则无实数根。本例恰好展示了两个根均为实数且不相同的中间情形。
For reference, the decimal approximations are x ≈ 5.65 and x ≈ 0.35. These are useful for sketching the quadratic curve y = x² − 6x + 2, which intercepts the x-axis at these two points.
作为参考，小数近似值为 x ≈ 5.65 和 x ≈ 0.35。这些近似值有助于绘制曲线 y = x² − 6x + 2 的草图，该曲线在这两点处与x轴相交。
9. Common Mistakes and Exam Tips | 常见错误与考试提示
There are several common mistakes that students make with this type of question. The most frequent is a sign error when b is negative: forgetting that −b = 6, not −6.
此类问题中，学生常犯的错误有几种。最常见的是当b为负数时的符号错误：忘记 −b = 6，而不是 −6。
- Sign errors: Always write −(−6) explicitly before simplifying. | 符号错误：化简前务必明确写出 −(−6)。
- Incorrect squaring: (−6)² is 36, not −36. The square of a negative number is positive. | 平方错误：(−6)² 等于36，而不是 −36。负数的平方为正。
- Incomplete simplification: Remember to divide both 6 and 2√7 by 2. Many candidates incorrectly leave the answer as (6 ± 2√7) ⁄ 2. | 化简不彻底：记得6和2√7都要除以2。许多考生错误地留下 (6 ± 2√7) ⁄ 2 作为最终答案。
- Sloppy surd work: Do not write √28 = 28² or confuse √(4 × 7) with √4 × √7 incorrectly. The rule √(ab) = √a × √b is valid only when a and b are positive. | 根式书写马虎：不要写 √28 = 28²，也不要混淆 √(4 × 7) 与 √4 × √7 的运用。规则 √(ab) = √a × √b 仅在a、b为正数时成立。
In an exam, you should always show the substitution step and the simplification of the surd separately. This earns method marks even if the final answer has a small error.
考试中，务必分别写出代入步骤和根式化简过程。这样即使最终答案有微小错误，也能获得方法分。
Also remember to reject no solutions here: both 3 + √7 and 3 − √7 are valid because each one satisfies the original equation.
还要注意，本例无需舍去任何解：3 + √7 和 3 − √7 均为有效解，因为二者都满足原方程。
10. Practice Questions | 练习题目
To master this skill, try the following similar questions. Each equation does not factorise, so use the quadratic formula and give answers in simplified surd form.
为掌握这一技能，请尝试以下类似题目。每个方程均无法因式分解，请使用二次公式并以最简根式形式作答。
| Question | 题目 | Answer | 答案 |
| 1. Solve x² + 4x − 1 = 0. | x = −2 ± √5 |
| 2. Solve 2x² − 4x − 3 = 0. | x = 1 ± √10 ⁄ 2 |
| 3. Solve x² + 6x + 3 = 0. | x = −3 ± √6 |
For question 2, note that after applying the formula you get (4 ± √40) ⁄ 4, and √40 = 2√10, so the fraction simplifies to 1 ± √10 ⁄ 2.
对于第2题，注意应用公式后得到 (4 ± √40) ⁄ 4，且 √40 = 2√10，因此分式化简为 1 ± √10 ⁄ 2。
Working through these questions repeatedly will help you internalise the quadratic formula and surd simplification, two of the most heavily tested topics in AQA A-Level Mathematics.
反复练习这些题目有助于你熟练掌握二次公式和根式化简，这是AQA A-Level数学中考查最频繁的两个知识点。
In summary, Example 2.8.2 demonstrates the full procedure for solving a quadratic equation with irrational roots: identifying coefficients, substituting them carefully into the quadratic formula, simplifying the discriminant, and reducing the surd to its simplest form. Mastery of this routine will save you valuable time in the exam and prevent avoidable arithmetic errors.
总而言之，示例2.8.2完整展示了求解具有无理根的二次方程的全过程：确定系数、小心代入二次公式、化简判别式，以及将根式化为最简形式。熟练掌握这一流程将在考试中节省宝贵时间，并避免不必要的计算错误。
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