📚 Example 2.6.2 Solving a Quadratic Equation Using the Quadratic Formula | 例 2.6.2：使用求根公式解二次方程
formula, and we leave the answers in exact surd form. Exact surd form, rather than rounded decimals, is the standard requirement for non-calculator questions and for any question that asks for an exact answer.
在这个 AQA A-Level 数学例题中，我们使用求根公式解二次方程 x² − 4x + 1 = 0，并把答案保留为精确根式形式。精确根式（而不是四舍五入的小数）是非计算器题目以及任何要求"精确答案"的题目的标准要求。
The example is deliberately small, but it contains every skill the examiner is testing: recognising that factorising fails, quoting the formula correctly, substituting with the right signs, simplifying a surd, and stating both roots. Each of these steps is examined separately in mark schemes, so we will treat them one at a time.
这道例题虽然简单，但它包含了考官考查的全部技能：判断因式分解行不通、正确写出公式、带正确符号代入、化简根式，以及写出两个根。每一步在评分标准中都是独立给分的，因此我们逐一来分析。
1. The Worked Example | 例题
g the answers in exact surd form. This means we cannot simply use a calculator to obtain rounded decimal answers and write those down; the answer must contain a square root symbol, and that square root must be in its simplest form.
本例题要求解方程 x² − 4x + 1 = 0，并用精确根式给出答案。这意味着我们不能只用计算器得到四舍五入的小数答案然后抄下来；答案中必须含有根号，而且这个根号必须是最简形式。
This type of question appears regularly in AQA AS and A-Level papers, especially where exact methods are tested rather than numerical approximation. The same equation can be solved by completing the square, and we will use that later as a check.
这类题目在 AQA AS 和 A-Level 试卷中经常出现，尤其是在考查精确方法而不是数值近似的部分。同一个方程也可以用配方法求解，我们稍后会用它来做检验。
A full solution should look like a short chain of reasoning rather than a single line of algebra. Writing the formula, the substitution and the simplification as separate steps makes it much easier for the examiner to award method marks, and much easier for you to find your own mistakes.
完整的解答应当是一小段推理链，而不是一行代数。把公式、代入和化简写成独立的步骤，既方便考官给出方法分，也方便你自己发现错误。
2. Why We Use the Quadratic Formula | 为什么使用求根公式
d up to 2, not −4, so there is no integer factorisation of x² − 4x + 1 into two brackets with integer terms.
处理任何二次方程的第一步都是检查它能否因式分解。这里常数项是 1，而 1 的唯一因子对是 1 × 1，它们相加得 2，而不是 −4，因此 x² − 4x + 1 无法分解为两个整数项括号的乘积。
The quadratic formula always works for any quadratic equation, so it is the safest method here. It does not depend on spotting a factor pair, and it handles irrational and complex roots equally well.
求根公式对任意二次方程都适用，因此这里是更稳妥的方法。它不依赖你是否能看出因子对，并且同样能处理无理根和复根。
Completing the square would also work, and we will check it later as an alternative. Seeing two methods agree is one of the best ways to be confident in an exact answer, because a surd cannot be "roughly right".
配方法同样可行，我们稍后会作为替代方法进行检验。看到两种方法得到一致结果，是确认精确答案最可靠的方式之一，因为根式答案不存在"差不多对"。
3. The Quadratic Formula | 求根公式
ratic equation of the form ax² + bx + c = 0, the roots are given by the quadratic formula:
对于任意形如 ax² + bx + c = 0 的二次方程，其根由求根公式给出：
x = (−b ± √b² − 4ac) / 2a
tions: one with a plus sign and one with a minus sign. This is why a quadratic can have two distinct roots, and why an answer that gives only one of them is incomplete.
符号 ± 通常给出两个解：一个取加号，一个取减号。这正是二次方程可以有两个不同根的原因，也是只写出其中一个根会被视为答案不完整的原因。
This formula is provided in the AQA formula booklet, so you are not expected to memorise it. However, you are expected to know exactly what a, b and c mean, and to substitute into it without sign errors — that is where almost all the lost marks come from.
该公式在 AQA 公式册中提供，因此不要求你背诵。但你必须清楚 a、b、c 分别代表什么，并能不出符号错误地代入——几乎所有的失分都出在这里。
Notice also that the whole numerator is divided by 2a, not just the square root term. Writing the fraction clearly, with a long fraction bar or with brackets, prevents one of the most common presentation errors in the exam.
还要注意，是整个分子除以 2a，而不是只有根号那一项除以 2a。把分数写清楚（长分数线或加括号），可以避免考试中最常见的书写错误之一。
4. Identifying a, b and c | 确定 a、b、c
² − 4x + 1 = 0 is already in standard form, so we can identify the three numbers directly.
代入之前，先把方程写成标准形式 ax² + bx + c = 0，然后读出各系数。我们的方程 x² − 4x + 1 = 0 已经是标准形式，因此可以直接确定这三个数。
- a = 1 (the coefficient of x², written as 1 when no number appears) | a = 1（x² 的系数，没有数字时就是 1）
- b = −4 (the coefficient of x, including its minus sign) | b = −4（x 的系数，必须带上负号）
- c = 1 (the constant term, including its sign) | c = 1（常数项，必须带上符号）
b = 4 instead of b = −4, then −b becomes −4 instead of +4 and the whole answer collapses.
这里最重要的习惯是把符号和数字一起写下来。如果你把 b 写成 4 而不是 −4，那么 −b 就变成 −4 而不是 +4，整个答案就错了。
g off a, b and c. Never substitute from a rearranged-in-your-head version of the equation.
如果题目给的是 x² + 1 = 4x，我们必须先整理成 x² − 4x + 1 = 0 再读出 a、b、c。绝不要凭脑子里"整理过"的方程去代入。
5. Substituting into the Formula | 代入公式
to the formula. The most reliable order is: compute the discriminant b² − 4ac first, then deal with −b, then divide by 2a.
现在把 a = 1、b = −4、c = 1 代入公式。最可靠的顺序是：先算判别式 b² − 4ac，再处理 −b，最后除以 2a。
For the discriminant we have b² = (−4)² = 16 and 4ac = 4 × 1 × 1 = 4, so:
对于判别式，b² = (−4)² = 16，4ac = 4 × 1 × 1 = 4，因此：
b² − 4ac = 16 − 4 = 12
Since −b = −(−4) = +4 and 2a = 2, the formula gives:
由于 −b = −(−4) = +4，2a = 2，公式给出：
x = (4 ± √12) / 2
At this stage the answer is correct but not yet in the required form. The surd √12 is not simplified, and the fraction has not been reduced. Both of these are needed for full marks.
此时答案虽然正确，但还没有达到要求的 form。根式 √12 尚未化简，分数也尚未约分，而这两点都是拿满分所必需的。
6. Simplifying the Surd | 化简根式
To simplify √12 we look for the largest square number that divides 12. Since 12 = 4 × 3, we can write √12 = √4 × √3 = 2√3. The number under the root is now square-free, so 2√3 is in its simplest surd form.
化简 √12 时，要找出能整除 12 的最大平方数。由于 12 = 4 × 3，可以写成 √12 = √4 × √3 = 2√3。此时根号下的数已不含平方因子，因此 2√3 就是最简根式。
Substituting back:
代回原式：
x = (4 ± 2√3) / 2
Both terms in the numerator share a factor of 2, and the denominator is 2, so the whole fraction simplifies:
分子中的两项都含有因子 2，分母也是 2，因此整个分数可以约分：
x = 2 ± √3
A common error is to cancel only part of the numerator, for example writing 4 ± 2√3 divided by 2 as "2 ± 2√3" or "4 ± √3". Every term on the top must be divided by the bottom; using brackets around the numerator prevents this mistake.
常见错误是只约掉分子的一部分，例如把 (4 ± 2√3) / 2 写成 "2 ± 2√3" 或 "4 ± √3"。分子上的每一项都必须除以分母；在分子上加括号可以避免这种错误。
7. The Two Exact Roots | 两个精确根
Reading the ± sign in both directions gives the two exact roots of the equation:
把 ± 号分别取正负两个方向，就得到方程的两个精确根：
x = 2 + √3 or x = 2 − √3
These are exact values. For a sense of scale, √3 approximately 3.732 and 0.268. Both are real and distinct, and both are irrational, since √3 is irrational.
这两个都是精确值。为了有个数量级上的感觉，√3 ≈ 1.732，所以两根约为 3.732 和 0.268。它们都是实数且互不相同，并且由于 √3 是无理数，两根都是无理数。
Note that the decimal values are useful for checking, but they are not the answer. If a question asks for exact surd form and you write "3.73", you have not answered the question, even if the arithmetic is perfect.
注意，小数可以用来检验，但它不是答案。如果题目要求精确根式而你写了 "3.73"，那么即使运算完全正确，你也没有回答题目所问。
8. Checking Your Answer | 检验答案
e calculation. The first uses the sum of the roots, which must equal −b/a = 4. Indeed (2 + √3) + (2 − √3) = 4, so the surd terms cancel as expected.
有两个快捷的精确检验方法，都不需要重做整个计算。第一种用两根之和，它应当等于 −b/a = 4。确实，(2 + √3) + (2 − √3) = 4，根式项按预期抵消。
The second uses the product of the roots, which must equal c/a = 1. Here (2 + √3)(2 − √3) = 4 − 3 = 1, which is exactly right and confirms both roots at once.
第二种用两根之积，它应当等于 c/a = 1。这里 (2 + √3)(2 − √3) = 4 − 3 = 1，完全正确，一次性确认了两个根。
You can also substitute x = 2 + √3 directly into the original equation. Since (2 + √3)² = 7 + 4√3, we get 7 + 4√3 − 8 − 4√3 + 1 = 0, which verifies the root explicitly.
你也可以把 x = 2 + √3 直接代入原方程。由于 (2 + √3)² = 7 + 4√3，得到 7 + 4√3 − 8 − 4√3 + 1 = 0，明确验证了这个根。
Checking is not optional in an exact-answer question. A single sign slip in the substitution produces an answer that looks perfectly plausible but is wrong, and only a check will reveal it.
在要求精确答案的题目中，检验不是可选项。代入时一个符号失误就会产生一个看起来完全合理、实际却是错误的答案，而只有检验才能把它揪出来。
9. Alternative Method: Completing the Square | 替代方法：配方法
1 = 0, we take half of the coefficient of x, which is −2, and write the square of (x − 2).
同一个方程也可以用配方法求解，这是一个有用的独立验证。从 x² − 4x + 1 = 0 出发，取 x 系数的一半，即 −2，写出 (x − 2) 的平方。
Since (x − 2)² = x² − 4x + 4, the equation becomes:
由于 (x − 2)² = x² − 4x + 4，方程变为：
(x − 2)² − 4 + 1 = 0 → (x − 2)² = 3
Taking square roots of both sides gives x − 2 = ±√3, so x = 2 ± √3. This matches the quadratic formula result exactly, which is a strong confirmation that the surd work was correct.
两边开平方得 x − 2 = ±√3，因此 x = 2 ± √3。这与求根公式的结果完全一致，有力证明了根式化简是正确的。
nowing both methods means you can pick the quicker one and still use the other to check, which is exactly the flexibility a high-scoring candidate needs.
当 a = 1 且 b 为偶数时（正如本题），配方法往往更快。掌握两种方法意味着你可以选更快的那个，再用另一个来检验，这正是高分考生所需要的灵活性。
10. The Discriminant and the Nature of the Roots | 判别式与根的性质
hich is positive, so the equation has two distinct real roots. This matches what we found.
表达式 b² − 4ac 称为判别式，通常记作 Δ。本题中 Δ = 12，是正数，所以方程有两个不同的实根。这与我们的结果一致。
Because 12 is not a perfect square, √12 is irrational, so the two roots are irrational surds rather than integers or fractions. If the discriminant had been a perfect square such as 16 or 25, the roots would have been rational, and the equation would have factorised after all.
由于 12 不是完全平方数，√12 是无理数，所以两个根是无理根式，而不是整数或分数。如果判别式是完全平方数（如 16 或 25），根将是有理数，方程其实就可以因式分解。
| Δ > 0 | Two distinct real roots; rational if Δ is a perfect square, irrational otherwise | 两个不同实根；Δ 为完全平方数时有理，否则无理 |
| Δ = 0 | One repeated real root, x = −b / 2a | 一个重根，x = −b / 2a |
| Δ < 0 | No real roots; the graph does not cross the x-axis | 无实根；图像不与 x 轴相交 |
Working out the discriminant before you finish the question is a good habit: it tells you what kind of answer to expect, so an unexpected integer or a negative under the root immediately signals an error.
在完成题目之前先算出判别式是个好习惯：它告诉你应该期待什么样的答案，因此一旦出现意料之外的整数或根号下的负数，就能立刻察觉有误。
11. Common Exam Mistakes | 常见考试错误
Most marks lost on this type of question come from a small set of repeated errors. Going through them once is worth more than solving ten more equations of the same kind.
这类题目丢的分大多来自少数几个反复出现的错误。把它们过一遍，比再解十道同类方程更有价值。
- Using b = 4 instead of b = −4, so that −b becomes −4 and the roots come out as negative. | 把 b 写成 4 而不是 −4，导致 −b 变成 −4，两根变成负数。
- Dividing only part of the numerator by 2a, for example writing (4 ± √12) / 2 as 4 ± √6. | 只把分子的一部分除以 2a，例如把 (4 ± √12) / 2 写成 4 ± √6。
- Leaving the answer as (4 ± √12) / 2 without simplifying to 2 ± √3. | 把答案留在 (4 ± √12) / 2 而不化简为 2 ± √3。
- Giving only one root and forgetting the minus branch of the ± sign. | 只写一个根，忘记 ± 号的减号分支。
- Quoting the formula with 2a in the wrong place, such as x = −b ± √b² − 4ac / 2a, which is ambiguous. | 公式写错位置，例如写成 x = −b ± √b² − 4ac / 2a，含义不清。
- Writing a rounded decimal when the question demands exact surd form. | 题目要求精确根式却写了四舍五入的小数。
Almost every one of these mistakes is prevented by two habits: bracketing the numerator and the denominator clearly, and simplifying the surd before doing anything else. Both take seconds and save whole marks.
以上几乎每一个错误都可以靠两个习惯避免：把分子和分母都用括号写清楚，以及在做其他事情之前先化简根式。两者都只花几秒，却能保住整道题的分数。
12. Exam Technique and Summary | 考试技巧与总结
A reliable routine for any "solve the quadratic, giving exact answers" question is: rearrange to standard form, try factorising, and if it fails, identify a, b and c with their signs, write the formula, compute the discriminant, substitute, simplify the surd, simplify the fraction, and state both roots.
对于任何"解二次方程并给出精确答案"的题目，可靠的流程是：整理成标准形式，尝试因式分解；若失败，带符号写出 a、b、c，写出公式，计算判别式，代入，化简根式，约分，最后写出两个根。
For our equation x² − 4x + 1 = 0 the complete solution is:
对于方程 x² − 4x + 1 = 0，完整解答为：
x = (4 ± √12) / 2 = (4 ± 2√3) / 2 = 2 ± √3
So x = 2 + √3 or x = 2 − √3 inct irrational roots. The same answer is obtained by completing the square, which makes the result very safe.
因此 x = 2 + √3 或 x = 2 − √3，而 Δ = 12 > 0 确认有两个不同的无理根。配方法得到同样的答案，使这个结果非常可靠。
The wider lesson is that the quadratic formula is a guaranteed method, but it rewards careful presentation. Write the formula, substitute in brackets, simplify the surd fully, and check with the sum and product of the roots. Do this consistently and this example becomes one of the most dependable sources of marks in the whole A-Level Mathematics paper.
更广泛的启示是：求根公式是一个保证可行的方法，但它奖励书写严谨的人。写出公式、带括号代入、彻底化简根式，并用两根之和与两根之积检验。坚持这样做，这道例题就会成为整个 A-Level 数学试卷中最稳定的得分点之一。
Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图