AQA A-Level Maths Paper 1 Question 2: Quadratic Discriminant and Inequalities | AQA A-Level 数学卷一第2题：二次判别式与不等式

📚 AQA A-Level Maths Paper 1 Question 2: Quadratic Discriminant and Inequalities | AQA A-Level 数学卷一第2题：二次判别式与不等式

In AQA A-Level Mathematics Paper 1, Question 2 often focuses on core algebra skills such as quadratic equations, the discriminant, and inequalities. This article breaks down a typical Question 2 topic: using the discriminant to determine the nature of roots and solve inequalities involving an unknown coefficient k.

在 AQA A-Level 数学卷一考试中，第 2 题通常考查核心代数技能，例如二次方程、判别式以及不等式。本文详细解析一个典型的第 2 题主题：利用判别式判断二次方程根的性质，并求解含未知系数 k 的不等式。

1. What Question 2 Usually Tests | 第2题通常考查什么

Question 2 in AQA A-Level Maths Paper 1 is designed to be accessible but still requires precise algebraic manipulation. It frequently tests the discriminant of a quadratic, the condition for real or equal roots, and the translation of these conditions into inequalities.

AQA A-Level 数学卷一第 2 题的设置通常较为基础，但仍要求考生具备严谨的代数运算能力。该题经常考查二次方程的判别式、有实根或重根的条件，以及如何将这些条件转化为不等式进行求解。

epends on k. The question then asks you to find the range of k for which the equation has two distinct real roots, one repeated root, or no real roots.

题目通常会给出一个形如 ax² + bx + c = 0 的二次方程，其中一个系数依赖于参数 k。然后要求你求出使该方程具有两个不同实根、一个重根或没有实根的 k 的取值范围。

Because Question 2 appears early in the paper, it should be answered efficiently but with great care to avoid sign errors and inequality mistakes.

由于第 2 题出现在试卷较前的位置，考生应当高效作答，但同时要格外小心，避免符号错误和不等式方向写反的问题。

2. Key Formula: The Discriminant | 核心公式：判别式

For any quadratic equation ax² + bx + c = 0, the discriminant Δ is given by the formula:

对于任意二次方程 ax² + bx + c = 0，判别式 Δ 由以下公式给出：

Δ = b² - 4ac

The value of Δ tells us the nature of the roots without solving the equation fully.

判别式 Δ 的值可以在不完全解方程的情况下告诉我们根的性质。

• If Δ > 0, the equation has two distinct real roots.
• If Δ = 0, the equation has one repeated real root, also called a double root.
• If Δ < 0, the equation has no real roots, only two complex conjugate roots.

• 若 Δ > 0，方程有两个不同的实数根。
• 若 Δ = 0，方程有一个重根，也称为二重根。
• 若 Δ < 0，方程没有实数根，只有两个共轭复数根。

These conditions are essential for Question 2, as they allow you to convert a statement about the number of roots into an algebraic inequality involving k.

这些条件是解答第 2 题的关键，因为它们能将关于根个数的陈述转化为含 k 的代数不等式。

3. Worked Example: Two Distinct Real Roots | 例题：两个不同实根

Consider the following typical Question 2 task:

请思考下面这道典型的第 2 题：

, find the set of values of k for which the equation f(x) = 0 has two distinct real roots.

已知 f(x) = x² + (k - 2)x + 4，求使得方程 f(x) = 0 有两个不同实数根的 k 的取值范围。

Here the quadratic is already in standard form with a = 1, b = k - 2, and c = 4.

这里的二次方程已经是标准形式，其中 a = 1，b = k - 2，c = 4。

To have two distinct real roots, the discriminant must be strictly greater than zero.

要有两个不同的实数根，判别式必须严格大于零。

4. Step-by-Step Solution | 分步解答

Step 1: Write down the discriminant expression using the given coefficients.

第 1 步：利用给定系数写出判别式表达式。

Δ = b² - 4ac = (k - 2)² - 4 × 1 × 4

This simplifies to:

化简后得到：

Δ = (k - 2)² - 16

Step 2: Apply the condition for two distinct real roots, which is Δ > 0.

第 2 步：应用两个不同实根的条件，即 Δ > 0。

(k - 2)² - 16 > 0

Step 3: Rearrange the inequality to isolate the squared term.

第 3 步：整理不等式，将平方项单独放在一边。

(k - 2)² > 16

Step 4: Take the square root of both sides. Remember that for an inequality of the form x² > a², the solution is x > a or x < -a.

第 4 步：对不等式两边开平方。请记住，对于形如 x² > a² 的不等式，解为 x > a 或 x < -a。

|k - 2| > 4

k - 2 > 4 or k - 2 < -4

Step 5: Solve each linear inequality separately.

第 5 步：分别解每一个一元一次不等式。

k > 6 or k < -2

Therefore the set of values of k is:

因此 k 的取值范围为：

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{k : k < -2} ∪ {k : k > 6}

This means any value of k smaller than -2 or greater than 6 will give the quadratic two distinct real roots.

这意味着任何小于 -2 或大于 6 的 k 值都会使该二次方程有两个不同的实数根。

5. Interpreting Inequalities Involving k | 解读含 k 的不等式

abola y = (k - 2)² - 16. The quadratic is positive outside the interval between its roots.

当你求解形如 (k - 2)² > 16 的不等式时，借助抛物线 y = (k - 2)² - 16 的图像会很有帮助。该二次函数在其两根之间的区间外取正值。

2 and k = 6. Testing intervals shows that the inequality holds when k < -2 or k > 6.

临界值为 k = -2 和 k = 6。通过区间测试可以发现，当 k < -2 或 k > 6 时不等式成立。

tinct real roots, the condition is strict, so the critical values themselves are excluded.

务必检查不等式是严格的（> 或 <）还是非严格的（≥ 或 ≤）。对于两个不同实根的情况，条件是严格的，因此临界值本身不包含在解集内。

6. Common Mistakes to Avoid | 常见错误与避免方法

xpanding (k - 2)² incorrectly. Remember that (k - 2)² = k² - 4k + 4, not k² + 4 or k² - 4.

一个常见错误是错误地展开 (k - 2)²。请记住 (k - 2)² = k² - 4k + 4，而不是 k² + 4 或 k² - 4。

Another common mistake is writing the solution to x² > 16 as -4 < x < 4. The correct solution for x² > 16 is x > 4 or x < -4. Many candidates forget the word 'or' and produce an incorrect interval.

另一个常见错误是将 x² > 16 的解写成 -4 < x < 4。实际上 x² > 16 的正确解为 x > 4 或 x < -4。许多考生会漏掉“或”字，从而得出错误的区间。

is the entire coefficient of x. In our example, b = k - 2, so b² is (k - 2)², not k² - 2².

此外，在代入判别式时，要确保 b 是 x 的完整系数。在例题中，b = k - 2，所以 b² 是 (k - 2)²，而不是 k² - 2²。

y. Here 4ac = 4 × 1 × 4 = 16, and a sign error in this step will change the entire answer.

最后，不要忘记正确计算 4ac。这里 4ac = 4 × 1 × 4 = 16，这一步的符号错误会改变整个答案。

7. Exam Technique for Question 2 | 第2题的考试技巧

Since Question 2 is an early question, aim to complete it in about five minutes. Write down the discriminant formula first, then substitute carefully.

由于第 2 题是试卷较前的题目，目标是在大约五分钟内完成。先写出判别式公式，然后仔细代入数值。

Always state the condition you are using, for example 'For two distinct real roots, Δ > 0'. This shows the examiner your reasoning and can earn method marks even if a later calculation slips.

一定要写出你所使用的条件，例如“对于两个不同的实数根，Δ > 0”。这样可以让阅卷老师看到你的推理过程，即使后面的计算出现小错，也能获得方法分。

Leave your final answer in set notation or as an inequality, depending on what the question asks. In our example, the answer is k < -2 or k > 6.

最后答案要根据题目要求写成集合形式或不等式形式。在例题中，答案为 k < -2 或 k > 6。

8. Practice Question with Mark Scheme | 练习题与评分标准

Try this similar question to consolidate your understanding.

请尝试下面这道类似的题目，以巩固理解。

8, find the set of values of k for which the equation g(x) = 0 has one repeated real root.

已知 g(x) = 2x² + kx + 8，求使得方程 g(x) = 0 有一个重根的 k 的取值范围。

Solution: Here a = 2, b = k, c = 8. The discriminant is Δ = k² - 4 × 2 × 8 = k² - 64.

解答：这里 a = 2，b = k，c = 8。判别式为 Δ = k² - 4 × 2 × 8 = k² - 64。

For one repeated root, the discriminant must equal zero.

对于重根，判别式必须等于零。

k² - 64 = 0

k² = 64

k = ±8

Thus the set of values is k = 8 or k = -8.

因此取值范围为 k = 8 或 k = -8。

write the final answer as k = ±8, showing both critical values from the squared equation.

在考试中，你应将最终答案写成 k = ±8，以展示由平方方程得到的两个临界值。

9. Summary and Key Takeaways | 总结与关键要点

The discriminant is a powerful tool for linking quadratic coefficients to the nature of roots. For Question 2, you must be able to form Δ, set it against zero, and solve the resulting inequality or equation correctly.

判别式是将二次项系数与根的性质联系起来的强大工具。对于第 2 题，你必须能够构造判别式 Δ，将其与零比较，并正确求解所得的不等式或方程。

Condition	Roots
Δ > 0	Two distinct real roots
Δ = 0	One repeated real root
Δ < 0	No real roots

Remember the critical step: for x² > a², the solution is x > a or x < -a, not a single interval. Mastering this will help you avoid common errors and secure full marks on Question 2.

请记住关键步骤：对于 x² > a²，解为 x > a 或 x < -a，而不是一个单独的区间。掌握这一点有助于避免常见错误，并在第 2 题上拿到满分。

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