Example 3.2.1: The Discriminant and Real Roots | 示例 3.2.1：判别式与实根

📚 Example 3.2.1: The Discriminant and Real Roots | 示例 3.2.1：判别式与实根

In this worked example, we will explore a classic AQA A-Level Mathematics question that uses the discriminant to determine the values of a parameter for which a quadratic equation has real roots. This example builds your pure mathematics skills in algebra, inequalities, and logical reasoning.

在这个工作示例中，我们将探讨一道经典的 AQA A-Level 数学题目，它利用判别式来确定二次方程何时具有实根的参数取值范围。这个示例将提升你在代数、不等式以及逻辑推理方面的纯数学技能。

1. Why This Example Matters | 这个示例为何重要

The discriminant is a powerful tool in A-Level Pure Mathematics. It not only tells you the nature of the roots of a quadratic equation but also appears in coordinate geometry, calculus, and even mechanics problems. Mastering Example 3.2.1 gives you a template for solving many similar exam questions.

判别式是 A-Level 纯数学中一个强大的工具。它不仅告诉你二次方程根的性质，还出现在坐标几何、微积分甚至力学问题中。掌握示例 3.2.1 会为你解决许多类似考试问题提供一个模板。

2. Recap: The Discriminant | 回顾：判别式

For a quadratic equation of the form \(ax^2 + bx + c = 0\), the discriminant is defined as \(D = b^2 – 4ac\). The value of \(D\) determines the nature of the roots:

对于形式为 \(ax^2 + bx + c = 0\) 的二次方程，判别式定义为 \(D = b^2 – 4ac\)。根据 \(D\) 的值可以判断根的性质：

Discriminant value	Nature of roots
\(D > 0\)	Two distinct real roots
\(D = 0\)	One repeated real root
\(D < 0\)	No real roots

For this example, we will use the condition \(D \geq 0\) to require that the equation has real roots (either distinct or repeated).

在本示例中，我们使用条件 \(D \geq 0\) 来要求方程具有实根（可以是两个不同实根或一个重根）。

3. The Problem Statement | 问题陈述

Find the range of values of \(k\) for which the equation

求参数 \(k\) 的取值范围，使得方程

x² + kx + (k + 3) = 0

has real roots.

具有实根。

4. Step 1: Write Down the Coefficients | 第一步：写出系数

Compare the given equation with the standard form \(ax^2 + bx + c = 0\). Here we identify:

将给定方程与标准形式 \(ax^2 + bx + c = 0\) 进行比较，我们可以确定：

\(a = 1\)
\(b = k\)
\(c = k + 3\)

Be careful with the sign of the constant term. In \(x^2 + kx + (k+3)\), all signs are positive.

请注意常数项的符号。在 \(x^2 + kx + (k+3)\) 中，所有符号均为正。

5. Step 2: Form the Discriminant | 第二步：构造判别式

Substitute \(a\), \(b\), and \(c\) into the discriminant formula \(D = b^2 – 4ac\):

将 \(a\)、\(b\)、\(c\) 代入判别式公式 \(D = b^2 – 4ac\)：

D = k² − 4 × 1 × (k + 3)

Simplify the expression step by step:

逐步化简表达式：

D = k² − 4k − 12

6. Step 3: Set Up the Inequality | 第三步：建立不等式

Since the equation must have real roots, we need the discriminant to be non-negative:

由于方程必须具有实根，我们需要判别式非负：

D ≥ 0 ⟹ k² − 4k − 12 ≥ 0

This is a quadratic inequality. Before solving, factorise the quadratic expression if possible.

这是一个二次不等式。在求解之前，尽可能对二次表达式进行因式分解。

7. Step 4: Solve the Inequality | 第四步：解不等式

Factorise \(k^2 – 4k – 12\):

对 \(k^2 – 4k – 12\) 进行因式分解：

(k − 6)(k + 2) ≥ 0

Identify the critical values where the expression equals zero: \(k = 6\) and \(k = -2\).

找到使表达式等于零的关键值：\(k = 6\) 和 \(k = -2\)。

To solve the inequality, sketch a graph of \(y = (k-6)(k+2)\) or use a sign table. The parabola opens upwards, so the expression is positive outside the interval between the roots and negative inside.

要解这个不等式，可以画出 \(y = (k-6)(k+2)\) 的图像或用符号表。抛物线开口向上，因此表达式在两根之外为正，在两根之间为负。

The solution is therefore:

因此解为：

k ≤ −2 or k ≥ 6

8. Step 5: Interpret the Result | 第五步：解释结果

The range of values of \(k\) for which the equation \(x^2 + kx + (k+3)=0\) has real roots is:

使得方程 \(x^2 + kx + (k+3)=0\) 具有实根的 \(k\) 的取值范围为：

\(k \in (-\infty, -2] \cup [6, \infty)\)

At the endpoints \(k = -2\) and \(k = 6\), the discriminant equals zero, meaning the equation has a repeated root.

在端点 \(k = -2\) 和 \(k = 6\) 处，判别式等于零，这意味着方程有一个重根。

9. Graphical Meaning | 图形含义

Imagine plotting the quadratic curve \(y = x^2 + kx + (k+3)\) for different values of \(k\). When \(k ≤ -2\) or \(k ≥ 6\), the curve crosses or touches the x-axis (real roots). When \(-2 < k < 6\), the curve lies entirely above or below the x-axis without touching it (no real roots).

想象一下，对于不同的 \(k\) 值绘制二次曲线 \(y = x^2 + kx + (k+3)\)。当 \(k ≤ -2\) 或 \(k ≥ 6\) 时，曲线与 x 轴相交或相切（实根）。当 \(-2 < k < 6\) 时，曲线完全位于 x 轴上方或下方且不相交（无实根）。

This visual interpretation helps you quickly verify your algebra and understand why the inequality gives two separate regions.

这种图形化的理解有助于你快速验证代数结果，并理解为什么不等式给出两个分离的区域。

10. Alternative Method: Completing the Square | 替代方法：配方法

We could also solve the problem by completing the square on the discriminant:

我们还可以通过对判别式配方来求解这个问题：

k² − 4k − 12 = (k − 2)² − 16

Then the inequality becomes \((k-2)^2 \ge 16\). Taking square roots gives \(|k-2| \ge 4\), which again leads to \(k \le -2\) or \(k \ge 6\).

于是不等式变为 \((k-2)^2 \ge 16\)。两边开平方得到 \(|k-2| \ge 4\)，这同样导出 \(k \le -2\) 或 \(k \ge 6\)。

This method is especially useful when the quadratic does not factorise nicely.

当二次式不易因式分解时，这种方法特别有用。

11. Common Mistakes to Avoid | 避免常见错误

Using the condition \(D > 0\) instead of \(D \ge 0\) when the question says “real roots”. Remember “real roots” includes a repeated root.
Forgetting to include the coefficient of \(x^2\) when writing \(a\). In this example \(a = 1\), but in others it may not be 1.
Misinterpreting the sign table: for a positive leading coefficient, the quadratic is positive outside the roots and negative between them.
Dropping the endpoints when writing interval notation. Since \(D = 0\) still gives a real root, the endpoints must be included.

不要把条件写成 \(D > 0\)，当题目要求“实根”时应使用 \(D \ge 0\)。请记住“实根”包含重根的情况。

在写 \(a\) 时不要忘记 \(x^2\) 的系数。这个例子中 \(a = 1\)，但在其他问题中可能不是 1。

不要误解符号表：当二次项系数为正时，二次式在两根之外为正，在两根之间为负。

在写区间时不要丢端点。由于 \(D = 0\) 仍能给出实根，端点必须包含在内。

12. Exam Tips from TutorHao | TutorHao 的考试提示

In AQA exams, questions on the discriminant often appear as part of a longer problem, for example, involving tangents to curves or intersections of lines and parabolas. Always start by writing down the standard form and the coefficients. Show your discriminant clearly before setting up the inequality. This earns method marks even if you make a small arithmetic slip later.

在 AQA 考试中，关于判别式的问题常常作为较长时间题目的一部分出现，例如与曲线的切线或直线与抛物线的交点相关。务必先写出标准形式和系数。清晰展示判别式，再建立不等式。这样即使之后有小的计算失误，也能获得方法分。

Practice sketching the parabola defined by the discriminant; it will help you solve inequalities quickly and accurately. Remember to use set notation for the final answer, and always check whether endpoints should be included.

练习画出以判别式定义的抛物线图像；这会帮助你快速而准确地解不等式。记住最终答案使用集合符号，并且始终检查端点是否应包含在内。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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Example 3.2.1: The Discriminant and Real Roots | 示例 3.2.1：判别式与实根