📚 Example 2.5.2: Mastering Quadratic Equations | 示例2.5.2：掌握二次方程
In the AQA A-Level Mathematics specification, quadratic equations form the bedrock of pure mathematics. Example 2.5.2 introduces a classic problem that tests your ability to solve a quadratic equation that cannot be factorised by inspection. The equation 2x² − 4x − 3 = 0 requires the use of completing the square or the quadratic formula, and it rewards students who understand the underlying structure of quadratics.
在AQA A-Level数学考试大纲中，二次方程是纯数学的基石。示例2.5.2引入了一个经典问题，考查你解一个无法通过直接观察因式分解的二次方程的能力。方程2x² − 4x − 3 = 0需要使用配方法或二次公式，并且它奖励那些理解二次方程内在结构的学生。
This example is carefully chosen because it demonstrates several important concepts at once: the use of the discriminant, the equivalence of algebraic methods, and the connection between algebra and graphs. By the end of this article, you will be able to solve such equations with confidence and avoid the common pitfalls that cost marks in the exam.
这个例子经过精心挑选，因为它同时展示了几个重要概念：判别式的使用、代数方法之间的等价性，以及代数与图形之间的联系。读完本文后，你将能够自信地解决此类方程，并避免在考试中失分的常见陷阱。
1. The Standard Form of a Quadratic Equation | 二次方程的标准形式
Every quadratic equation can be written in the standard form ax² + bx + c = 0, where a, b and c are real constants and a ≠ 0. The coefficient a determines the width and orientation of the parabola: if a > 0 the curve opens upward, and if a < 0 it opens downward.
每个二次方程都可以写成标准形式 ax² + bx + c = 0，其中a、b、c是实数常数，且a ≠ 0。系数a决定抛物线的宽窄和开口方向：若a > 0则曲线开口向上，若a < 0则开口向下。
In Example 2.5.2, we encounter the equation 2x² − 4x − 3 = 0. Here a = 2, b = −4 and c = −3. Because a ≠ 1 and the discriminant b² − 4ac = 40 is not a perfect square, this quadratic cannot be factorised using integer coefficients. We therefore turn to two algebraic techniques: completing the square and the quadratic formula.
在示例2.5.2中，我们遇到方程2x² − 4x − 3 = 0。这里a = 2，b = −4，c = −3。由于a ≠ 1且判别式b² − 4ac = 40不是完全平方数，这个二次方程无法用整数系数因式分解。因此我们转而使用两种代数技巧：配方法和二次公式。
Before choosing a method, always check whether the equation is already in standard form. If it is not—for example, if there are brackets to expand or terms to collect—simplify it first. This simple habit prevents many careless errors.
在选择方法之前，始终检查方程是否已经是标准形式。如果不是——例如有括号需要展开或有项需要合并——先化简它。这个简单的习惯可以避免许多粗心错误。
2. Method 1: Try Factorisation First | 方法一：优先尝试因式分解
Factorisation is often the quickest method when the quadratic has simple integer roots. The general strategy is to find two numbers that multiply to give ac and add to give b. For 2x² − 4x − 3 = 0, we would need two numbers with product 2 × (−3) = −6 and sum −4. The possible factor pairs of −6 are:
当二次方程具有简单的整数根时，因式分解通常是最快的方法。一般策略是找到两个数，使其乘积为ac、和为b。对于2x² − 4x − 3 = 0，我们需要两个数，其乘积为2 × (−3) = −6，和为−4。−6的可能因数对有：
- −1 and 6, whose sum is 5
- −2 and 3, whose sum is 1
- 1 and −6, whose sum is −5
- 2 and −3, whose sum is −1
−1和6，和为5；−2和3，和为1；1和−6，和为−5；2和−3，和为−1。
None of these pairs sums to −4, which confirms that this quadratic does not factorise over the integers. However, factorisation remains your first instinct in an exam because, when it works, it is the fastest method. In Section 9 you will meet a practice question that does factorise neatly.
这些数对的和都不等于−4，这证实了该二次方程在整数范围内无法因式分解。然而，在考试中因式分解仍然是你的第一直觉，因为当它可行时，它是最快的方法。在第9节中，你将遇到一个可以整齐因式分解的练习题。
3. Method 2: Completing the Square | 方法二：配方法
Completing the square is a powerful method that works for every quadratic, and it also reveals the vertex of the parabola. The key idea is to rewrite x² + bx as (x + b/2)² − (b/2)². When the coefficient of x² is not 1, we must factor it out of the first two terms first.
配方法是一种强大的方法，适用于所有二次方程，同时它还能揭示抛物线的顶点坐标。关键思想是将x² + bx改写为(x + b/2)² − (b/2)²。当x²的系数不是1时，我们必须先将它从前两项中提取出来。
Applying this to 2x² − 4x − 3 = 0, we proceed step by step:
将其应用于2x² − 4x − 3 = 0，我们逐步进行：
2x² − 4x − 3 = 0
2(x² − 2x) − 3 = 0
2[(x − 1)² − 1] − 3 = 0
2(x − 1)² − 2 − 3 = 0
2(x − 1)² − 5 = 0
Now isolate the squared term and solve:
现在分离出平方项并求解：
2(x − 1)² = 5
(x − 1)² = 5/2
x − 1 = ±√(5/2)
x = 1 ± √(5/2)
We can simplify the surd using √(5/2) = √10/2, so the exact roots are x = 1 ± √10/2. Written as two separate solutions: x = 1 + √10/2 and x = 1 − √10/2.
我们可以使用√(5/2) = √10/2来化简根式，因此精确根为x = 1 ± √10/2。写成两个单独的解：x = 1 + √10/2 和 x = 1 − √10/2。
Notice that completing the square also gives us the vertex of the parabola. From the form 2(x − 1)² − 5, the vertex is (1, −5). This is a key exam skill: the completed square form is often requested in part (a) of a question, with part (b) asking you to solve the equation or sketch the graph.
请注意，配方法还给出了抛物线的顶点坐标。从形式2(x − 1)² − 5来看，顶点为(1, −5)。这是一项关键的考试技能：完成平方的形式通常是题目第(a)问的要求，而第(b)问则要求你求解方程或绘制图形。
4. Method 3: The Quadratic Formula | 方法三：二次公式
The quadratic formula is a universal tool. For any quadratic ax² + bx + c = 0, the solutions are given by:
二次公式是一个通用工具。对于任何二次方程ax² + bx + c = 0，解由下式给出：
x = [−b ± √(b² − 4ac)] / 2a
Substituting a = 2, b = −4 and c = −3 into the formula, we obtain:
将a = 2，b = −4，c = −3代入公式，我们得到：
x = [4 ± √((−4)² − 4 × 2 × (−3))] / (2 × 2)
x = [4 ± √(16 + 24)] / 4
x = [4 ± √40] / 4
x = [4 ± 2√10] / 4
x = 1 ± √10/2
This matches exactly the answer obtained by completing the square. When the question asks for answers to three significant figures, we evaluate the surds:
这与配方法得到的答案完全一致。当题目要求答案保留到三位有效数字时，我们计算根式的值：
x = 1 + √10/2 ≈ 2.581
x = 1 − √10/2 ≈ −0.581
Always keep the exact surd form in your working, and only round the final answers. Substituting the rounded values back into the original equation is a quick check: 2(2.581)² − 4(2.581) − 3 ≈ 0 and 2(−0.581)² − 4(−0.581) − 3 ≈ 0. This verification takes seconds and catches most algebraic slips.
在计算过程中始终保留精确的根式形式，只在最终答案中四舍五入。将四舍五入后的值代回原方程是一个快速检查：2(2.581)² − 4(2.581) − 3 ≈ 0，2(−0.581)² − 4(−0.581) − 3 ≈ 0。这个检验只需几秒钟，能捕捉到大多数代数错误。
5. The Discriminant and the Nature of Roots | 判别式与根的性质
Inside the quadratic formula, the expression b² − 4ac is called the discriminant, often denoted Δ. It determines the nature of the roots without solving the equation. This is a frequently tested concept in AQA A-Level papers.
在二次公式中，表达式b² − 4ac称为判别式，常用Δ表示。它可以在不解方程的情况下判断根的性质。这是AQA A-Level考试中经常考查的概念。
| Value of Δ | Nature of Roots | Δ的值 | 根的性质 |
| Δ > 0 and a perfect square | Two distinct rational roots | Δ > 0且为完全平方数 | 两个不同的有理数根 |
| Δ > 0 but not a perfect square | Two distinct irrational roots | Δ > 0但不是完全平方数 | 两个不同的无理数根 |
| Δ = 0 | One repeated real root | Δ = 0 | 一个重根（实根） |
| Δ < 0 | No real roots | Δ < 0 | 没有实数根 |
For Example 2.5.2, Δ = (−4)² − 4 × 2 × (−3) = 16 + 24 = 40. Since 40 > 0 but is not a perfect square, the equation has two distinct irrational roots. This agrees with our earlier solutions involving √10.
对于示例2.5.2，Δ = (−4)² − 4 × 2 × (−3) = 16 + 24 = 40。由于40 > 0但不是完全平方数，该方程有两个不同的无理数根。这与我们之前包含√10的解一致。
Exam questions often ask you to ‘find the range of values of k for which the equation has real roots’. In such problems, set the discriminant ≥ 0 and solve the resulting inequality. This is a direct application of the discriminant and appears frequently in AS papers.
考试题目经常要求你“求k的取值范围，使方程有实数根”。在这种问题中，令判别式 ≥ 0并解所得的不等式。这是判别式的直接应用，在AS试卷中经常出现。
6. Graphical Interpretation of the Solution | 解的图形解释
The quadratic equation 2x² − 4x − 3 = 0 corresponds to the parabola y = 2x² − 4x − 3. The roots we found are precisely the x-coordinates where this curve crosses the x-axis. Using the completed square form y = 2(x − 1)² − 5, we can extract the full picture of the graph.
二次方程2x² − 4x − 3 = 0对应于抛物线y = 2x² − 4x − 3。我们找到的根正是这条曲线与x轴交点的x坐标。利用完成平方的形式y = 2(x − 1)² − 5，我们可以得到图形的完整信息。
- The vertex is at (1, −5), the minimum point of the parabola.
- The line of symmetry is x = 1.Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
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