📚 Example 2.4.1: Solving Quadratics with the Quadratic Formula | 例题 2.4.1：用求根公式解二次方程
ines several key skills: identifying coefficients, using the quadratic formula correctly, simplifying surds, and interpreting the discriminant.
在本节 AQA A-Level 数学课程中，我们详细讲解例题 2.4.1：解二次方程 2x² − 5x + 1 = 0。该例题综合了多个核心技能，包括确定系数、正确代入求根公式、化简根式以及解读判别式。
1. What the Example Asks | 例题要求
t be factorised easily because the discriminant is not a perfect square. We are required to give exact solutions, not decimal approximations, unless stated otherwise.
方程为 2x² − 5x + 1 = 0。由于判别式不是完全平方数，该方程不易因式分解。题目要求给出准确解，除非另有说明，否则不写近似小数。
exact surd form, correct substitution, and clear simplification. Leaving an answer as (5 ± √17) ÷ 4 is preferred over 1.78 or 0.72.
在 AQA 考试中，准确根式形式、正确代入以及清晰化简通常都有给分。将答案写成 (5 ± √17) ÷ 4 比写成 1.78 或 0.72 更符合得分要求。
2. AQA Specification Links | AQA 考纲衔接
This topic appears in AQA AS and A-Level Pure Mathematics under algebra and functions. You need to know how to solve quadratics by factorising, completing the square, and the quadratic formula.
本主题属于 AQA AS 与 A-Level 纯数学中的代数与函数部分。你需要掌握因式分解、配方法和求根公式三种解二次方程的方法。
The quadratic formula is essential when a quadratic does not factorise easily. It also leads naturally to discriminant questions and to roots of more general polynomial equations in Year 2.
当二次方程不易因式分解时，求根公式是基本工具。它也自然引向判别式问题，并衔接第二年课程中更一般多项式方程的根。
3. The Quadratic Formula | 求根公式
any quadratic equation ax² + bx + c = 0, where a ≠ 0, the solutions are given by x = (−b ± √b² − 4ac) ÷ 2a.
对于任意二次方程 ax² + bx + c = 0，其中 a ≠ 0，其解为 x = (−b ± √b² − 4ac) ÷ 2a。
x = (−b ± √b² − 4ac) ÷ 2a
uare root part determines how far the two roots sit from the axis of symmetry x = −b ÷ 2a.
由于带有 ± 号，该公式一次给出两个根。平方根部分决定两个根距离对称轴 x = −b ÷ 2a 有多远。
4. Discriminant and Nature of Roots | 判别式与根的性质
The expression b² − 4ac is called the discriminant. If it is positive, there are two distinct real roots; if zero, one repeated root; if negative, no real roots.
表达式 b² − 4ac 称为判别式。若判别式大于零，则有两个不同的实根；若等于零，则有一个重根；若小于零，则没有实根。
| b² − 4ac > 0 | two distinct real roots | 两个不同实根 |
| b² − 4ac = 0 | one repeated real root | 一个重根 |
| b² − 4ac < 0 | no real roots | 无实根 |
In Example 2.4.1, the discriminant will be 17, which is positive. We therefore expect two distinct real roots in exact surd form.
在例题 2.4.1 中，判别式为 17，是正数。因此我们预期会得到两个不同实根，并以准确根式形式表示。
5. Step 1: Identify Coefficients | 步骤一：确定系数
+ 1 = 0 with ax² + bx + c = 0. We get a = 2, b = −5, c = 1. Be careful with the sign of b.
将 2x² − 5x + 1 = 0 与 ax² + bx + c = 0 对照，可得 a = 2，b = −5，c = 1。特别注意 b 的符号。
Writing down the coefficients first reduces sign errors later. Many students lose marks because they treat b as 5 instead of −5.
先把系数写下来可以减少后续符号错误。许多学生因为把 b 当作 5 而不是 −5 而丢分。
6. Step 2: Substitute into the Formula | 步骤二：代入公式
itute the values into the quadratic formula: x = (−(−5) ± √((−5)² − 4 × 2 × 1)) ÷ (2 × 2).
将数值代入求根公式：x = (−(−5) ± √((−5)² − 4 × 2 × 1)) ÷ (2 × 2)。
x = (5 ± √25 − 8) ÷ 4
Notice how −(−5) becomes +5, and (−5)² is 25. The denominator 2a becomes 4 because a = 2.
注意 −(−5) 变为 +5，(−5)² 为 25。分母 2a 变为 4，因为 a = 2。
7. Simplify the Discriminant | 化简判别式
Compute 25 − 8 = 17, so x = (5 ± √17) ÷ 4. Since 17 is prime, √17 cannot be simplified further.
计算 25 − 8 = 17，因此 x = (5 ± √17) ÷ 4。由于 17 是质数，√17 无法继续化简。
x = (5 ± √17) ÷ 4
Do not convert √17 to a decimal at this stage. The exact surd form carries full marks and avoids rounding errors.
此阶段不要将 √17 转换为小数。准确根式形式可得满分，并能避免舍入误差。
8. Final Exact Roots | 最终准确根
The two exact roots are x = (5 + √17) ÷ 4 and x = (5 − √17) ÷ 4. Written as a set: x = {(5 + √17)/4, (5 − √17)/4}.
两个准确根为 x = (5 + √17) ÷ 4 和 x = (5 − √17) ÷ 4。写成解集：x = {(5 + √17)/4, (5 − √17)/4}。
x = (5 + √17) ÷ 4 or x = (5 − √17) ÷ 4
These roots are approximately 2.28 and 0.22, but the exact form is the official mathematical answer. Always keep surds unless the question explicitly requests decimals.
这两个根大约为 2.28 和 0.22，但准确形式才是正式数学答案。除非题目明确要求小数，否则应始终保留根式。
9. Checking Your Answers | 检验答案
Substitute one root back into the original equation. For x = (5 + √17) ÷ 4, the expression 2x² − 5x + 1 should simplify to 0. This is good exam practice.
将其中一个根代回原方程。例如 x = (5 + √17) ÷ 4 时，2x² − 5x + 1 应化简为 0。这是良好的考试习惯。
A quick check can catch arithmetic slips, especially in the sign of b or in the denominator 2a. Use exact algebra rather than decimal approximations when verifying.
快速检验可以发现运算错误，尤其是 b 的符号或分母 2a 的错误。验证时应使用准确代数，而非小数近似。
10. Common Mistakes and Exam Tips | 常见错误与应试技巧
Forgetting that b = −5 leads to (−5 ± √17) ÷ 4 instead of (5 ± √17) ÷ 4.
忘记 b = −5 会导致写成 (−5 ± √17) ÷ 4，而不是 (5 ± √17) ÷ 4。
Using 2a incorrectly is another common error. Since a = 2, the denominator is 4, not 2.
错误使用 2a 是另一个常见错误。由于 a = 2，分母是 4，而不是 2。
Rounding too early loses accuracy. Always give the exact surd form unless the question asks for a decimal approximation.
过早四舍五入会失去准确性。除非题目要求小数近似，否则应始终给出准确根式形式。
s only one root. The quadratic formula produces two roots unless the discriminant is zero.
忘记 ± 号只会得到一个根。除非判别式为零，否则求根公式会产生两个根。
Finally, show your substitution clearly. AQA mark schemes often reward correct substitution even if the final simplification has a minor slip.
最后，清晰地展示代入过程。AQA 评分标准通常对正确代入给予分数，即使最终化简有轻微失误。
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