📚 Solving Quadratic Equations with the Quadratic Formula – Example 1.7.3 | 用求根公式解二次方程——示例 1.7.3
In AQA A Level Mathematics, quadratic equations appear in pure mathematics, coordinate geometry, mechanics and statistics. Example 1.7.3 focuses on solving a quadratic equation using the quadratic formula, especially when factorising is not straightforward. This article works through the method step by step and connects the formula to the discriminant, surd simplification, significant figures and typical AQA exam techniques.
在 AQA A Level 数学中，二次方程广泛出现在纯数学、坐标几何、力学和统计中。示例 1.7.3 重点讲解如何用求根公式解二次方程，尤其是在因式分解不容易进行的情况下。本文逐步讲解该方法，并将求根公式与判别式、根式化简、有效数字以及 AQA 典型考试技巧联系起来。
1. What Quadratic Equations Are | 什么是二次方程
A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of x is 2, so the graph of y = ax² + bx + c is a parabola.
二次方程是指任何可以写成 ax² + bx + c = 0 的方程，其中 a、b、c 为常数且 a ≠ 0。x 的最高次数为 2，因此 y = ax² + bx + c 的图像是一条抛物线。
Solving a quadratic equation means finding the values of x that make the equation true. These values are called the roots of the equation, and they are also the x-coordinates where the parabola crosses the x-axis.
解二次方程就是求出使方程成立的 x 值。这些值称为方程的根，它们也是抛物线与 x 轴交点的 x 坐标。
2. Standard Form and Coefficients | 标准形式与系数
Before using the quadratic formula, you must write the equation in standard form ax² + bx + c = 0. In this form:
在使用求根公式之前，你必须先把方程写成标准形式 ax² + bx + c = 0。在这种形式下：
- a is the coefficient of x²
- b is the coefficient of x
- c is the constant term
a 是 x² 的系数；b 是 x 的系数；c 是常数项。
For example, in the equation 2x² + 3x − 4 = 0, we have a = 2, b = 3 and c = −4. Always check the signs carefully before substituting, because sign errors are one of the most common mistakes in AQA exams.
例如，在方程 2x² + 3x − 4 = 0 中，我们有 a = 2、b = 3、c = −4。代入前一定要仔细检查符号，因为符号错误是 AQA 考试中最常见的错误之一。
3. Three Main Methods of Solution | 三种主要解法
There are three standard ways to solve a quadratic equation at A Level:
A Level 阶段解二次方程有三种标准方法：
- Factorising: write the quadratic as a product of two linear factors and set each factor to zero.
- Completing the square: rewrite the quadratic in the form a(x + p)² + q = 0 and then solve for x.
- Quadratic formula: substitute a, b and c into x = (−b ± √(b² − 4ac)) / (2a).
因式分解法：把二次式写成两个一次因式的乘积，然后令每个因式等于零。配方法：把二次式改写为 a(x + p)² + q = 0，然后解出 x。求根公式法：把 a、b、c 代入 x = (−b ± √(b² − 4ac)) / (2a)。
| Method | Best for | Limitation |
|---|---|---|
| Factorising | Simple quadratics with integer roots | Not always possible |
| Completing the square | Finding vertex form and proof | Can be slower for messy coefficients |
| Quadratic formula | Always works, including surd answers | Needs careful substitution |
Example 1.7.3 uses the quadratic formula because the equation does not factorise easily and the answer involves a surd.
示例 1.7.3 使用求根公式，因为该方程不容易因式分解，且答案含有根式。
4. Discriminant: The Key to the Formula | 判别式：公式的关键
The expression inside the square root in the quadratic formula, b² − 4ac, is called the discriminant. It tells you how many real roots the quadratic equation has before you complete the calculation.
求根公式中平方根内部的部分 b² − 4ac 称为判别式。它能在你完成计算之前告诉你二次方程有多少个实数根。
- If b² − 4ac > 0, the equation has two distinct real roots.
- If b² − 4ac = 0, the equation has one repeated real root.
- If b² − 4ac < 0, the equation has no real roots.
如果 b² − 4ac > 0，方程有两个不同的实数根。如果 b² − 4ac = 0，方程有一个重复的实数根。如果 b² − 4ac < 0，方程没有实数根。
AQA questions often ask you to interpret the discriminant or use it to decide whether a line and a curve intersect. In Example 1.7.3, the discriminant is positive, so we expect two distinct roots.
AQA 考题经常要求你解释判别式的含义，或利用它判断一条直线与一条曲线是否相交。在示例 1.7.3 中，判别式为正，因此我们预期有两个不同的实根。
5. Worked Example 1.7.3 | 示例 1.7.3 详解
Use the quadratic formula to solve 2x² + 3x − 4 = 0. Give your answers in exact surd form.
用求根公式解方程 2x² + 3x − 4 = 0。答案用精确根式表示。
Step 1: Identify a, b and c from the standard form ax² + bx + c = 0.
步骤 1：从标准形式 ax² + bx + c = 0 中确定 a、b、c。
We have a = 2, b = 3 and c = −4.
我们有 a = 2、b = 3、c = −4。
Step 2: Substitute these values into the quadratic formula.
步骤 2：将这些值代入求根公式。
x = (−b ± √(b² − 4ac)) / (2a)
This gives:
得到：
x = (−3 ± √(3² − 4 × 2 × (−4))) / (2 × 2)
Step 3: Simplify the discriminant first.
步骤 3：先化简判别式。
b² − 4ac = 9 − 4 × 2 × (−4) = 9 + 32 = 41
So the formula becomes:
因此公式变为：
x = (−3 ± √41) / 4
The exact roots are:
精确根为：
x = (−3 + √41) / 4 or x = (−3 − √41) / 4
Since √41 cannot be simplified further, these are the exact surd answers. A decimal approximation would give x ≈ 0.851 and x ≈ −2.35 to 3 significant figures.
由于 √41 无法进一步化简，这些就是精确的根式答案。小数近似值为 x ≈ 0.851 和 x ≈ −2.35，保留三位有效数字。
6. Step-by-Step Substitution | 逐步代入法
Many AQA candidates lose marks because they rush the substitution step. Always use brackets around negative values and work out the discriminant b² − 4ac before touching the fraction.
许多 AQA 考生因为在代入步骤中过于着急而丢分。在计算分数之前，一定要用括号括住负值，并先计算出判别式 b² − 4ac。
A reliable routine is:
一个可靠的步骤是：
- Write down a, b and c.
- Write the formula with empty spaces or brackets.
- Substitute each value carefully.
- Simplify the discriminant.
- Simplify the denominator.
- Separate the two roots using the plus and minus signs.
写出 a、b、c。写出带有空位或括号的公式。仔细代入每个值。化简判别式。化简分母。用正负号分成两个根。
This method reduces sign errors and makes your working clear to the examiner, which is important for method marks in AQA mark schemes.
这种方法可以减少符号错误，并使你的解题过程对考官清晰可见，这对于 AQA 评分方案中的方法分非常重要。
7. Simplifying Surd Answers | 化简根式答案
If the discriminant is not a perfect square, the answer will often involve a surd such as √41. AQA often expects the answer in exact surd form unless the question asks for a decimal approximation.
如果判别式不是完全平方数，答案通常包含根式，例如 √41。AQA 通常要求答案以精确根式表示，除非题目要求给出小数近似值。
Sometimes the surd can be simplified. For example, if b² − 4ac = 40, then √40 = √(4 × 10) = 2√10. You should always look for square factors under the square root.
有时根式可以进一步化简。例如，如果 b² − 4ac = 40，那么 √40 = √(4 × 10) = 2√10。你应该总是检查平方根下是否有平方因数。
Example: Solve 2x² − 4x − 3 = 0 using the quadratic formula.
示例：用求根公式解方程 2x² − 4x − 3 = 0。
Here a = 2, b = −4, c = −3. The discriminant is (−4)² − 4 × 2 × (−3) = 16 + 24 = 40. So:
这里 a = 2、b = −4、c = −3。判别式为 (−4)² − 4 × 2 × (−3) = 16 + 24 = 40。因此：
x = (4 ± √40) / 4 = (4 ± 2√10) / 4 = 1 ± (√10) / 2
The exact roots are x = 1 + (√10)/2 and x = 1 − (√10)/2. This shows how simplifying surds can make the final answer cleaner.
精确根为 x = 1 + (√10)/2 和 x = 1 − (√10)/2。这说明化简根式可以使最终答案更简洁。
8. Significant Figures and Decimal Answers | 有效数字与小数答案
When a question asks for answers to 3 significant figures, you should keep more figures during the calculation and only round at the final step. Use your calculator’s memory or store the surd value.
当题目要求答案保留三位有效数字时，你应在计算过程中保留更多位数，只在最后一步进行四舍五入。可以使用计算器的记忆功能或存储根式值。
For Example 1.7.3, √41 ≈ 6.403124. One root is (−3 + 6.403124) / 4 = 3.403124 / 4 = 0.850781, which rounds to 0.851. The other root is (−3 − 6.403124) / 4 = −9.403124 / 4 = −2.350781, which rounds to −2.35.
对于示例 1.7.3，√41 ≈ 6.403124。一个根为 (−3 + 6.403124) / 4 = 3.403124 / 4 = 0.850781，四舍五入为 0.851。另一个根为 (−3 − 6.403124) / 4 = −9.403124 / 4 = −2.350781，四舍五入为 −2.35。
Do not write the exact surd answer and the decimal answer together unless the question asks for both. AQA questions will clearly state the required form.
除非题目同时要求两种形式，否则不要同时写出精确根式答案和小数答案。AQA 题目会明确说明所需的形式。
9. Completing the Square Connection | 配方法的联系
The quadratic formula is derived by completing the square on the general quadratic ax² + bx + c = 0. This means the two methods are mathematically equivalent.
求根公式是通过对一般二次式 ax² + bx + c = 0 进行配方法推导出来的。这意味着这两种方法在数学上是等价的。
If you complete the square on 2x² + 3x − 4 = 0, you can obtain the same roots as in Example 1.7.3, but the algebra is longer. The formula is simply a shortcut that automates this process.
如果你对 2x² + 3x − 4 = 0 进行配方，你可以得到与示例 1.7.3 相同的根，但代数运算更长。求根公式只是自动化这一过程的捷径。
Understanding this connection helps you check your answer: the roots derived from the formula must also satisfy the completed square form, and their symmetry is represented by the ± sign.
理解这一联系有助于你检验答案：由求根公式得到的根也必须满足配方法的形式，其对称性由 ± 号表示。
10. Common
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