📚 Attempt to use correct formula (with values for a, b and c) | 正确使用二次公式：代入a、b、c的值
ost not because the formula itself is remembered incorrectly, but because students misidentify the values of a, b and c or mishandle signs during substitution. This guide focuses entirely on the skill of attempting to use the correct formula by carefully extracting and substituting the right constants. We will move step by step from rewriting an equation into standard form, through handling negative coefficients, to applying the formula accurately and checking your discriminant.
在 A-Level 数学中，二次公式是求解形如 ax² + bx + c = 0 的方程的基本工具。然而，丢分往往不是因为记错了公式本身，而是因为学生在代入时错误地识别了 a、b 和 c 的值，或者处理符号时出现了失误。本指南专注于通过仔细提取并代入正确常数来正确使用公式的技能。我们将从将方程改写为标准形式开始，逐步学习如何处理负系数，直至准确应用公式并检查判别式。
1. The Standard Quadratic Form | 二次方程的标准形式
Before any formula can be used, the equation must be rearranged into the form ax² + bx + c = 0. All terms need to be on one side, with zero on the other, and the x² term must have a positive coefficient where possible, although the formula works for any sign.
在使用任何公式之前，必须将方程整理成 ax² + bx + c = 0 的形式。所有项都要移到一边，另一边为零，并且 x² 项的系数最好为正，尽管公式适用于任何符号。
ten as −3x² + 5x − 2 = 0, or better (multiplying by −1) as 3x² − 5x + 2 = 0. The values of a, b and c are then taken from this final standard form.
例如，5x − 3x² = 2 应首先改写为 −3x² + 5x − 2 = 0，或更好（乘以 −1）化为 3x² − 5x + 2 = 0。然后从这个最终的标准形式中提取 a、b 和 c 的值。
2. Identifying a, b and c Correctly | 正确识别 a、b 和 c
The coefficient a is the number in front of x², b is the coefficient of x, and c is the constant term. Crucially, the signs attached to these coefficients must travel with them into the formula.
系数 a 是 x² 前面的数，b 是 x 的系数，c 是常数项。关键是，这些系数所带的符号必须随着系数一起代入公式。
Consider 2x² − 7x + 1 = 0. Here, a = 2, b = −7, c = 1. If the equation were 4x − x² + 6 = 0, rewriting as −x² + 4x + 6 = 0 gives a = −1, b = 4, c = 6. The sign of a can be negative, and the formula handles this automatically.
考虑 2x² − 7x + 1 = 0。这里 a = 2，b = −7，c = 1。如果方程是 4x − x² + 6 = 0，重写为 −x² + 4x + 6 = 0 得到 a = −1，b = 4，c = 6。a 可以为负值，公式会自动处理。
3. Extracting Values from Non‑Standard Equations | 从非标准方程中提取数值
st. For instance, (x + 3)(x − 2) = 5 becomes x² + x − 6 = 5, and then x² + x − 11 = 0. Now a = 1, b = 1, c = −11.
当方程包含括号、分数或两边都有项时，首先展开并化简。例如，(x + 3)(x − 2) = 5 变成 x² + x − 6 = 5，然后得到 x² + x − 11 = 0。此时 a = 1，b = 1，c = −11。
− 1) = 3x + 4, expand: 2x² − 2x = 3x + 4, then collect all terms: 2x² − 5x − 4 = 0. Here, a = 2, b = −5, c s check that the x² term is visible; if it cancels, the equation is linear, not quadratic.
对于 2x(x − 1) = 3x + 4，展开得 2x² − 2x = 3x + 4，然后合并所有项：2x² − 5x − 4 = 0。这里 a = 2，b = −5，c = −4。务必检查 x² 项是否可见；如果抵消了，该方程就是一次方程，而非二次方程。
4. The Quadratic Formula | 二次公式
The formula itself is x = (−b ± √b² − 4ac) / (2a) shows that there are generally two solutions, one using the plus and one using the minus.
公式本身是 x = (−b ± √b² − 4ac) / (2a)。它给出 ax² + bx + c = 0 的解。符号 ± 表明通常有两个解，一个使用加号，一个使用减号。
x = (−b ± √b² − 4ac) / (2a)
You must know this formula exactly, but equally important is the ability to slot in the correct signed values for a, b and c without algebraic mistakes.
你必须准确记住这个公式，但同样重要的是能够将带符号的 a、b 和 c 值正确代入，而不出现代数错误。
5. Substituting with Negative Coefficients | 代入负系数
The most common error occurs when b is negative. Because the formula already contains −b, two negatives interact to form a positive. For example, if b = −5, then −b = −(−5) = +5.
最常见的错误发生在 b 为负数时。由于公式中已经包含 −b，两个负号相互作用得到正号。例如，若 b = −5，则 −b = −(−5) = +5。
Take 3x² − 8x + 2 = 0. Here a = 3, b = −8, c = 2. The numerator becomes −(−8) ± √((−8)² − 4×3×2) = 8 ± √64 − 24. Many students incorrectly keep −8, losing marks. Practise writing out the substitution as a separate line: −b = −(−8) = 8.
以 3x² − 8x + 2 = 0 为例。这里 a = 3，b = −8，c = 2。分子变为 −(−8) ± √((−8)² − 4×3×2) = 8 ± √64 − 24。许多学生错误地保留 −8，因而丢分。请练习将代入过程单独写一行：−b = −(−8) = 8。
When c is negative, the expression inside the square root becomes b² − 4a(negative) = b² + 4a|c|, so the plus sign is vital. For 2x² + 3x − 5 = 0, a = 2, b = 3, c = −5, giving discriminant 3² − 4×2×(−5) = 9 + 40 = 49.
当 c 为负时，平方根内的表达式变为 b² − 4a(负数) = b² + 4a|c|，因此加号至关重要。对于 2x² + 3x − 5 = 0，a = 2，b = 3，c = −5，判别式为 3² − 4×2×(−5) = 9 + 40 = 49。
6. Understanding the Discriminant | 理解判别式
D < 0, no real solutions (but two complex ones). Before finishing your calculation, it is wise to compute the discriminant to check for arithmetic mistakes.
判别式 D = b² − 4ac 决定了根的性质。若 D > 0，有两个相异实根；若 D = 0，有一个重根；若 D < 0，没有实数解（但有两个复数解）。在完成计算之前，先计算判别式来检查运算错误是一个明智的做法。
es D = (−6)² − 4×1×9 = 36 − 36 = 0, so a repeated root x = 3. If you had accidentally used b the root would be x = (−6 ± 0)/2 = −3, which is incorrect. Always double-check the sign of b.
例如，x² − 6x + 9 = 0 得到 D = (−6)² − 4×1×9 = 36 − 36 = 0，因此有一个重根 x = 3。如果你不小心用了 b = 6，D 仍为 36 − 36 = 0，但根将是 x = (−6 ± 0)/2 = −3，这是错误的。务必再次检查 b 的符号。
7. Worked Example 1: Positive b, Negative c | 示例1：正 b，负 c
Solve 2x² + 5x − 3 = 0. Identify a = 2, b = 5, c = −3. Write the formula substitution carefully:
解 2x² + 5x − 3 = 0。识别 a = 2，b = 5，c = −3。仔细写出公式的代入过程：
x = ( −5 ± √(5² − 4×2×(−3)) ) / (2×2)
Simplify inside the root: 25 + 24 = 49, so √49 7) / 4. The two solutions are x = (−5 + 7)/4 = 2/4 = 1/2, and x = (−5 − 7)/4 = −12/4 = −3.
化简根号内：25 + 24 = 49，于是 √49 = 7。因此 x = (−5 ± 7) / 4。两个解为 x = (−5 + 7)/4 = 2/4 = 1/2，以及 x = (−5 − 7)/4 = −12/4 = −3。
Notice that the −b part is −5, not +5, because b = +5. The minus signs in c turn 4ac into adding a positive amount, which is why the discriminant became 49.
注意 −b 部分是 −5，而不是 +5，因为 b = +5。c 中的负号使得 4ac 变为加上一个正数，这就是判别式变为 49 的原因。
8. Worked Example 2: Negative b and Negative c | 示例2：负 b 且负 c
Solve 4x² − 4x − 3 = 0. Here a = 4, b = −4, c erator −b becomes −(−4) = 4. The discriminant: b² − 4ac = (−4)² − 4×4×(−3) = 16 + 48 = 64.
解 4x² − 4x − 3 = 0。这里 a = 4，b = −4，c = −3。分子 −b 变为 −(−4) = 4。判别式：b² − 4ac = (−4)² − 4×4×(−3) = 16 + 48 = 64。
x = (4 ± √64) / (2×4) = (4 ± 8) / 8
Thus x = (4+8)/8 = 12/8 = 3/2, and x = (4−8)/8 = −4/8 = −1/2.
因此 x = (4+8)/8 = 12/8 = 3/2，以及 x = (4−8)/8 = −4/8 = −1/2。
This example demonstrates that a negative b flips the sign at the front, while a negative c increases the discriminant. Always use brackets when substituting negative numbers to avoid sign errors.
此例表明，负的 b 会翻转前面的符号，而负的 c 会增大判别式。代入负数时始终使用括号以避免符号错误。
9. Common Mistakes and How to Avoid Them | 常见错误及避免方法
Mistake 1: forgetting to set the equation to zero. If you apply the formula to 2x² + 3x = 5, you might incorrectly take c = 5. Instead, rewrite as 2x² + 3x − 5 = 0, so c = −5.
错误1：忘记将方程设为零。 如果你对 2x² + 3x = 5 使用公式，可能会错误地取 c = 5。实际上应改写为 2x² + 3x − 5 = 0，因此 c = −5。
Mistake 2: mishandling the '−b' part. When b is positive, −b is negative; when b is negative, −b becomes positive. Write −(b) explicitly if it helps.
错误2：对 '−b' 部分处理不当。 当 b 为正时，−b 为负；当 b 为负时，−b 变为正。如果有帮助，可以明确写出 −(b)。
Mistake 3: division by 2a applies to the entire numerator. Write the fraction carefully, or use brackets: ( −b ± √discriminant ) ÷ (2a). Never divide only the discriminant by 2a.
错误3：除以 2a 应用于整个分子。仔细写出分数，或使用括号：( −b ± √判别式 ) ÷ (2a)。绝不要只将判别式除以 2a。
Mistake 4: ignoring the sign of a. If a is negative, the whole denominator is negative. This can change the direction of inequality in later reasoning, but for the formula it simply scales the solution.
错误4：忽略 a 的符号。 如果 a 为负，整个分母即为负。这在后续的不等式推理中可能会改变方向，但对公式而言，它只是按比例缩放解。
10. Table for Quick Identification | 快速识别的表格
The table below shows some equations and the corresponding correct a, b, c. Use it to train your eye to pick out the values instantly.
下表展示了一些方程及其对应的正确 a、b、c 值。用它来训练你快速识别数值的能力。
| Equation / 方程 | a | b | c |
|---|---|---|---|
| x² − 5x + 6 = 0 | 1 | −5 | 6 |
| −2x² + 4x − 3 = 0 | −2 | 4 | −3 |
| 3x² − 7 = 0 | 3 | 0 | −7 |
| 4x − x² = 0 (rewrite: −x²+4x=0) | −1 | 4 | 0 |
| (x+2)(x−1) = 4 → x²+x−2=4 → x²+x−6=0 | 1 | 1 | −6 |
Notice that when no x term is visible, b = 0, and when there is no constant, c = 0. These cases are often missed, so check carefully.
注意，当看不到 x 项时，b = 0；当没有常数项时，c = 0。这些情况经常被忽略，因此要仔细检查。
11. Practice Drill: Spot the Error | 练习：找出错误
Each of the following attempted substitutions contains a mistake. Identify the error and write the correct a, b, c.
以下每个尝试代入的过程都包含一个错误。找出错误并写出正确的 a、b、c。
- Equation: 5x − 2x² = 3. Student writes a = 5, b = −2, c = 3.
方程：5x − 2x² = 3。 学生写成 a = 5, b = −2, c = 3。 - Equation: x² + 4x = 0. Student writes a = 1, b = 4, c = 1.
方程：x² + 4x = 0。 学生写成 a = 1, b = 4, c = 1。 - Equation: 3x² − 6x + 2 = 0, but student uses b = 6.
方程：3x² − 6x + 2 = 0，但学生用了 b = 6。
c = 0; (iii) b = −6, not 6. Practising with such drills sharpens your attention to signs.
正确的值应为：(i) 重写为 −2x² + 5x − 3 = 0 或 2x² − 5x + 3 = 0，则 a = 2，b = −5，c = 3（或 a=−2, b=5, c=−3）；(ii) a = 1，b = 4，c = 0；(iii) b = −6，而不是 6。通过此类练习可以提高你对符号的注意力。
12. Summary and Exam Tips | 总结与考试技巧
Every time you reach for the quadratic formula, follow this checklist: Step 1: Rearrange to ax² + bx + c = 0. Step 2: Write down a = … , b = … , c = … including signs. Step 3: Work out −b and the discriminant b² − 4ac separately. Step 4: Substitute into x = (−b ± √D) / (2a) and simplify.
每当你需要使用二次公式时，请遵循以下清单：第1步： 整理成 ax² + bx + c = 0。第2步： 写下 a = … , b = … , c = …，包括符号。第3步： 分别计算 −b 和判别式 b² − 4ac。第4步： 代入 x = (−b ± √D) / (2a) 并化简。
In the exam, showing the substitution line with the correct signed numbers explicitly can earn method marks even if a later arithmetic slip occurs. Clear layout and systematic working are therefore just as important as the final answer.
在考试中，明确展示带正确符号数字的代入过程，即使之后出现运算错误，也可以获得方法分。因此，清晰的布局和系统化的步骤与最终答案同样重要。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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