📚 Mastering Quadratics: Workbook G-5 Exercise 221 | 掌握二次方程：練習冊 G-5 第221題
Welcome to this comprehensive guide on quadratic equations, tailored specifically to the problems found in Workbook G-5, Exercise 221. This exercise is designed to test your understanding of quadratics, including factorisation, the quadratic formula, completing the square, and practical applications. We will go through each type of problem step by step, providing clear explanations in both English and Chinese. Make sure to have a pen and paper ready to follow along!
歡迎閱讀這份關於二次方程的全面指南，專為練習冊 G-5 第221題的練習而編寫。這練習旨在測試你對二次方程的理解，包括因式分解、二次公式、配方法及實際應用。我們將逐步詳細講解每類問題，提供中英雙語的清晰解釋。請準備好紙筆，跟隨我們一起學習！
1. Understanding Quadratic Equations | 理解二次方程
A quadratic equation is a polynomial equation of degree 2, typically written in the form ax² + bx + c = 0, where a ≠ 0. The coefficients a, b, and c are real numbers, and x represents an unknown variable. The graph of a quadratic function y = ax² + bx + c is a parabola, which opens upward if a > 0 and downward if a < 0.
二次方程是次數為2的多項式方程，一般形式為 ax² + bx + c = 0，其中 a ≠ 0。係數 a、b、c 是實數，x 代表未知數。二次函數 y = ax² + bx + c 的圖像是拋物線，當 a > 0 時開口向上，當 a < 0 時開口向下。
Solving a quadratic equation means finding all values of x that make the equation true. These values are called the roots or solutions. Depending on the discriminant, a quadratic equation can have two distinct real roots, one repeated real root, or no real roots (but two complex roots).
解二次方程就是找出所有使方程成立的 x 值。這些值稱為根或解。根據判別式，二次方程可能有兩個不相等的實根、一個重根，或沒有實根（但有兩個複數根）。
2. Expanding and Factorising Review | 展開與因式分解複習
Before solving quadratics, it is essential to master expansion and factorisation. Expanding involves removing brackets by multiplying out, for example: (x + 3)(x – 2) = x² + x – 6. Factorisation is the reverse process: x² + 5x + 6 = (x + 2)(x + 3). These skills are the foundation for solving many quadratic equations.
在解二次方程之前，必須熟練掌握展開和因式分解。展開是去除括號並相乘，例如： (x + 3)(x – 2) = x² + x – 6。因式分解是相反的過程：x² + 5x + 6 = (x + 2)(x + 3)。這些技能是解許多二次方程的基礎。
For a quadratic in the form ax² + bx + c, when a = 1, we look for two numbers that multiply to c and add to b. When a ≠ 1, we may use methods such as grouping or trial and error. It is crucial to check your factorisation by expanding the product.
對於形式為 ax² + bx + c 的二次式，當 a = 1 時，我們尋找兩個數，它們相乘等於 c，相加等於 b。當 a ≠ 1 時，可以使用分組法或試錯法。務必通過展開乘積來檢查因式分解是否正確。
3. Solving by Factorisation | 用因式分解求解
The simplest way to solve a quadratic equation is by factorisation. If we can write the equation as (px + q)(rx + s) = 0, then by the zero-product property, either px + q = 0 or rx + s = 0. For example, solve x² − 5x + 6 = 0. We factor to get (x − 2)(x − 3) = 0, so x = 2 or x = 3.
解二次方程最簡單的方法是因式分解。如果能將方程寫成 (px + q)(rx + s) = 0，則根據零乘積性質，要麼 px + q = 0，要麼 rx + s = 0。例如，解 x² − 5x + 6 = 0。因式分解得 (x − 2)(x − 3) = 0，所以 x = 2 或 x = 3。
Not all quadratics can be factorised over the integers. In such cases, we must rely on other methods. However, factorisation is often the quickest method when it works, so we should always attempt it first.
並非所有二次式都能在整數範圍內因式分解。在這種情況下，我們必須依靠其他方法。然而，因式分解在可行時往往是最快的方法，因此我們應始終先嘗試因式分解。
4. Solving Using the Quadratic Formula | 用二次公式求解
The quadratic formula solves any quadratic equation of the form ax² + bx + c = 0. It is derived from the method of completing the square and states that:
x = (−b ± √(b² − 4ac)) / (2a)
The symbol ± means that there are typically two solutions: one with the plus sign and one with the minus sign. To use it, simply substitute the values of a, b, and c into the formula and simplify.
二次公式可以求解任何形式為 ax² + bx + c = 0 的二次方程。它源於配方法，公式為：
x = (−b ± √(b² − 4ac)) / (2a)
± 符號表示通常有兩個解：一個取加號，一個取減號。使用時只需將 a、b、c 的值代入公式並化簡。
For example, solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, c = −2. Substituting gives x = (−3 ± √(9 + 16)) / 4 = (−3 ± 5) / 4, so x = 0.5 or x = −2.
例如，解 2x² + 3x − 2 = 0。這裡 a = 2，b = 3，c = −2。代入得 x = (−3 ± √(9 + 16)) / 4 = (−3 ± 5) / 4，所以 x = 0.5 或 x = −2。
5. Completing the Square | 配方法
Completing the square is a technique that transforms a quadratic expression into a perfect square trinomial plus a constant. For a quadratic in the form x² + bx, we add (b/2)² to complete the square. The result is (x + b/2)² − (b/2)². This method is useful for solving equations and for deriving the quadratic formula.
配方法是將二次式轉化為完全平方三項式加上一個常數的技術。對於形式為 x² + bx 的二次式，我們加上 (b/2)² 來完成平方。結果是 (x + b/2)² − (b/2)²。此方法適用於解方程和推導二次公式。
Example: solve x² + 6x + 1 = 0 by completing the square. First rewrite as x² + 6x = −1. Add (6/2)² = 9 to both sides, giving (x + 3)² = 8. Then take square roots: x + 3 = ±2√2, so x = −3 ± 2√2.
例子：用配方法解 x² + 6x + 1 = 0。首先改寫為 x² + 6x = −1。兩邊加上 (6/2)² = 9，得到 (x + 3)² = 8。然後開平方：x + 3 = ±2√2，所以 x = −3 ± 2√2。
For quadratics with a leading coefficient not equal to 1, we first factor out a from the first two terms, then complete the square inside the parentheses.
當二次項係數不等於 1 時，我們先從前兩項提取 a，然後在括號內完成平方。
6. Discriminant and Nature of Roots | 判別式與根的性質
The discriminant, denoted by Δ, is the part of the quadratic formula under the square root: Δ = b² − 4ac. Its sign determines the nature of the roots:
- If Δ > 0, there are two distinct real roots.
- If Δ = 0, there is exactly one repeated real root (a double root).
- If Δ < 0, there are no real roots; the equation has two complex conjugate roots.
判別式，通常以 Δ 表示，是二次公式中平方根下的部分：Δ = b² − 4ac。它的正負決定了根的性質：
- 若 Δ > 0，有兩個不相等的實根。
- 若 Δ = 0，恰好有一個重根（二重根）。
- 若 Δ < 0，沒有實根；方程有兩個共軛複數根。
This concept is extremely useful for determining whether a quadratic equation can be solved without actually solving it. For instance, if Δ is a perfect square and Δ > 0, then the roots are rational and the equation can be factorised.
這個概念非常有用，無需實際求解就能判斷二次方程是否有解。例如，若 Δ > 0 且 Δ 是完全平方數，則根為有理數，方程可以因式分解。
7. Forming Quadratic Equations from Roots | 由根構造二次方程
If we know the roots α and β of a quadratic equation, we can form the equation. The general form is x² − (α + β)x + αβ = 0. In other words, the sum of the roots gives the coefficient of x (with a negative sign), and the product of the roots gives the constant term.
如果已知二次方程的根 α 和 β，我們可以構造該方程。一般形式為 x² − (α + β)x + αβ = 0。換句話說，根的和給出 x 項的係數（帶負號），根的積給出常數項。
For example, if the roots are 2 and −3, then the equation is x² − (2 + (−3))x + (2 × −3) = x² + x − 6 = 0. This works because (x − 2)(x + 3) expands to x² + x − 6.
例如，若根為 2 和 −3，則方程為 x² − (2 + (−3))x + (2 × −3) = x² + x − 6 = 0。這是因為 (x − 2)(x + 3) 展開後就是 x² + x − 6。
If the roots are irrational or complex, the same relationship holds, though the coefficients may not be integers. This method is essential for solving problems where the roots are given.
即使根是無理數或複數，上述關係仍然成立，只是係數可能不是整數。當已知根時，這個方法非常重要。
8. Solving Quadratic Inequalities | 解二次不等式
Quadratic inequalities such as ax² + bx + c > 0 or ax² + bx + c ≤ 0 are solved by first finding the roots of the corresponding equation, then testing intervals on a number line. The graph of the quadratic is a parabola, and the inequality asks for the range of x where the parabola is above or below the x-axis.
二次不等式，如 ax² + bx + c > 0 或 ax² + bx + c ≤ 0，首先要求出對應方程的根，然後在數軸上測試區間。二次函數的圖像是拋物線，不等式要求的是拋物線在 x 軸上方或下方的 x 範圍。
For example, solve x² − 4 > 0. The roots are x = ±2. Since the parabola opens upward, the expression is positive when x < −2 or x > 2. The solution set is (−∞, −2) ∪ (2, ∞).
例如，解 x² − 4 > 0。根為 x = ±2。因為拋物線開口向上，當 x < −2 或 x > 2 時表達式為正。解集為 (−∞, −2) ∪ (2, ∞)。
Remember to include or exclude the roots depending on whether the inequality is strict or non-strict. A sign diagram can be helpful to avoid mistakes.
記住根據不等式是嚴格還是不嚴格來決定是否包含根。畫符號圖可以幫助避免錯誤。
9. Graphs of Quadratic Functions | 二次函數的圖像
The graph of y = ax² + bx + c is a parabola. Key features include the vertex (turning point), the axis of symmetry, and the y-intercept. The x-intercepts are the roots of the equation. The vertex can be found by completing the square or using the formula x = −b/(2a).
函數 y = ax² + bx + c 的圖像是拋物線。關鍵特徵包括頂點（轉折點）、對稱軸和 y 截距。x 截距就是方程的根。頂點可以通過配方法或公式 x = −b/(2a) 求得。
When sketching a quadratic, it is useful to plot the y-intercept (0, c), the roots (if real), and the vertex. The axis of symmetry is the vertical line passing through the vertex. If a > 0, the vertex is a minimum point; if a < 0, it is a maximum.
繪製二次函數草圖時，標出 y 截距 (0, c)、根（如果為實數）以及頂點很有幫助。對稱軸是通過頂點的垂直線。若 a > 0，頂點為最小值點；若 a < 0，則為最大值點。
Graphing is not just a visual tool; it helps in solving inequalities and understanding the behaviour of functions. In Exercise 221, you may be asked to match graphs with equations or to find the coordinates of the vertex.
作圖不僅是視覺工具；它有助於解不等式和理解函數的行為。在第221題中，你可能會被要求將圖像與方程匹配，或求出頂點坐標。
10. Applied Word Problems | 應用題
Quadratic equations often appear in real-world contexts, such as projectile motion, area problems, and revenue optimisation. These word problems typically translate into a quadratic equation that we then solve. It is crucial to define the unknown variable and interpret the solutions in the context of the problem.
二次方程常出現在實際情境中，如拋體運動、面積問題和收入優化。這些應用題通常轉化為二次方程來求解。關鍵是定義未知數，並在題目情境中解釋解。
For example, the area of a rectangle is 24 m², and the length is 2 m longer than the width. Let the width be x. Then length = x + 2, and x(x + 2) = 24. This expands to x² + 2x − 24 = 0, which factors as (x + 6)(x − 4) = 0. Since width cannot be negative, x = 4 m.
例如，長方形的面積為 24 平方公尺，長比寬長 2 公尺。設寬為 x。則長 = x + 2，且 x(x + 2) = 24。展開得 x² + 2x − 24 = 0，因式分解為 (x + 6)(x − 4) = 0。因為寬不能為負，所以 x = 4 公尺。
Always check whether the solutions make sense physically. Discard any negative or unreasonable answers, and state your final result clearly with appropriate units.
始終檢查解在物理上是否合理。捨去負數或不合理的答案，並以適當的單位清楚寫出最終結果。
11. Common Mistakes and How to Avoid Them | 常見錯誤與避免方法
When working on quadratic equations, students often make several recurring mistakes. One common error is forgetting to set the equation to zero before factorising or applying the quadratic formula. Another is misidentifying the coefficients a, b, and c, especially when the quadratic is not in standard form.
在解二次方程時，學生常犯幾個重複出現的錯誤。一個常見的錯誤是在因式分解或使用二次公式前忘記把方程化為零。另一個是誤認 a、b、c 的係數，尤其是當二次方程不是標準形式時。
To avoid these pitfalls, always rewrite the equation as ax² + bx + c = 0 first. When using the quadratic formula, check the signs of b and c carefully. Also, remember that √(x²) = |x|, not just x, when solving by square roots.
為了避免這些陷阱，務必先將方程改寫為 ax² + bx + c = 0。使用二次公式時，仔細檢查 b 和 c 的符號。同時記住，當用開平方法求解時，√(x²) = |x|，而不只是 x。
Another common mistake is incorrect sign handling in inequalities. Always draw a quick sketch or sign diagram to visualise the solution intervals, and test a point from each interval to confirm the sign.
另一個常見錯誤是不等式中的符號處理不當。建議畫簡圖或符號圖來可視化解區間，並從每個區間取一個點測試符號。
12. Full Solutions to Exercise 221 | 第221題的完整解答
Now let’s apply everything we have learned to a set of typical problems from Exercise 221. The following examples are representative of the exercise. We will solve them step by step, explaining each action.
現在，我們將所學到的內容應用到第221題中的典型問題。以下例子代表了該練習的題型。我們將逐步求解並解釋每個步驟。
Problem 1: Solve x² − 7x + 10 = 0.
問題1： 解 x² − 7x + 10 = 0。
We look for two numbers that multiply to 10 and add to −7. These are −2 and −5. Therefore, (x − 2)(x − 5) = 0, so x = 2 or x = 5.
我們尋找兩個數，它們相乘等於 10，相加等於 −7。這兩個數是 −2 和 −5。因此 (x − 2)(x − 5) = 0，所以 x = 2 或 x = 5。
Problem 2: Solve 2x² + 5x − 3 = 0 using the quadratic formula.
問題2： 使用二次公式解 2x² + 5x − 3 = 0。
Here a = 2, b = 5, c = −3. Substituting into the formula gives:
x = (−5 ± √(25 + 24)) / 4 = (−5 ± 7) / 4
Thus x = 0.5 or x = −3.
這裡 a = 2，b = 5，c = −3。代入公式得：
x = (−5 ± √(25 + 24)) / 4 = (−5 ± 7) / 4
所以 x = 0.5 或 x = −3。
Problem 3: Write down the discriminant and determine the nature of roots for x² + 4x + 5 = 0.
問題3： 寫出判別式並判斷 x² + 4x + 5 = 0 的根的性質。
Δ = 4² − 4(1)(5) = 16 − 20 = −4 < 0. Therefore, there are no real roots.
Δ = 4² − 4(1)(5) = 16 − 20 = −4 < 0。因此，沒有實根。
Problem 4: Solve 3x² − 2x − 1 = 0 by completing the square.
問題4： 用配方法解 3x² − 2x − 1 = 0。
First divide by 3: x² − (2/3)x − 1/3 = 0. Then x² − (2/3)x = 1/3. Completing the square: (x − 1/3)² − 1/9 = 1/3, so (x − 1/3)² = 4/9. Taking roots: x − 1/3 = ±2/3, so x = 1 or x = −1/3.
首先除以 3：x² − (2/3)x − 1/3 = 0。然後 x² − (2/3)x = 1/3。配方：(x − 1/3)² − 1/9 = 1/3，所以 (x − 1/3)² = 4/9。開方：x − 1/3 = ±2/3，所以 x = 1 或 x = −1/3。
Problem 5: The sum of the squares of two consecutive integers is 365. Find the integers.
問題5： 兩個連續整數的平方和為 365，求這兩個整數。
Let the integers be n and n + 1. Then n² + (n + 1)² = 365. Expand: n² + n² + 2n + 1 = 365 → 2n² + 2n − 364 = 0 → n² + n − 182 = 0. Factorise: (n + 14)(n − 13) = 0, so n = 13 (since n = −14 gives the same pair). The integers are 13 and 14.
設整數為 n 和 n + 1。則 n² + (n + 1)² = 365。展開：n² + n² + 2n + 1 = 365 → 2n² + 2n − 364 = 0 → n² + n − 182 = 0。因式分解：(n + 14)(n − 13) = 0，所以 n = 13（因為 n = −14 給出同一對整數）。這兩個整數是 13 和 14。
These problems illustrate the various techniques we have discussed. Practice them on your own, and then compare with the solutions above. If you made a mistake, go back and review the corresponding section.
這些問題展示了我們討論過的各種技巧。請自己先練習，然後與上面的解答比較。如果出錯，請回到對應的小節複習。
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