作業13：二次方程應用題 — 5個完整示例
作業13二次方程應用題是許多代數學生首次發現解決 x² + 5x + 6 = 0 只是工作的一半的地方。更難的一半是首先從一段英文段落建立方程式。應用題需要一個翻譯步驟，將真實場景轉換為二次方程式模型，而該翻譯步驟比代數本身的練習要少得多。本指南涵蓋了從最常見的作業13二次方程應用題類型中提取的五個完整計算示例 — 面積、射彈運動、數字關係、收入和距離-速率-時間 — 顯示每個計算，以便您可以跟隨並在自己的問題上重複使用該方法。
目錄
什麼是二次方程應用題以及為什麼出現在作業13上？
二次方程應用題是任何應用問題，其數學模型包含帶有二次變數項 (x²)。與線性應用題不同，線性應用題中量之間的關係是成比例的，圖形是直線，二次方程應用題對兩個量相乘的情況進行建模 — 矩形的長度和寬度、拋出物體的時間和初始速度、售出物品的數量和單價。作業13二次方程應用題通常在學生掌握了代數求解二次方程之後出現，因此該作業旨在測試您是否能夠識別故事中的二次關係。最常出現的五個類別是：面積和幾何問題、射彈運動問題、連續數字問題、收入和優化問題，以及速度變化的距離-速率-時間問題。每個類別都有一個標準的設置模式，一旦您知道這些模式，翻譯步驟就會變得更加系統化。
A quadratic word problem always contains a quantity multiplied by itself or two related quantities multiplied together — look for area, products of unknowns, or squared terms in any formula given.
任何二次方程應用題的4步框架
無論問題是關於飛行球還是矩形花園，每個作業13二次方程應用題都遵循相同的四步翻譯和求解過程。跳過第1步 — 清楚地定義變數 — 是最大的錯誤來源，因為學生要麼忘記 x 代表什麼，要麼選擇 x 作為使代數不必要複雜的量。每次都要按順序進行這四個步驟。
1. Step 1 — Define your variable precisely
選擇一個未知數稱為 x，並明確寫下來：「令 x = 花園的寬度（米）。」如果出現第二個量，請用 x 表示 — 例如，「長度 = x + 3」。當您可以用另一個表示時，永遠不要使用兩個單獨的變數；這樣可以將問題保持為一個未知數中的單個方程式。
2. Step 2 — Build the equation from the word problem
識別問題所述的關係（面積 = l × w，或距離 = 速率 × 時間，或兩個數字的乘積 = 給定值），替換第1步的表達式，並設置方程式。大多數二次應用題都會給您乘積等於的數值 — 那就是您的方程式。展開任何括號，以便您可以看到 x² 項。
3. Step 3 — Solve the quadratic equation
重新排列為標準形式 ax² + bx + c = 0，然後選擇您的方法：如果數字友好則進行因數分解，如果首項係數為1則完成平方，或對任何方程式使用二次公式 x = (−b ± √(b² − 4ac)) / (2a)。您通常會得到兩個解 — 那是正常的。
4. Step 4 — Interpret the answer and reject impossible values
詢問：這個解在上下文中有意義嗎？負長度、拋出前的負秒數或負人數在數學上都是二次方程式的有效解，但在物理上是不可能的答案。拒絕負的（或其他無意義的）根，並以問題要求的單位陳述您的最終答案。然後通過代入原始應用題描述 — 而不只是您寫的方程式 — 來驗證。
Always write 'Let x = ____' before writing any equation. Students who skip this step almost always end up confused about which root to keep.
面積問題：最常見的二次方程應用題類型
面積問題是最常見的二次方程應用題類型，因為它們自然源於公式面積 = 長 × 寬。當長度和寬度都用同一變數表示時，將它們相乘會產生 x² 項。標準設置是：一個維度定義為 x，另一個為 x 加上（或減去）某個常數，面積作為已知數字給出，您必須找到兩個維度。以下是此問題類型的完整解答示例。
1. Problem
A rectangular garden has a length that is 3 meters longer than its width. The area of the garden is 40 m². Find the width and length.
2. Step 1 — Define the variable
Let x = the width of the garden in meters. Then the length = x + 3 meters.
3. Step 2 — Build the equation
Area = length × width, so (x + 3)(x) = 40. Expanding: x² + 3x = 40.
4. Step 3 — Solve
Rearrange to standard form: x² + 3x − 40 = 0. Factor: look for two numbers that multiply to −40 and add to +3. Those numbers are +8 and −5. So: (x + 8)(x − 5) = 0. Set each factor to zero: x + 8 = 0 → x = −8, or x − 5 = 0 → x = 5.
5. Step 4 — Interpret
Width cannot be negative, so reject x = −8. Width = 5 m, Length = 5 + 3 = 8 m. Check: 5 × 8 = 40 m² ✓. The garden is 5 meters wide and 8 meters long.
For area problems: always set up Area = length × width using your variable expressions, expand, move everything to one side, and factor.
射彈運動應用題：高度和時間
射彈運動問題是作業13二次方程問題集中的第二主要類別。它們依賴於物理公式 h = −(g/2)t² + v₀t + h₀，其中 h 是高度，t 是時間，v₀ 是初始上升速度，h₀ 是初始高度，g 是重力加速度（公制約10 m/s²或英制約32 ft/s²）。大多數作業版本都是預先簡化的，所以您只需使用給定的公式，並在 h = 0（地面）或 h = 某個目標高度時求解 t。以下是一個清晰的示例，使用整數讓您可以進行因數分解而不是使用公式。
1. Problem
A ball is thrown upward from ground level with an initial velocity of 20 m/s. Its height after t seconds is h = −5t² + 20t. At what times is the ball at ground level?
2. Step 1 — Define the variable
t = time in seconds after the ball is thrown. Ground level means h = 0.
3. Step 2 — Build the equation
Set h = 0: −5t² + 20t = 0.
4. Step 3 — Solve
Factor out −5t: −5t(t − 4) = 0. Set each factor to zero: −5t = 0 → t = 0, or t − 4 = 0 → t = 4.
5. Step 4 — Interpret
t = 0 is the moment the ball is thrown (it starts at ground level). t = 4 is when it returns to the ground. The ball is at ground level at t = 0 seconds (launch) and t = 4 seconds (landing). Check: h(4) = −5(16) + 20(4) = −80 + 80 = 0 ✓.
6. Extension: When does the ball reach maximum height?
The maximum height occurs at the midpoint between the two roots: t = (0 + 4)/2 = 2 seconds. Maximum height = −5(2²) + 20(2) = −20 + 40 = 20 m. This is a useful fact many homework 13 projectile problems ask for as a follow-up.
For projectile problems: set h = 0 to find when the object hits the ground. The two roots are launch time and landing time. Maximum height occurs at the vertex, t = −b/(2a).
使用二次方程的數字關係問題
數字關係問題要求您根據兩個未知數的和、差或乘積找到它們。當問題給您兩個數的乘積時，您幾乎總是會得到一個二次方程。最常見的版本涉及連續整數（如8和9，或7和−8）、連續奇數（如5和7）或具有規定差的兩個數。這些問題看起來很簡單，但它們需要仔細的設置 — 第二個數必須用 x 表示，然後才能寫出方程。
1. Problem
The product of two consecutive positive integers is 72. Find the integers.
2. Step 1 — Define the variable
Let x = the smaller integer. Then the next consecutive integer = x + 1.
3. Step 2 — Build the equation
Product of the two integers = 72: x(x + 1) = 72. Expanding: x² + x = 72.
4. Step 3 — Solve
Rearrange: x² + x − 72 = 0. Factor: find two numbers that multiply to −72 and add to +1. Those are +9 and −8. So: (x + 9)(x − 8) = 0. Solutions: x = −9 or x = 8.
5. Step 4 — Interpret
The problem says positive integers, so reject x = −9. x = 8, and x + 1 = 9. The integers are 8 and 9. Check: 8 × 9 = 72 ✓.
6. Variation: Consecutive odd integers
If the problem said 'two consecutive odd integers whose product is 63', let x = first odd integer and x + 2 = second odd integer (odd integers differ by 2). Then x(x + 2) = 63 → x² + 2x − 63 = 0 → (x + 9)(x − 7) = 0 → x = 7. The integers are 7 and 9. Check: 7 × 9 = 63 ✓.
Consecutive integers differ by 1: use x and x + 1. Consecutive odd or even integers differ by 2: use x and x + 2. Write this at the top of every number problem before doing anything else.
收入和定價：商業二次方程應用題
收入問題在作業13二次方程問題集中出現頻繁，因為收入 = 價格 × 售出數量，當價格和數量彼此線性相關（提高價格會減少售出數量）時，它們的乘積是一個二次方程。這些問題通常要求最大化收入的價格，這意味著找到拋物線的頂點。y = ax² + bx + c 的頂點出現在 x = −b/(2a) 處。以下是一個完整示例。
1. Problem
A theater charges $8 per ticket and sells 200 tickets per show. For each $1 increase in ticket price, 10 fewer tickets are sold. What ticket price produces the maximum revenue? What is the maximum revenue?
2. Step 1 — Define the variable
Let x = the number of $1 price increases. Then ticket price = (8 + x) dollars and tickets sold = (200 − 10x).
3. Step 2 — Build the revenue equation
Revenue R = price × tickets sold = (8 + x)(200 − 10x). Expand: R = 1600 − 80x + 200x − 10x² = −10x² + 120x + 1600.
4. Step 3 — Find the vertex
R = −10x² + 120x + 1600 is a downward parabola (a = −10 < 0), so the vertex is the maximum. x = −b/(2a) = −120 / (2 × −10) = −120 / −20 = 6. So the optimal number of price increases is 6.
5. Step 4 — Interpret
Optimal price = 8 + 6 = $14. Tickets sold = 200 − 10(6) = 140. Maximum revenue = 14 × 140 = $1,960. Check using the formula: R = −10(36) + 120(6) + 1600 = −360 + 720 + 1600 = $1,960 ✓.
For revenue maximization: write R = (price)(quantity), expand to get ax² + bx + c, then find the vertex at x = −b/(2a). The vertex gives the input that produces maximum (or minimum) revenue.
導致二次方程的距離、速率和時間問題
距離-速率-時間問題通常產生線性方程（d = rt），但當問題涉及行程的兩段以不同速度行進且相互相關時，或當您添加具有不同分母的兩個時間表達式且分母包含 x 時，它們變成二次方程。關鍵公式是時間 = 距離 ÷ 速率。當您的分子中有 x，並通過乘以清除分母時，您會產生一個二次方程。這類問題在作業13二次方程問題集中出現頻繁，因為它結合了兩項技能：有理表達式和二次方程。
1. Problem
A motorboat travels 24 km upstream and then 24 km back downstream. The river current flows at 3 km/h. If the total trip takes 6 hours, find the boat's speed in still water.
2. Step 1 — Define the variable
Let v = the boat's speed in still water (km/h). Upstream speed = v − 3 km/h (fighting the current). Downstream speed = v + 3 km/h (helped by the current).
3. Step 2 — Build the equation
Time = distance ÷ rate. Upstream time = 24 / (v − 3). Downstream time = 24 / (v + 3). Total time = 6 hours: 24/(v − 3) + 24/(v + 3) = 6.
4. Step 3 — Clear the denominators
Multiply every term by (v − 3)(v + 3): 24(v + 3) + 24(v − 3) = 6(v − 3)(v + 3). Expand the left side: 24v + 72 + 24v − 72 = 48v. Expand the right side: 6(v² − 9) = 6v² − 54. Equation: 48v = 6v² − 54.
5. Step 4 — Solve
Rearrange: 6v² − 48v − 54 = 0. Divide by 6: v² − 8v − 9 = 0. Factor: (v − 9)(v + 1) = 0. Solutions: v = 9 or v = −1.
6. Step 5 — Interpret
Speed cannot be negative, so reject v = −1. The boat's speed in still water is 9 km/h. Check: upstream time = 24/6 = 4 h, downstream time = 24/12 = 2 h, total = 6 h ✓.
Distance-rate-time problems become quadratic when you add two fractions (time = d/r) with x in both denominators and clear them by cross-multiplying. Always check that the denominator does not equal zero for your answer.
學生在作業13二次問題中常犯的錯誤
作業13二次方程應用題有可預測的失敗點。大多數錯誤發生在任何代數寫下之前 — 在設置階段。以下是造成大多數錯誤答案的六個錯誤，以及避免每個錯誤的具體方法。
1. Mistake 1: Not defining the variable before writing the equation
Jumping straight to writing an equation without stating 'Let x = ___' leads to confusion when two solutions appear. You won't know which quantity x represents or why one answer should be rejected. Fix: always write 'Let x = [specific quantity and units]' as the first line of your solution.
2. Mistake 2: Keeping both roots without checking context
Quadratic equations produce two solutions. Students sometimes report both without checking which makes sense in the problem. A rectangle cannot have a negative width. A ball cannot land before it is thrown. Fix: after solving, ask 'does each root make physical sense?' and reject the one that does not.
3. Mistake 3: Forgetting to move everything to one side
After expanding, students try to factor something like x² + 3x = 40 instead of x² + 3x − 40 = 0. Factoring only works reliably when one side is zero. Fix: always rearrange to ax² + bx + c = 0 before factoring or applying the quadratic formula.
4. Mistake 4: Sign errors when expanding (a + b)(a − b) vs (a − b)²
In revenue problems, expanding (8 + x)(200 − 10x) produces a mix of positive and negative terms. Students commonly drop a minus sign. Fix: write out every multiplication step explicitly and circle the sign of each term before combining.
5. Mistake 5: Using the wrong formula for projectile problems
Some textbooks use h = −16t² + v₀t + h₀ (feet, g = 32 ft/s²) and others use h = −5t² + v₀t + h₀ (meters, approximate). Using the wrong constant produces a completely wrong answer. Fix: read the problem to see whether it gives the formula explicitly, or note the units — feet usually means −16, meters usually means −5 or −4.9.
6. Mistake 6: Not checking the answer in the original word problem
Students check their answer in the equation they wrote, but if they set up the equation wrong, a correct algebraic check still gives a wrong answer to the word problem. Fix: after finding x, substitute back into the original problem description (the English sentences) and verify that the stated condition is satisfied.
The setup step takes less than two minutes but eliminates the majority of errors. Writing 'Let x = ___' and rearranging to standard form before anything else is worth more than speed.
五個完整解答的二次方程應用題練習
使用這五個問題在提交您的作業之前測試該框架。它們從簡單到更複雜的排列。隱蓋解答，自己嘗試該問題，然後逐步比較您的工作。
1. Practice Problem 1 — Area
A rectangle's length is twice its width. Its area is 98 cm². Find the dimensions. Solution: Let x = width. Length = 2x. Equation: x(2x) = 98 → 2x² = 98 → x² = 49 → x = 7 (reject −7). Width = 7 cm, Length = 14 cm. Check: 7 × 14 = 98 ✓.
2. Practice Problem 2 — Number relationship
Two positive numbers differ by 5. Their product is 84. Find the numbers. Solution: Let x = smaller number. Larger = x + 5. Equation: x(x + 5) = 84 → x² + 5x − 84 = 0. Factor: (x + 12)(x − 7) = 0 → x = 7 (reject −12). Numbers are 7 and 12. Check: 7 × 12 = 84, 12 − 7 = 5 ✓.
3. Practice Problem 3 — Projectile
A rocket is fired upward. Its height in feet after t seconds is h = −16t² + 96t. When does it reach a height of 128 feet? Solution: Set h = 128: −16t² + 96t = 128 → −16t² + 96t − 128 = 0. Divide by −16: t² − 6t + 8 = 0. Factor: (t − 2)(t − 4) = 0 → t = 2 or t = 4. The rocket reaches 128 feet at 2 seconds (on the way up) and again at 4 seconds (on the way down). Both answers are valid and both should be stated.
4. Practice Problem 4 — Revenue
A store sells 300 units per week at $5 each. For every $0.50 price increase, it sells 20 fewer units. What price maximizes revenue? Solution: Let x = number of $0.50 increases. Price = 5 + 0.5x, Units = 300 − 20x. Revenue R = (5 + 0.5x)(300 − 20x) = 1500 − 100x + 150x − 10x² = −10x² + 50x + 1500. Vertex: x = −50/(2 × −10) = 2.5 increases. Price = 5 + 0.5(2.5) = $6.25. Units = 300 − 20(2.5) = 250. Revenue = 6.25 × 250 = $1,562.50.
5. Practice Problem 5 — Distance-rate-time
A cyclist rides 30 km to a town. On the return trip she rides 5 km/h faster and takes 1 hour less. Find her speed on the outward trip. Solution: Let v = speed going out (km/h). Return speed = v + 5. Time out = 30/v, Time back = 30/(v + 5). Difference = 1: 30/v − 30/(v + 5) = 1. Multiply by v(v + 5): 30(v + 5) − 30v = v(v + 5) → 30v + 150 − 30v = v² + 5v → 150 = v² + 5v → v² + 5v − 150 = 0. Factor: (v + 15)(v − 10) = 0 → v = 10 (reject −15). Speed going out = 10 km/h. Check: Time out = 3 h, time back = 30/15 = 2 h, difference = 1 h ✓.
更快求解二次方程應用題的策略和快捷方式
一旦您識別出二次應用題的類別，設置就變得幾乎是自動的。這些策略幫助您有效地完成任何分配上的二次應用題，而不會犧牲準確性。
1. Identify the category first
Before writing anything, classify the problem: area (look for 'rectangular', 'dimensions', 'area = '), projectile (look for 'thrown', 'height', 'falls', 'seconds'), number relationship (look for 'product', 'consecutive', 'two numbers'), revenue (look for 'price', 'sold', 'revenue', 'profit'), or distance-rate-time (look for 'upstream', 'downstream', 'faster', 'slower', 'trip'). Each category has a known equation structure, so classification saves time.
2. Try factoring before the quadratic formula
Factoring is faster when the discriminant b² − 4ac is a perfect square. Quickly compute b² − 4ac: if it is 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, etc., the equation factors cleanly. If not, go directly to the quadratic formula x = (−b ± √(b² − 4ac)) / (2a) and save the factoring attempt.
3. Keep units throughout every step
Write the units on every quantity: x meters, v km/h, t seconds. If the units in your equation don't make sense (e.g., adding meters to meters² without noticing), that's an early signal that your setup has an error. Catching this at Step 2 is much better than catching it after a full solution.
4. Use the discriminant to predict solution type
For ax² + bx + c = 0, compute Δ = b² − 4ac. If Δ > 0: two real solutions (most word problems). If Δ = 0: exactly one solution (the ball just touches the ground, the dimensions are equal, etc.). If Δ < 0: no real solution, which means either the problem has no physical answer or you set up the equation incorrectly — go back and recheck.
5. For optimization problems, skip the quadratic formula
Revenue and area maximization problems ask for the vertex, not the roots. Use x = −b/(2a) directly — no need to set the equation to zero and solve. Compute x, substitute back to get the maximum or minimum value, and interpret in context.
Δ = b² − 4ac tells you everything before you solve: positive means two roots, zero means one, negative means recheck your setup.
關於作業13二次方程應用題的常見問題
當學生首次解決作業13二次方程應用題時，這些問題會重複出現。答案解決了最常見的混淆之處。
1. When should I use the quadratic formula vs. factoring?
Use factoring when the discriminant b² − 4ac is a perfect square, because the roots will be rational numbers and factoring is faster. Use the quadratic formula when the discriminant is not a perfect square, when the leading coefficient is large, or when you're unsure whether it factors. The formula always works; factoring only sometimes works quickly.
2. What if both roots are positive — which one do I use?
When both roots are positive, both may be valid mathematical answers, but usually the problem context rules one out. For example, if the problem says 'the smaller integer', take the smaller root. If the problem asks for 'dimensions' and both give valid positive dimensions, check which one satisfies any additional constraint (like 'the width is less than 10'). If neither constraint rules one out, both are valid and you should state both.
3. How do I know what x should represent?
Define x as the quantity the problem asks you to find. If the problem asks 'find the width', let x = the width. If the problem asks 'find both numbers', let x = the smaller number. Choosing x as the quantity you want makes interpreting the final answer trivial — you just read off x = [answer].
4. My equation doesn't factor — did I set it up wrong?
Not necessarily. Many real quadratic equations do not factor over integers, especially distance-rate-time problems and some projectile problems. Compute the discriminant: if Δ > 0, use the quadratic formula and leave the answer in simplified radical form or as a decimal. If Δ < 0, recheck your setup — that usually means an error in the equation.
5. How should I check my final answer?
Plug your value of x back into the original word problem sentence, not just the equation. For the garden problem: 'Does a garden of width 5 m and length 8 m have an area of 40 m²? Yes, 5 × 8 = 40.' For the boat problem: 'Does a boat going 9 km/h upstream (speed 6 km/h) cover 24 km in 4 hours and then 24 km downstream (speed 12 km/h) in 2 hours, totaling 6 hours? Yes.' This two-sentence check catches setup errors that algebraic substitution misses.
6. What is the hardest type of quadratic word problem?
Most students find distance-rate-time problems the most difficult because they require building two fractions (time = d/r), adding them, and then clearing denominators before any quadratic algebra begins. The two extra steps — fraction setup and denominator clearing — make errors more likely. Practice these specifically: write time = d/r for each leg, add the expressions, set equal to total time, and multiply both sides by the LCD.
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