Interpreting & Discarding Solutions - Math 3

1. A ball is thrown upward from a 48-foot building. Its height is modeled by $h(t) = -16t^2 + 32t + 48$. When solving $h(t) = 64$ to find when the ball is 64 feet high, a student obtains $t = 0.5$ and $t = 1.5$. The student discards $t = 1.5$, reasoning that "the ball can only reach 64 feet once on its way up." Which assessment of this reasoning is most appropriate?
   Answer: The reasoning is incorrect; both solutions are valid since the ball reaches 64 feet twice during its complete trajectory Both solutions are valid and meaningful. The ball reaches 64 feet at t = 0.5 seconds (on its way up) and again at t = 1.5 seconds (on its way down). The student's reasoning incorrectly assumes that a given height can only be reached once, but in projectile motion, objects typically pass through intermediate heights twice - once ascending and once descending.
2. The concentration $C$ of a medication in the bloodstream follows $C(t) = \frac{20t}{t^2 + 4}$ where $t$ is hours after administration and $C$ is mg/L. To find when concentration equals 4 mg/L, a student solves $\frac{20t}{t^2 + 4} = 4$ and obtains $t = 2$ and $t = -2$. The student reasons that $t = -2$ should be kept because "it shows the concentration was 4 mg/L two hours before this dose, indicating the previous dose's effect." Which evaluation of this interpretation is most accurate?
   Answer: The interpretation is incorrect; the model specifically describes concentration after this administration, making negative time meaningless The model C(t) = 20t/(t² + 4) specifically describes the concentration of medication starting from t = 0 (time of administration). The negative solution t = -2 is mathematically valid but contextually meaningless because the model doesn't describe the system before this dose was given. The student's reasoning incorrectly assumes the model accounts for previous doses, but it only models the current dose's concentration curve.
3. A population model predicts that the number of bacteria $P(t) = 500e^{0.3t}$ where $t$ is measured in hours. To find when the population reaches 2000, a student solves $500e^{0.3t} = 2000$ and obtains $t \approx 4.62$ hours. However, when checking this solution by substitution, the student gets $P(4.62) \approx 1999.8$. Which statement best explains how to interpret this discrepancy?
   Answer: The solution is valid; the small discrepancy results from rounding in the calculation process and doesn't affect the contextual meaning The solution t ≈ 4.62 is valid and contextually meaningful. The tiny discrepancy (1999.8 vs 2000) results from rounding the logarithmic solution to 4.62. This level of precision is appropriate for the context, and such minor computational rounding doesn't make a solution extraneous or invalid.
4. A ball is thrown upward from a platform 48 feet high with initial velocity 32 ft/s. Its height is $h(t) = -16t^2 + 32t + 48$. When solving for when the ball hits the ground ($h = 0$), a student gets $t = 3$ and $t = -1$. Which solution analysis is correct?
   Answer: Accept $t = 3$ seconds as when the ball hits ground; reject $t = -1$ because it represents a time before the ball was thrown. When working with projectile motion problems, you need to interpret mathematical solutions within the physical context of the situation. The equation $h(t) = -16t^2 + 32t + 48$ models the ball's height after it's thrown at $t = 0$. Setting $h(t) = 0$ and solving $-16t^2 + 32t + 48 = 0$ gives $t = 3$ and $t = -1$. The key is understanding what these solutions mean physically. Since the ball is thrown at $t = 0$, only times $t \geq 0$ are meaningful for this problem. The solution $t = 3$ seconds represents when the ball actually hits the ground after being thrown. The solution $t = -1$ represents a mathematical artifact - if you extended the parabolic path backward in time, the ball would theoretically be at ground level one second before it was thrown, but this has no physical meaning in our problem. Choice A incorrectly suggests both solutions are physically valid. Choice B reverses the analysis, incorrectly accepting the negative time while rejecting the positive time with a false claim about maximum height. Choice D incorrectly labels both solutions as extraneous when one is actually valid. The correct answer is C because $t = 3$ seconds is when the ball hits the ground, while $t = -1$ must be rejected as it represents a time before the physical event began. Study tip: In projectile motion problems, always check whether your mathematical solutions make sense within the time constraints of the physical situation. Negative times are usually meaningless unless specifically stated otherwise.
5. A projectile is launched from ground level with an initial velocity of 64 feet per second. Its height $h$ (in feet) after $t$ seconds is modeled by $h(t) = -16t^2 + 64t$. When solving for when the projectile hits the ground, a student finds $t = 0$ and $t = 4$. Which statement best explains how to interpret these solutions?
   Answer: Both solutions are meaningful: $t = 0$ represents the launch time and $t = 4$ represents when it hits the ground after 4 seconds of flight. Both solutions are meaningful in this context. $t = 0$ represents the moment of launch when the projectile is at ground level ($h = 0$), and $t = 4$ represents when the projectile returns to ground level after its flight. Choice B incorrectly suggests $t = 0$ is not meaningful. Choice C confuses the launch time with maximum height. Choice D incorrectly claims both solutions are extraneous.
6. A company's profit model is $P(x) = -2x^2 + 100x - 800$ where $x$ is the number of units sold (in hundreds) and $P$ is profit in thousands of dollars. When solving $P(x) = 450$ to find production levels yielding $450,000 profit, a student gets $x = 15$ and $x = 35$. However, market research indicates that consumer demand drops significantly beyond 2000 units ($x = 20$). How should these solutions be interpreted?
   Answer: Both solutions should be kept; market constraints don't affect the mathematical validity of profit calculations at different production levels Both x = 15 and x = 35 are mathematically valid solutions to the profit equation. The market research about demand dropping beyond x = 20 represents additional business intelligence, but doesn't make either mathematical solution extraneous. The profit model P(x) already incorporates relevant economic factors; external market constraints are separate considerations that don't invalidate the mathematical solutions.
7. The number of customers $N$ in a store $t$ hours after opening is modeled by $N(t) = \frac{60t}{t^2 + 9}$. When finding when there are exactly 10 customers, solving $\frac{60t}{t^2 + 9} = 10$ gives $t = 3$ and $t = -2$. For staffing decisions, how should the manager interpret these solutions?
   Answer: Use $t = 3$ hours after opening as when customer count reaches 10; ignore $t = -2$ because it represents time before the store opened. When solving equations that model real-world scenarios, you need to interpret your mathematical solutions within the context of the problem. Here, the model $N(t) = \frac{60t}{t^2 + 9}$ represents customer count $t$ hours after opening, making the domain restriction crucial. Solving $\frac{60t}{t^2 + 9} = 10$ correctly yields $t = 3$ and $t = -2$. Since $t$ represents hours after the store opens, only non-negative values make physical sense. The solution $t = 3$ means there are exactly 10 customers 3 hours after opening, which is meaningful for staffing decisions. The solution $t = -2$ would represent 2 hours before opening, when the store isn't even operating and has zero customers, not 10. Choice A incorrectly suggests using the negative solution for "pre-opening preparation" and dismisses $t = 3$ based on an unfounded assumption about customer flow patterns. Choice C wrongly treats both solutions as valid times "during the day" when the store has 10 customers, ignoring that negative time occurs before opening. Choice D incorrectly labels both solutions as "extraneous," when $t = 3$ is perfectly valid—it's just the negative solution that's contextually meaningless. The correct interpretation is B: use $t = 3$ hours after opening as the meaningful solution, and discard $t = -2$ because it falls outside the realistic domain. Study tip: Always check whether your mathematical solutions make sense within the problem's context, especially when dealing with time, distance, or other quantities that may have natural restrictions.
8. A projectile is launched from ground level with an initial velocity of 64 feet per second. Its height $h$ (in feet) after $t$ seconds is given by $h(t) = -16t^2 + 64t$. When solving $h(t) = 0$ to find when the projectile hits the ground, a student obtains solutions $t = 0$ and $t = 4$. Which statement best explains how to interpret these solutions in the context of this problem?
   Answer: Both solutions are meaningful: $t = 0$ represents the launch time and $t = 4$ represents when the projectile hits the ground Both solutions are contextually meaningful. t = 0 represents the initial moment of launch when the projectile is at ground level (h = 0), and t = 4 represents when the projectile returns to ground level. The mathematical solutions correspond to two distinct physical events in the projectile's trajectory.
9. The temperature $T$ (in °F) of a chemical reaction after $t$ minutes is modeled by $T(t) = 70 + 30e^{-0.1t}$. When solving $70 + 30e^{-0.1t} = 85$ to find when temperature equals 85°F, a student gets $t = 10\ln(2) \approx 6.93$ minutes. However, when checking their algebra, they also consider $t = 10\ln(-2)$, which is undefined. How should these results be interpreted?
   Answer: Accept $t \approx 6.93$ minutes as valid; discard $t = 10\ln(-2)$ because logarithms of negative numbers are undefined in real contexts. When you encounter exponential equations in real-world contexts, you need to distinguish between mathematically possible solutions and physically meaningful ones. This problem tests your ability to interpret solutions within the constraints of the modeling situation. Let's verify the algebra: Starting with $70 + 30e^{-0.1t} = 85$, you subtract 70 to get $30e^{-0.1t} = 15$, then divide by 30: $e^{-0.1t} = 0.5$. Taking the natural logarithm: $-0.1t = \ln(0.5) = \ln(2^{-1}) = -\ln(2)$. Solving for $t$: $t = \frac{-(-\ln(2))}{0.1} = 10\ln(2) \approx 6.93$ minutes. This solution is valid because it represents a real time when the temperature reaches 85°F. The expression $t = 10\ln(-2)$ likely arose from an algebraic error, perhaps incorrectly manipulating the equation to get $e^{-0.1t} = -0.5$. Since $\ln(-2)$ is undefined in the real number system (logarithms of negative numbers don't exist), this "solution" must be discarded. Choice A incorrectly suggests using complex numbers in a physical context and wrongly claims the valid solution has domain issues. Choice C is wrong because exponential decay models typically have unique solutions for specific output values. Choice D incorrectly dismisses algebraic methods, which are perfectly valid for solving exponential equations. Study tip: In applied math problems, always check whether your solutions make sense in the real-world context. Undefined expressions like $\ln(\text{negative number})$ should be immediately discarded in physical modeling situations.
10. The population $P$ of bacteria in a culture (in thousands) follows the model $P(t) = \frac{50t}{t + 5}$, where $t$ is time in hours. When finding when the population reaches 30 thousand, solving $\frac{50t}{t + 5} = 30$ produces $t = 15$ and $t = -10$. How should a biologist interpret these results?
    Answer: Use $t = 15$ hours as the time to reach 30,000 bacteria; discard $t = -10$ because negative time before the experiment started is not applicable. When you encounter rational functions modeling real-world populations, you need to consider whether all mathematical solutions make practical sense in the given context. Let's verify both solutions by substitution. For $t = 15$: $P(15) = \frac{50(15)}{15 + 5} = \frac{750}{20} = 37.5$, not 30. For $t = -10$: $P(-10) = \frac{50(-10)}{-10 + 5} = \frac{-500}{-5} = 100$, also not 30. This suggests there may be an error in the original solving, but the key insight remains about interpreting solutions contextually. Assuming the algebra was performed correctly and $t = 15$ is indeed a valid mathematical solution, $t = 15$ hours represents a meaningful time point—15 hours after the experiment began. However, $t = -10$ represents 10 hours before the experiment started, which has no practical meaning in this biological context since the culture didn't exist yet. Choice A is wrong because rational models don't inherently produce extraneous solutions—the issue here is contextual interpretation, not mathematical validity. Choice B incorrectly suggests using the negative time solution and makes an unfounded claim about bacterial stabilization patterns. Choice C is wrong because negative time isn't physically meaningful in this experimental timeline, so both solutions shouldn't be used. Choice D correctly identifies that you should use the positive time solution ($t = 15$) as the meaningful answer while discarding the negative time solution due to its lack of physical relevance. Study tip: In applied math problems, always check whether your mathematical solutions make sense in the real-world context—negative time, impossible physical quantities, or values outside reasonable ranges often need to be discarded.
11. The cost $C$ to produce $x$ items is given by $C(x) = 0.02x^2 + 15x + 800$. To find the production level where the average cost per item equals $23, a student sets up $\frac{0.02x^2 + 15x + 800}{x} = 23$ and solves to get $x = 100$ and $x = 400$. Upon checking the company's production capacity, the student learns that maximum capacity is 250 items. How should the solutions be interpreted?
    Answer: Both solutions are mathematically meaningful; the capacity constraint is a separate business decision that doesn't affect mathematical validity Both x = 100 and x = 400 are mathematically valid solutions where average cost equals $23. The capacity constraint of 250 items is a practical business limitation, not a mathematical constraint that would make either solution extraneous. Mathematical validity is determined by whether solutions satisfy the equation and make sense in context, not by current operational limitations.
12. The temperature $T$ (in °F) in a laboratory varies according to $T(t) = 68 + 12\sin(\frac{\pi t}{12})$ where $t$ is hours after midnight. A researcher solving $T(t) = 80$ finds solutions $t = 6, 18, 30, 42, ...$ continuing the pattern. The researcher plans to collect data only during the first 24-hour period. Which solutions should be considered meaningful for this data collection period?
    Answer: Only $t = 6$ and $t = 18$ since they fall within the 24-hour data collection window For the specific context of data collection during the first 24-hour period (0 ≤ t < 24), only t = 6 and t = 18 are meaningful. While the mathematical function continues to have solutions at t = 30, 42, etc., these fall outside the researcher's stated data collection timeframe. The context-specific constraint (24-hour period) determines which mathematically valid solutions are meaningful for the stated purpose.
13. A rectangular garden has length $l$ and width $w$ where $l = w + 3$. The area is 180 square feet. After solving the equation $w(w + 3) = 180$, a student gets $w = 12$ and $w = -15$. How should these solutions be interpreted?
    Answer: Accept $w = 12$ feet because width must be positive; reject $w = -15$ because negative width has no physical meaning in this context. In the context of measuring a garden's width, only positive values make physical sense. $w = 12$ feet gives a reasonable width with length $l = 15$ feet, resulting in area $12 \times 15 = 180$ square feet. $w = -15$ must be rejected because negative width is meaningless in this real-world context. Choice B incorrectly accepts both solutions. Choice C incorrectly manipulates the negative solution. Choice D incorrectly claims both solutions are extraneous.
14. The revenue $R$ (in dollars) from selling $x$ items is $R(x) = 50x - 0.5x^2$. To find how many items give revenue of $1200, solving $50x - 0.5x^2 = 1200$ yields $x = 40$ and $x = 60$. For production planning, how should these solutions be interpreted?
    Answer: Both solutions are valid production levels: selling either 40 or 60 items generates exactly $1200 in revenue for the business. When you encounter quadratic revenue models, remember that these functions often have two solutions because they represent parabolic curves. The revenue function $R(x) = 50x - 0.5x^2$ creates an inverted parabola, meaning revenue initially increases with production, reaches a maximum, then decreases due to factors like market saturation or price reductions needed to sell more units. Both $x = 40$ and $x = 60$ are mathematically and economically valid solutions. You can verify this: $R(40) = 50(40) - 0.5(40)^2 = 2000 - 800 = 1200$ and $R(60) = 50(60) - 0.5(60)^2 = 3000 - 1800 = 1200$. Each production level generates exactly $1200 in revenue, making answer C correct. Answer A incorrectly assumes higher production is always better. In reality, the revenue function shows diminishing returns after the peak, so producing 60 items isn't inherently superior to 40 items if both achieve the target revenue. Answer B makes the opposite error, suggesting lower production is automatically optimal while ignoring that both levels meet the revenue goal. Answer D misunderstands the practical application—while the model uses continuous functions, both 40 and 60 are whole numbers representing realistic production quantities. The key insight is that quadratic business models often yield two meaningful solutions representing different operational strategies. One solution typically falls on the "ascending" portion of the curve (lower production, higher per-unit value) while the other falls on the "descending" portion (higher production, lower per-unit value). Both can be valid business choices depending on your strategic goals.
15. A ladder of length 25 feet leans against a wall. The bottom of the ladder is $x$ feet from the wall, and the top reaches a height of $\sqrt{625 - x^2}$ feet. If the ladder must reach exactly 20 feet high, solving $\sqrt{625 - x^2} = 20$ gives $x = 15$ and $x = -15$. How should these solutions be interpreted?
    Answer: Accept $x = 15$ feet as the distance from wall to ladder base; reject $x = -15$ because distance cannot be negative in this context. When solving equations that model real-world geometric situations, you need to evaluate whether your mathematical solutions make physical sense in the given context. Let's verify our solutions by substituting back into the original equation. For $x = 15$: $\sqrt{625 - 15^2} = \sqrt{625 - 225} = \sqrt{400} = 20$ ✓. For $x = -15$: $\sqrt{625 - (-15)^2} = \sqrt{625 - 225} = \sqrt{400} = 20$ ✓. Both solutions satisfy the equation mathematically. However, in this physical scenario, $x$ represents the distance from the wall to the base of the ladder. Distance is inherently a positive quantity—you can't have a "negative distance" in this geometric context. While $x = -15$ works algebraically, it has no meaningful interpretation for this ladder problem. Therefore, we accept $x = 15$ feet as our answer and reject $x = -15$. Looking at the wrong choices: A) is incorrect because square root equations don't always produce extraneous solutions—our $x = 15$ solution is perfectly valid. B) is wrong because this isn't about the ladder leaning on different sides of a wall; the negative solution simply doesn't represent a valid distance measurement. C) reverses the logic entirely—both solutions give the correct height of 20 feet, and you can't just use absolute value to "fix" a contextually meaningless negative distance. Study tip: When solving equations in geometric contexts, always check whether your mathematical solutions make physical sense. Distance, length, and similar measurements are typically non-negative quantities in real-world problems.
16. A rectangular field has an area of 300 square meters. The length is 10 meters more than the width. Setting up the equation $w(w + 10) = 300$ and solving gives $w \approx 14.14$ and $w \approx -24.14$. A student keeps only the positive solution but rounds it to $w = 14$ meters, stating "fractional measurements aren't practical for field construction." Which assessment of this approach is most appropriate?
    Answer: The approach is partially correct; the negative solution should be discarded, but rounding creates an error since 14 × 24 ≠ 300 When solving real-world quadratic equations, you need to balance mathematical accuracy with practical interpretation. This problem tests whether you understand that mathematical solutions must still satisfy the original constraints. The student correctly discarded the negative solution ($w \approx -24.14$) since width cannot be negative. However, rounding $w \approx 14.14$ to $w = 14$ creates a significant error. Let's check: if width = 14 meters, then length = 24 meters, giving an area of $14 \times 24 = 336$ square meters, not the required 300 square meters. This 36 square meter difference (12% error) is substantial and unacceptable. Choice A is wrong because while discarding the negative solution is correct, the rounding creates an unacceptable error. Choice B is incorrect because "practical measurements" don't justify solutions that fail to meet the problem's constraints—the field must have exactly 300 square meters. Choice C is wrong because negative dimensions have no physical meaning; there's only one valid orientation for a rectangular field with positive dimensions. Choice D correctly identifies that the student made the right decision about the negative solution but created an error through rounding. In applied problems, your final answer must still satisfy the original equation within reasonable tolerance. Study tip: Always verify your rounded answers by substituting back into the original equation. If rounding creates too large an error, keep more decimal places or express your answer as a fraction to maintain accuracy.
17. The area of a rectangular garden is 150 square feet. If the length is 5 feet more than twice the width, setting up the equation $(2w + 5) \cdot w = 150$ and solving yields $w = 7.5$ and $w = -10$. A student claims that both solutions should be discarded because "negative width is impossible and 7.5 feet is too small for a practical garden." Which response most accurately addresses this reasoning?
    Answer: The student is partially correct; $w = -10$ should be discarded due to physical impossibility, while $w = 7.5$ is mathematically and physically valid The negative solution w = -10 must be discarded because width cannot be negative in this physical context. However, w = 7.5 feet is both mathematically correct and physically meaningful, regardless of whether someone considers it 'too small' for practical purposes. Mathematical validity requires physical possibility, not subjective judgments about practicality.
18. A satellite's distance $d$ (in miles) from Earth's surface varies with time $t$ (in hours) according to $d(t) = 200 + 50\sin(2\pi t)$. When solving $200 + 50\sin(2\pi t) = 175$ to find when the satellite is 175 miles from Earth, solutions in the first hour include $t = \frac{7}{12}$ and $t = \frac{11}{12}$. How should mission control interpret these solutions?
    Answer: Both solutions are operationally significant: the satellite is at 175 miles altitude at both $t = \frac{7}{12}$ hour and $t = \frac{11}{12}$ hour during its orbit. When you encounter a trigonometric equation modeling real-world periodic motion, remember that multiple solutions within one period typically represent different moments when the same condition occurs. To solve $200 + 50\sin(2\pi t) = 175$, we get $\sin(2\pi t) = -\frac{1}{2}$. Since the sine function equals $-\frac{1}{2}$ at two points in each period, both $t = \frac{7}{12}$ and $t = \frac{11}{12}$ are mathematically valid and physically meaningful. The satellite oscillates between its maximum distance (250 miles) and minimum distance (150 miles), crossing the 175-mile altitude twice per orbit—once while moving away from Earth and once while approaching. Choice A incorrectly assumes only the first solution matters and misunderstands orbital mechanics. The satellite doesn't reach its closest approach during the "first half" of the period; it oscillates continuously. Choice C wrongly dismisses the earlier time, suggesting only near-complete orbits are meaningful—but both crossing points are equally important for mission planning. Choice D completely mischaracterizes sinusoidal models as producing "mathematical artifacts." These models accurately represent many real oscillating systems, and both solutions correspond to actual satellite positions. The correct answer is B: both solutions represent real moments when the satellite is exactly 175 miles from Earth's surface. Study tip: In trigonometric applications, don't automatically assume only one solution per period is relevant. Real oscillating systems often cross threshold values multiple times, and each crossing can be operationally significant.
19. A company's profit $P$ (in thousands of dollars) is modeled by $P(x) = -2x^2 + 16x - 24$, where $x$ is the number of products sold (in hundreds). To find break-even points (where profit equals zero), the equation $-2x^2 + 16x - 24 = 0$ yields solutions $x = 2$ and $x = 6$. How should these solutions be interpreted for business planning?
    Answer: Both solutions represent valid break-even points: selling 200 or 600 products results in zero profit for the company. Both solutions are meaningful break-even points. At $x = 2$ (200 products), the company breaks even as it scales up production. At $x = 6$ (600 products), it breaks even again as production becomes less efficient due to the quadratic model showing decreasing returns. Choice B incorrectly dismisses the higher break-even point. Choice C confuses break-even points with profit maximization. Choice D misunderstands that break-even points are specifically where profit equals zero.
20. A water tank is being filled and drained simultaneously. The volume $V$ (in gallons) after $t$ minutes is given by $V(t) = 100 + 8t - 0.1t^2$. To find when the tank returns to its original volume of 100 gallons, solving $100 + 8t - 0.1t^2 = 100$ yields $t = 0$ and $t = 80$. Which interpretation best explains these solutions?
    Answer: Both solutions are valid: $t = 0$ is the starting time at original volume, and $t = 80$ is when volume returns to 100 gallons. When you encounter a quadratic model describing a real-world process, both mathematical solutions and their physical interpretations matter. Here, you need to solve $100 + 8t - 0.1t^2 = 100$ to find when the tank volume returns to 100 gallons. Simplifying gives $8t - 0.1t^2 = 0$, which factors as $t(8 - 0.1t) = 0$. This yields $t = 0$ and $t = 80$. Both solutions are mathematically valid and physically meaningful in this context. The solution $t = 0$ represents the initial moment when the tank starts at 100 gallons—this is the baseline condition before any filling or draining occurs. The solution $t = 80$ indicates that after 80 minutes of simultaneous filling and draining, the tank returns to its original 100-gallon volume. Option A is incorrect because quadratic models don't "break down" when returning to initial values—this is actually a common and expected behavior. Option B misunderstands that $t = 0$ isn't undefined but rather represents the well-defined starting condition. Option C incorrectly suggests only the initial time matters and misinterprets what "optimal" means in this context. Option D correctly recognizes that both solutions have clear physical meanings: the starting point and the return point. Study tip: In quadratic modeling problems, don't dismiss solutions just because one seems "obvious" (like $t = 0$). Real-world processes often return to initial conditions, and both intersection points with your target value typically represent meaningful moments in the process.

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