Extraneous Solutions in Systems - Math 2

1. When solving $25 + 30\ln(t) = 85$ for the stopping time, a student gets $t = e^2 \approx 7.39$ minutes. However, the reaction chamber can only maintain controlled conditions for a maximum of 6 minutes. What is the most accurate interpretation of this mathematical solution?
   Answer: The solution is extraneous because it falls outside the domain of safe operation for this experimental setup. The mathematical solution t = e² ≈ 7.39 minutes is correct for the equation 25 + 30ln(t) = 85. Solving: 30ln(t) = 60, so ln(t) = 2, giving t = e². However, since the reaction chamber can only maintain controlled conditions for 6 minutes maximum, any solution beyond t = 6 is outside the valid domain for this physical system. Therefore, t ≈ 7.39 is extraneous in this context because it represents a time when the controlled reaction conditions no longer exist.
2. To find when the population reaches 750 bacteria, a student solves $100e^{0.1t} = 750$ and obtains $t = 10\ln(7.5) \approx 20.1$ hours. What is the most appropriate interpretation of this solution?
   Answer: The solution is extraneous because the model becomes invalid once the population reaches the 500-bacteria limit. The mathematical solution t = 10ln(7.5) ≈ 20.1 hours is algebraically correct. However, the problem states that the population cannot exceed 500 bacteria due to nutrient depletion. To find when this limit is reached: $100e^{0.1t}$ = 500, so $e^{0.1t}$ = 5, giving t = 10ln(5) ≈ 16.1 hours. Since the exponential model N(t) = $100e^{0.1t}$ is only valid until nutrient depletion occurs at t ≈ 16.1 hours, any solution beyond this point (like t ≈ 20.1) is extraneous because the model no longer applies to the physical system.
3. A student solving $|t^2 - 8t + 12| = 20$ to find when the storm is 20 km away obtains four potential solutions after considering both cases of the absolute value. Given the equipment's range limitation, which conclusion is most appropriate?
   Answer: All solutions are extraneous because 20 km exceeds the equipment's 15 km maximum detection range. When working with mathematical models in real-world contexts, you must always consider the practical limitations of the scenario, not just the mathematical solutions. The equation $|t^2 - 8t + 12| = 20$ does indeed have mathematical solutions when you solve both cases: $t^2 - 8t + 12 = 20$ and $t^2 - 8t + 12 = -20$. However, the key insight is recognizing that the weather station's equipment can only detect storms within 15 km. Since we're trying to find when the storm is 20 km away, and 20 > 15, the equipment simply cannot detect the storm at this distance. This makes all mathematical solutions irrelevant to the physical situation—they're all extraneous because they describe a scenario the equipment cannot measure. Choice A incorrectly assumes all mathematical solutions have physical meaning. Choice B focuses on the sign of the expression inside the absolute value, but misses that the absolute value already ensures distance is positive—the real issue is the range limitation. Choice D suggests only the first occurrence matters, but this ignores that none of the solutions are practically meaningful due to the equipment's limitations. The correct answer is C because when a distance exceeds the detection range, the equipment cannot provide meaningful data, regardless of whether mathematical solutions exist. Study tip: In applied math problems, always check whether your mathematical solutions make sense within the given constraints and physical limitations of the scenario.
4. When solving $\frac{100}{x^2 + 1} = 8$ to find where concentration equals 8 ppm, a student obtains $x = \pm\sqrt{11.5} \approx \pm 3.39$ km. Considering the regulatory constraint, which analysis is correct?
   Answer: Both solutions are extraneous because 8 ppm violates the 5 ppm regulatory limit beyond 2 km. When analyzing solutions to real-world mathematical models, you must consider both the mathematical validity and the physical constraints of the problem. This question tests whether you can distinguish between mathematically correct solutions and practically meaningful ones within given parameters. The student correctly solved $\frac{100}{x^2 + 1} = 8$ by cross-multiplying to get $100 = 8(x^2 + 1)$, then $100 = 8x^2 + 8$, so $92 = 8x^2$, giving $x^2 = 11.5$ and $x = \pm\sqrt{11.5} \approx \pm 3.39$. The mathematics is sound. However, the regulatory constraint states that concentration must stay below 5 ppm beyond the 2 km mark. Since both $x \approx 3.39$ km and $x \approx -3.39$ km are beyond 2 km from the source (in absolute distance), and 8 ppm exceeds the 5 ppm limit, both solutions represent violations of environmental regulations. This makes both solutions extraneous to the practical problem, confirming answer C. Answer A incorrectly focuses only on the negative distance issue while ignoring the regulatory violation. Answer B wrongly suggests the positive solution meets regulatory requirements when 8 ppm actually exceeds the 5 ppm limit. Answer D correctly identifies both as regulatory violations but incorrectly calls them "meaningful" when they're actually extraneous given the constraints. Study tip: In applied math problems, always check whether your mathematical solutions satisfy all given constraints and conditions—not just the equation you solved.
5. When solving $-4.9t^2 + 24.5t = 29.4$ to find when the projectile hits the target, a student obtains $t = 2$ seconds and $t = 3$ seconds. However, the projectile explodes 2.8 seconds after launch. Which statement correctly interprets these mathematical solutions?
   Answer: Only t = 2 is meaningful because t = 3 occurs after the projectile explodes and no longer exists as a physical object. When solving quadratic equations in real-world contexts, you must always consider the physical constraints of the problem, not just the mathematical solutions. The equation $-4.9t^2 + 24.5t = 29.4$ gives you two mathematically valid times when the projectile would be at the target height, but physics limits which solutions are meaningful. The projectile follows a parabolic path, reaching the target height twice: once on its way up (t = 2 seconds) and once on its way down (t = 3 seconds). However, since the projectile explodes at 2.8 seconds, it ceases to exist as a physical object after that point. Therefore, only t = 2 seconds represents a real intersection with the target, while t = 3 seconds describes what would happen in an ideal world where the projectile continued its trajectory uninterrupted. Choice A is wrong because t = 3 occurs after the 2.8-second explosion time, not before. Choice B incorrectly calls both solutions extraneous—t = 2 is perfectly valid since it occurs before the explosion. Choice C misunderstands the physics; there's nothing about "maximum effectiveness" that makes the upward trajectory less meaningful, and t = 3 would actually be the meaningful solution if both times were physically possible. When working with projectile motion problems, always check your mathematical solutions against the given constraints. Time limits, physical barriers, or other real-world factors can eliminate otherwise valid mathematical solutions. The math gives you candidates; the physics tells you which ones make sense.
6. To find when the drug becomes ineffective, a student solves $20te^{-0.2t} = 2$ and finds two solutions: $t \approx 0.11$ hours and $t \approx 18.5$ hours. Which interpretation of these solutions is most appropriate?
   Answer: Only t ≈ 18.5 is meaningful because t ≈ 0.11 occurs before the drug has been fully absorbed into the system. The equation $20te^{-0.2t}$ = 2 simplifies to $te^{-0.2t}$ = 0.1. This equation has two solutions because C(t) = $20te^{-0.2t}$ increases initially then decreases. At t ≈ 0.11 hours (about 6.6 minutes), the concentration is rising and briefly equals 2 mg/L, but this occurs during the absorption phase before peak effectiveness. At t ≈ 18.5 hours, the concentration has decreased back to 2 mg/L during elimination. Only t ≈ 18.5 is meaningful for determining when the drug becomes ineffective, as t ≈ 0.11 represents a time when the drug is still becoming effective.
7. A physicist models wave interference using $|x - 3| + |y + 2| = 7$ and $x^2 + y^2 = 25$. When solving this system by cases based on the absolute value expressions, which solutions require the most careful verification?
   Answer: Solutions that lie exactly on the boundaries where expressions equal zero The absolute value equation requires case analysis. The critical boundaries are $x = 3$ and $y = -2$. When $x = 3$: $|y + 2| = 7$, so $y = 5$ or $y = -9$. Checking in $x^2 + y^2 = 25$: $(3)^2 + (5)^2 = 34 \neq 25$ and $(3)^2 + (-9)^2 = 90 \neq 25$. When $y = -2$: $|x - 3| = 7$, so $x = 10$ or $x = -4$. Checking: $(10)^2 + (-2)^2 = 104 \neq 25$ and $(-4)^2 + (-2)^2 = 20 \neq 25$. Solutions on these boundaries are most prone to errors because students often assume the case conditions are automatically satisfied, but they must still satisfy both original equations. These boundary cases require verification in both the case condition and the other constraint.
8. An engineer models the intersection of two pipes using the system $x^2 + y^2 = 25$ and $y = x^2 - 7$, where measurements are in centimeters. Due to manufacturing constraints, only solutions where both $x \geq 0$ and $y \geq -3$ are physically realizable. How many of the mathematical solutions represent actual intersection points?
   Answer: Exactly two mathematical solutions are physically realizable intersection points Substituting $y = x^2 - 7$ into $x^2 + y^2 = 25$: $x^2 + (x^2 - 7)^2 = 25$. This gives $x^2 + x^4 - 14x^2 + 49 = 25$, or $x^4 - 13x^2 + 24 = 0$. Let $u = x^2$: $u^2 - 13u + 24 = 0$, so $(u-8)(u-5) = 0$. Thus $x^2 = 8$ or $x^2 = 5$, giving $x = \pm 2\sqrt{2}$ or $x = \pm \sqrt{5}$. For $x = 2\sqrt{2}$: $y = 8 - 7 = 1 \geq -3$ ✓. For $x = \sqrt{5}$: $y = 5 - 7 = -2 \geq -3$ ✓. The negative $x$ values violate $x \geq 0$. Therefore, exactly two solutions are physically realizable.
9. A biologist models population interactions between two species using $x^2 - y^2 = 16$ and $xy = 15$, where $x$ and $y$ represent populations in hundreds. Given that both populations must be positive for the ecosystem model to be valid, what can be concluded about the mathematical solutions?
   Answer: Exactly half of the mathematical solutions are biologically meaningful ecosystem states From $xy = 15$, we have $y = \frac{15}{x}$. Substituting into $x^2 - y^2 = 16$: $x^2 - \frac{225}{x^2} = 16$. Multiplying by $x^2$: $x^4 - 16x^2 - 225 = 0$. Let $u = x^2$: $u^2 - 16u - 225 = 0$. Using the quadratic formula: $u = \frac{16 \pm \sqrt{256 + 900}}{2} = \frac{16 \pm \sqrt{1156}}{2} = \frac{16 \pm 34}{2}$. So $u = 25$ or $u = -9$. Since $u = x^2 \geq 0$, we have $x^2 = 25$, giving $x = \pm 5$. When $x = 5$, $y = 3$; when $x = -5$, $y = -3$. Since populations must be positive, only $(5,3)$ is biologically meaningful. This is exactly half of the two mathematical solutions.
10. A student finds that the particle's velocity $v(t) = 3t^2 - 12t + 9 = 0$ when $t = 1$ and $t = 3$ seconds. Upon further analysis, which conclusion about these solutions is most accurate?
    Answer: Both solutions are meaningful as they represent times when the particle momentarily stops before changing direction. The velocity equation v(t) = 3t² - 12t + 9 = 0 factors as 3(t² - 4t + 3) = 3(t-1)(t-3) = 0, confirming t = 1 and t = 3. To verify these represent direction changes, we check the sign of velocity: v(0) = 9 > 0, v(2) = 3(4) - 12(2) + 9 = -3 < 0, and v(4) = 3(16) - 12(4) + 9 = 21 > 0. The particle moves forward until t = 1, then backward from t = 1 to t = 3, then forward again after t = 3. Both solutions represent genuine times when the particle stops and changes direction.
11. To find when the population reaches 800 fish, a student solves the equation and gets $t = 2\ln(4) \approx 2.77$ years. What is the most appropriate interpretation of this solution?
    Answer: The solution is extraneous because it assumes continuous population growth without accounting for the 4-month seasonal absence period. When working with mathematical models that have domain restrictions, you must carefully consider whether your algebraic solution falls within the valid range of the model. The student correctly solved $\frac{1000}{1 + 4e^{-0.5t}} = 800$ to get $t = 2\ln(4) \approx 2.77$ years. However, this model only applies during 8 months of each year (March through October), meaning the fish are absent for 4 months annually due to migration. The solution $t = 2.77$ years assumes the population grows continuously for the entire duration, but the model explicitly states it's only valid for 8 months per year. During the 4-month absence periods, the model doesn't apply at all. This makes the solution extraneous because it ignores a fundamental constraint of the problem. Answer choice A incorrectly treats the solution as valid without considering the domain restriction. Choice B focuses on a general claim about logistic model accuracy that isn't supported by the passage. Choice C suggests the solution represents a hypothetical scenario without migration, but the student wasn't asked to solve a modified problem—they were working with the given seasonal model. Choice D correctly identifies that the solution is extraneous because it assumes continuous growth while ignoring the 4-month seasonal absence periods that are built into the model's validity constraints. Study tip: Always check whether your algebraic solution respects all constraints given in the problem, especially domain restrictions for real-world models.
12. Setting up the equation $w(2w + 3) = 54$ where w is the width, a student solves to get $w = 4.5$ feet and $w = -6$ feet. Considering the zoning restriction, which analysis is correct?
    Answer: Only w = 4.5 is meaningful since negative width is impossible and the resulting length of 12 feet satisfies zoning requirements. Expanding w(2w + 3) = 54 gives 2w² + 3w - 54 = 0. Using the quadratic formula: w = (-3 ± √(9 + 432))/4 = (-3 ± 21)/4, so w = 4.5 or w = -6. Since width cannot be negative, w = -6 is extraneous. For w = 4.5, the length is 2(4.5) + 3 = 12 feet. Both dimensions (4.5 and 12 feet) are less than the 15-foot zoning limit, so w = 4.5 is a meaningful solution that satisfies all constraints.
13. An economist models market equilibrium using $p = \sqrt{100 - q^2}$ and $p^2 + 2pq = 75$, where $p$ is price and $q$ is quantity. Both must be non-negative for economic validity. Which mathematical solution(s) should be rejected as economically meaningless?
    Answer: Any solution where substitution back into the first equation yields a contradiction From $p = \sqrt{100 - q^2}$, we get $p^2 = 100 - q^2$. Substituting into $p^2 + 2pq = 75$: $100 - q^2 + 2pq = 75$, so $2pq = q^2 - 25$. Since $p = \sqrt{100 - q^2}$, we have $2q\sqrt{100 - q^2} = q^2 - 25$. If $q \neq 0$, then $2\sqrt{100 - q^2} = q - \frac{25}{q}$. Squaring both sides: $4(100 - q^2) = (q - \frac{25}{q})^2 = q^2 - 50 + \frac{625}{q^2}$. This leads to a quartic equation. However, after solving, we must substitute solutions back into $p = \sqrt{100 - q^2}$ to verify they actually satisfy this constraint, as squaring can introduce extraneous solutions.

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