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It would take Charlene and Gina 18 minutes to organize Charlene's comic book collection if they work together. If Gina works alone, it would take her 15 minutes longer than it would take Charlene working alone.



When $x$ represents the number of minutes it takes Charlene to organize the comic book collection working alone, the situation is represented by this equation:

$\frac{1}{x}+\frac{1}{x+15}=\frac{1}{18}$



Which value is a solution of the rational equation that must be discarded because of the context?


A$x=0$
B$x=-9$
C$x=-15$
D$x=-30$

Verified Answer :

  • Simplify the equation \(\frac{1}{x} + \frac{1}{x+15} = \frac{1}{18}\) to obtain the quadratic form \(x^2 - 21x - 270 = 0\).
  • Solve the quadratic equation to find two roots: \(x = 30\) and \(x = -9\).
  • Determine that while \(x = -9\) satisfies the algebraic equation, it is invalid because time must be positive.
  • Identify the discarded value as \(\boxed{-9}\).

Explanation

  1. Analyze the Equation and Find the Quadratic Form

The given rational equation is \(\frac{1}{x} + \frac{1}{x+15} = \frac{1}{18}\), where \(x\) represents the time (in minutes) it takes Charlene to organize the collection working alone. Because \(x\) represents a duration of time, it must be a positive value (\(x > 0\)). To find the solutions, we first clear the fractions by multiplying the entire equation by the common denominator \(18x(x+15)\):
\[18(x+15) + 18x = x(x+15)\]
Expanding the terms, we have:
\[18x + 270 + 18x = x^2 + 15x\]
\[36x + 270 = x^2 + 15x\]
Moving all terms to one side to form a quadratic equation gives:
\[x^2 - 21x - 270 = 0\]

  1. Solve the Quadratic Equation

To solve the quadratic equation \(x^2 - 21x - 270 = 0\), we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=1\), \(b=-21\), and \(c=-270\):
\[x = \frac{21 \pm \sqrt{(-21)^2 - 4(1)(-270)}}{2(1)}\]
\[x = \frac{21 \pm \sqrt{441 + 1080}}{2}\]
\[x = \frac{21 \pm \sqrt{1521}}{2}\]
\[x = \frac{21 \pm 39}{2}\]
The two possible solutions for \(x\) are:
\[x_1 = \frac{21 + 39}{2} = \frac{60}{2} = 30\]
\[x_2 = \frac{21 - 39}{2} = \frac{-18}{2} = -9\]

  1. Evaluate Solutions and Identify the Discarded Value

In the context of the problem, \(x\) represents the time it takes Charlene to organize the collection. Since a duration of time cannot be negative, \(x = -9\) is an extraneous solution that must be discarded. Comparing this with the given options, we identify that \(x = -9\) is indeed a solution to the algebraic equation but is invalid as a physical time duration. Thus, the value to be discarded is \(-9\).

Examples

This problem demonstrates how to model real-world rates and times using rational equations. Just as we must discard negative time values in this task, engineers often use rational models to analyze system throughput, such as pipe flow or data transmission rates, where physical constraints like non-negative flow or processing time must be respected. Recognizing and discarding mathematically valid but physically impossible roots is a crucial skill in applied mathematics.

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