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It takes Greg and Nikki a total of 40 minutes to paint a room if they work together. If Greg works alone, he takes 18 minutes longer to paint the room than Nikki takes if she works alone.



When $x$ represents the number of minutes it takes Nikki to paint the room working alone, the situation is represented by this equation:

$\frac{1}{x}+\frac{1}{x+18}=\frac{1}{40}$



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

A$x=0$
B$x=-10$
C$x=-18$
D$x=-72$

Verified Answer :

  • The given rational equation is simplified to the quadratic equation \(x^2 - 62x - 720 = 0\).
  • Solving this quadratic equation yields two values: \(x = 72\) and \(x = -10\).
  • Because \(x\) represents time, it must be positive; therefore, \(x = -10\) is a solution to the equation that must be discarded due to the problem's context.
  • The correct discarded solution is \(\boxed{-10}\).

Explanation

  1. Analyze the Equation and Context

To solve the problem, we first consider the given rational equation:
\[\frac{1}{x} + \frac{1}{x+18} = \frac{1}{40}\]
where \(x\) represents the time in minutes it takes Nikki to paint the room alone. Since this is a time duration, the context requires \(x > 0\).

First, we clear the denominators by multiplying both sides of the equation by the common denominator \(40x(x+18)\):
\[40(x+18) + 40x = x(x+18)\]
Expanding both sides:
\[40x + 720 + 40x = x^2 + 18x\]
\[80x + 720 = x^2 + 18x\]
Rearranging into a standard quadratic form \(ax^2 + bx + c = 0\):
\[x^2 - 62x - 720 = 0\]

  1. Solve the Quadratic Equation

Next, we solve the quadratic equation \(x^2 - 62x - 720 = 0\). Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[x = \frac{62 \pm \sqrt{(-62)^2 - 4(1)(-720)}}{2(1)}\]
\[x = \frac{62 \pm \sqrt{3844 + 2880}}{2}\]
\[x = \frac{62 \pm \sqrt{6724}}{2}\]
Since \(\sqrt{6724} = 82\), the potential solutions are:
\[x = \frac{62 \pm 82}{2}\]
The two solutions are:

  1. \(x_1 = \frac{144}{2} = 72\)
  2. \(x_2 = \frac{-20}{2} = -10\)
  1. Identify the Discarded Solution

We have found two mathematical solutions: \(x = 72\) and \(x = -10\).
Given the context that \(x\) represents a duration of time, it must be positive (\(x > 0\)). Therefore, the solution \(x = 72\) is valid for this scenario, while \(x = -10\) is a valid mathematical solution to the equation but must be discarded because it is negative and cannot represent a duration of time.

Comparing this to the provided options (0, -10, -18, -72), the value \(-10\) is the only one that is both a mathematical solution to the equation and must be discarded due to the context of the problem.

Examples

In many real-world scenarios, such as engineering, finance, or physics, mathematical modeling often leads to equations that describe a physical situation. However, the resulting equations may yield solutions that, while mathematically correct, do not make sense in reality (e.g., negative time, negative mass, or negative distance). By identifying and discarding these "extraneous" or contextually impossible solutions, we ensure that our final answers accurately reflect the physical world. For example, if you are modeling the time it takes for a resource to be consumed, a negative solution for time would be discarded as it falls outside the feasible domain defined by the problem.

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