585 views
3 votes
When the system of equations is solved for x, the two solutions are x = ? and x = ?. Which of those solutions will make the width of the garage, x + 15, negative, and which of those solutions will make the garage length, 2x – 19, negative? Part C question: What are the dimensions of the parking garage?

User Nhoxbypass
by
7.8k points

1 Answer

5 votes

Final answer:

In a quadratic equation, sometimes only one of the two solutions for x is feasible for real-world applications. The feasible solution is the one that does not result in negative dimensions when substituted into the expressions for the garage's width and length.

Step-by-step explanation:

When solving quadratic equations, often two solutions for the variable x are found. However, not both solutions may be feasible in practical scenarios, such as dimensions of real-world objects like a parking garage. If the two solutions for x are given, and we need to determine which solution will make the width (x + 15) and length (2x - 19) of a garage negative, we must consider each solution separately.

If the first solution for x when substituted into x + 15 yields a negative result, this solution is not feasible as a physical width cannot be negative. Similarly, if the second solution for x when substituted into 2x - 19 yields a negative result, this solution is not feasible for a physical length of a garage.

For the dimensions of the parking garage, we simply substitute the feasible solution for x into the width and length expressions to get the positive dimensions of the garage as the length and width can only be positive in real-life applications.

User Maazza
by
7.7k points