SOLUTION: Could you please help me with an application using a quadratic equation? The question is: The width of a rectangular parking lot is 51 ft less than its length. Determine the d

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Could you please help me with an application using a quadratic equation? The question is: The width of a rectangular parking lot is 51 ft less than its length. Determine the d      Log On


   

Question 111414This question is from textbook
: Could you please help me with an application using a quadratic equation?
The question is: The width of a rectangular parking lot is 51 ft less than its length. Determine the dimensions of the parking lot if it measures 250 ft diagonally.
What I've got so far:
x^2+(x-51)^2=250^2
x^2+x^2-102x+2601=62500
2x^2-102x-59899=0
factor out 2
x^2-51x-29949.5=0
I stopped there because I got a decimal number (29949.5) and that seems wrong.
Please help!
Thank you in advance!
Brenda This question is from textbook

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Your work so far is good!
Looking at your quadratic equation:2x%5E2-102x-59899+=+0, I don't think that you can factor it so you have to use the quadratic formula to solve:
x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
It looks quite formidable with the numbers you have, but it's easy enough with a good calculator.
x+=+%28-%28-102%29%2B-sqrt%28%28-102%292-4%282%29%28-59899%29%29%29%2F2%282%29 and you'll get:
x+=+200.43
x+=+-149.43
Take only the positive answer since lengths cannot be negative.
The length (x) is 200.43 feet.
The width (x-51) is 149.43 feet.