SOLUTION: A room's length is 3 feet less than twice its width. The area of the room is 135 sq feet. What are the room's dimensions?
This is what I have done so far
A= LxW
135 = (2w-3
Algebra ->
Polynomials-and-rational-expressions
-> SOLUTION: A room's length is 3 feet less than twice its width. The area of the room is 135 sq feet. What are the room's dimensions?
This is what I have done so far
A= LxW
135 = (2w-3
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Question 185102This question is from textbook
: A room's length is 3 feet less than twice its width. The area of the room is 135 sq feet. What are the room's dimensions?
This is what I have done so far
A= LxW
135 = (2w-3)(W)
135 = (2w squared - 3W) This question is from textbook
You can put this solution on YOUR website! A room's length is 3 feet less than twice its width. The area of the room is 135 sq feet. What are the room's dimensions?
This is what I have done so far
A= LxW
135 = (2w-3)(W)
135 = (2w squared - 3W)
135 = 2w^2-3w
2w^2-3w-135 = 0
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Use the quadratic formula to get:
w = [3 +- sqrt(9 - 4*2*-135)]/4
Positive solution:
w = [3 + sqrt(1089)/4
w = [3 + 33]/4
w = 9 ft
L = 2W-3 = 2*9-3 = 15 ft.
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Cheers,
Stan H.
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