SOLUTION: The area of a rectangle is 52 ft sq., and the length of the rectangle is 5 ft less than twice the width. Find the dimensions of the rectangle. Please help, I am so confused on this

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Question 805479: The area of a rectangle is 52 ft sq., and the length of the rectangle is 5 ft less than twice the width. Find the dimensions of the rectangle. Please help, I am so confused on this problem.
Found 2 solutions by ankor@dixie-net.com, josmiceli:
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
Write an equation for exactly what it says.
:
The area of a rectangle is 52 ft sq.,
L * W = 52
:
and the length of the rectangle is 5 ft less than twice the width.
L = (2W-5)
In the first equation, replace L with (2W-5), and we have
(2W-5)*W = 52
multiply by W inside the brackets
2W^2 - 5W = 52
Subtract 52 from both sides, we have a quadratic equation
2W^2 - 5W - 52 = 0
You can use the quadratic formula to find w, but this will factor to
(2W-13)(W+4) = 0
The positive solution is all we want her
2W = 13
W = 13/2
W = 6.5 ft is the width
We know
L = 2W-5
L = 2(6.5) - 5
L = 13 - 5
L = 8 ft is the length
:
Check this by finding the area with these values
8 * 6.5 = 52
:
Do you understand this now? C

Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website!
Let = the length
Let = the width
Let = the area
-------------------
For a rectangle,
You are given that ft2, so
(1)
You are also given that
(2)
-----------------
Substitute (2) into (1)
(1)
(1)
(1)
Use the quadratic formula

( I can't use the minus square root of )

and, since
(2)
(2)
(2)
(2)
The dimensions are 6.5' x 8'
check:
(1)
(1)
(1)
OK