Asked by Marysvoice
The width of a rectangle is 2ft less than its length. The area is 8ft^2. Find the length and width.
I can't seem to come up with anything that equals 64 sq ft.
Do I start with area 64sqft.=W x L-2.
I feel like I should be coming up with something around 7 times 9 but I can't seem to come up with 64, ideas?
I can't seem to come up with anything that equals 64 sq ft.
Do I start with area 64sqft.=W x L-2.
I feel like I should be coming up with something around 7 times 9 but I can't seem to come up with 64, ideas?
Answers
Answered by
Damon
In the first statement you say area is 8 ft^2
Then you say it is 64 ft^2
If it is eight, it is obvious by inspection
2*4 = 8
If it is 64, we better try to do it right
W * (W+2) = 64
W^2 + 2 W - 64 = 0
W = -2/2 +/- (1/2)sqrt (4 + 4*64)
W = -1 +/- (1/2) sqrt ( 4*65)
W = -1 +/- sqrt(65)
W = -1 - sqrt 65 forget this solution, no negative width
W = 8.06 - 1 = 7.06
W + 2 = 9.06
so indeed very close to 7 and 9
Then you say it is 64 ft^2
If it is eight, it is obvious by inspection
2*4 = 8
If it is 64, we better try to do it right
W * (W+2) = 64
W^2 + 2 W - 64 = 0
W = -2/2 +/- (1/2)sqrt (4 + 4*64)
W = -1 +/- (1/2) sqrt ( 4*65)
W = -1 +/- sqrt(65)
W = -1 - sqrt 65 forget this solution, no negative width
W = 8.06 - 1 = 7.06
W + 2 = 9.06
so indeed very close to 7 and 9
Answered by
Reiny
x(x+2) = 64
If you are studying quadratic equations, you should be able to solve this one.
It comes out as a radical
(7.0622 and 9.0622)
If you are studying quadratic equations, you should be able to solve this one.
It comes out as a radical
(7.0622 and 9.0622)
Answered by
Marysvoice
You are right. I should sleep and try again later. I am starting to make silly mistakes.
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