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SOLUTION: When the area of a rectangle is 3 times the length and the lenght is 1 more than twice the width, what are the demensions? I know somthing is missing in my formula. Please help
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-> SOLUTION: When the area of a rectangle is 3 times the length and the lenght is 1 more than twice the width, what are the demensions? I know somthing is missing in my formula. Please help
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Question 395429
:
When the area of a rectangle is 3 times the length and the lenght is 1 more than twice the width, what are the demensions?
I know somthing is missing in my formula. Please help? Thanks.
A = L*W
W = x
L = 2x + 1
x(2x + 1) = 3(2x + 1)
2x^2 + x = 6x + 3
2x^2 + 5x + 3
(2x + 3)(x+1)
Answer by
stanbon(69061)
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When the area of a rectangle is 3 times the length and the lenght is 1 more than twice the width, what are the demensions?
I know somthing is missing in my formula. Please help? Thanks.
A = L*W
W = x
L = 2x + 1
x(2x + 1) = 3(2x + 1)
2x^2 + x = 6x + 3
---
You missed the signs at this point:
2x^2 - 5x - 3 = 0
2x^2 -6x+x -3 = 0
---
2x(x-3)+(x-3) = 0
-----------------
(x-3)(2x+1) = 0
---
Positive solution:
x = 3 (width)
2x+1 = 7 (length)
=============
Cheers,
Stan H.
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