w(2w+3) = 77
now just solve for w
The length of a rectangle is 3 m more than double the width, and the area of the rectangle is 77 m^2. Find the dimensions of the rectangle.
4 answers
Expanding the left side of the equation, we get:
2w^2 + 3w - 77 = 0
Using the quadratic formula:
w = (-3 ± sqrt(3^2 - 4(2)(-77))) / (2(2))
w = (-3 ± sqrt(617)) / 4
Ignoring the negative solution (since width cannot be negative):
w = (-3 + sqrt(617)) / 4 ≈ 4.216
Then, using the given relationship between length and width:
length = 2w + 3 ≈ 11.432
Therefore, the dimensions of the rectangle are approximately 4.216 m by 11.432 m.
2w^2 + 3w - 77 = 0
Using the quadratic formula:
w = (-3 ± sqrt(3^2 - 4(2)(-77))) / (2(2))
w = (-3 ± sqrt(617)) / 4
Ignoring the negative solution (since width cannot be negative):
w = (-3 + sqrt(617)) / 4 ≈ 4.216
Then, using the given relationship between length and width:
length = 2w + 3 ≈ 11.432
Therefore, the dimensions of the rectangle are approximately 4.216 m by 11.432 m.
The bot is wrong again.
oobleck got the start correct, and it was a simple thing to finish it,
but the bot has problem with basic arithmetic
2w^2 + 3w - 77 = 0 factors to
(2w - 11)(w + 7) = 0
w = 5.5 or w = -7
rejecting the negative width, we have
width = 5.5 and the length is 14
check: 5.5*14 = 77
oobleck got the start correct, and it was a simple thing to finish it,
but the bot has problem with basic arithmetic
2w^2 + 3w - 77 = 0 factors to
(2w - 11)(w + 7) = 0
w = 5.5 or w = -7
rejecting the negative width, we have
width = 5.5 and the length is 14
check: 5.5*14 = 77
I apologize for my mistake in the previous response. You are correct in your calculation, and the dimensions of the rectangle are indeed 5.5 m by 14 m. Thank you for catching that error.
2 answers