SOLUTION: The length of a rectangle is 3 cm more than 4 times its width. If the area of the rectangle is 79 cm^2, find the dimensions of the rectangle to the nearest thousandth. To start

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Question 46778: The length of a rectangle is 3 cm more than 4 times its width. If the area of the rectangle is 79 cm^2, find the dimensions of the rectangle to the nearest thousandth.
To start should it be - using x for the width and l for the length
l = s(4) + 3
Is there anyone who can help and thank you.
Found 2 solutions by Nate, Earlsdon:
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
width = w
length = 4w + 3
lw+=+AREA
w%284w+%2B+3%29+=+79
4w%5E2+%2B+3w+-+79+=+0
w+=+%28-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
w+=+%28-3+%2B-+sqrt%28+9+-+4%2A4%2A%28-79%29+%29%29%2F%282%2A4%29+
width: %28-3+%2B-+sqrt%28+1273+%29%29%2F%288%29+ can not have negative measurement
width: %28-3+%2B+sqrt%28+1273+%29%29%2F%288%29+
length %28-3+%2B+sqrt%28+1273+%29%29%2F%282%29+%2B+3+

Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
Let the length of the rectangle be L and its width be W. Acoording to the story:
L = 4W+3 cm and the area is:
L X W = 79 cm^2 Substitute the L in the second equation with its equivalent (4W+3) and solve for W.
(4W+3) X W = 79 cm^2 Simplify and solve for W.
4W%5E2%2B3W+=+79 Subtract 79 from both sides of the equation.
4W%5E2%2B3W-79+=+0 Use the quadratic formula to solve for W:W+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
W+=+%28-3%2B-sqrt%283%5E2-4%284%29%28-79%29%29%29%2F2%284%29
W+=+%28-3%2B-sqrt%289%2B1264%29%29%2F8
W+=+%28-3%2B-sqrt%281273%29%29%2F8 Only the positive answer is meaningful for the width.
W+=+4.09 and:
L+=+4W%2B3
L+=+4%284.09%29%2B3
L+=+16.36%2B3
L+=+19.36