SOLUTION: The length of a rectangle is 1 cm more than 5 times its width. If the area of the rectangle is 79 cm2, find the dimensions of the rectangle to the nearest thousandth. I get so

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Question 120938: The length of a rectangle is 1 cm more than 5 times its width. If the area of the rectangle is 79 cm2, find the dimensions of the rectangle to the nearest thousandth.
I get so confused on these questions. I am just lost on this. Please help!!!
Found 2 solutions by checkley71, stanbon:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
LENGTH=5W+1
AREA=LW
79=(5W+1)W
79=5W^2+W=0
5W^2+W-79=0
USING THE QUADRATIC EQUATION x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ WE GET.
W=(-1+-SQRT[1^2-4*5*-79])/2*5
W=(-1+-SQRT[1+1580])/10
W=(-1+-SQRT1581)/10
W=(-1+-39.76)/10
W=(-1+39.76)/10
W=38.76/10
W=3.876 ANSWER.
W=(-1-39.76)/10
W=-40.76/10
W=-4.076 ANSWER.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 1 cm more than 5 times its width. If the area of the rectangle is 79 cm2, find the dimensions of the rectangle to the nearest thousandth.
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Formula: Area = (length)*(width)
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Let the width be "x" cm. ; Then length = "5x+1" cm.
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Your Problem:
79 cm^2 = (5x+1)*x
79 = 5x^2+x
5x^2+x-79=0
x = [-1 +- sqrt(1-4*5*-79)]/10
x = [-1 +- sqrt(1581)]/10
x = [-1 +- 39.7618]/10
x = 3.8762 cm (The width of the rectangle)
5x+1 = 20.3809 cm (The length of the rectangle)
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Cheers,
Stan H.
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