SOLUTION: The length of a rectangle is 5 cm more than 4 times its width. If the area of the rectangle is 80 cm2, find the dimensions of the rectangle to the nearest thousandth.
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Question 95532: The length of a rectangle is 5 cm more than 4 times its width. If the area of the rectangle is 80 cm2, find the dimensions of the rectangle to the nearest thousandth.
Please Help! Found 2 solutions by stanbon, checkley71:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The length of a rectangle is 5 cm more than 4 times its width. If the area of the rectangle is 80 cm2, find the dimensions of the rectangle to the nearest thousandth.
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Let the width be "x" cm
Then the length is "4x+5" cm
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EQUATION:
x(4x+5) = 80 cm^2
4x^2+5x-80 = 0
Use the quadratic formula:
x = [-5 +- sqrt(5^2-4*4*-80)]/8
x = [-5 +- sqrt(1305)]/8
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Positive answer:
x = [-5 + 36.12478..]/8
x = 3.890597..cm (width of the rectangle)
4x+5 = 20.56239..cm (length of the rectangle)
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Cheers,
Stan H.
You can put this solution on YOUR website! width=x
length=5+4x
(x)(5+4x)=80
5x+4x^2-80=0
4x^2+5x-80=0
using the quadratic equation we get;
x=(-5+-sqrt[5^2-4*4*-80])/2*4
x=(-5+-sqrt[25+1280])/8
x=(-5+-sqrt[1305])/8
x=(-5+-36.125)/8
x=(-5+36.125)/8
x=(-5+36.125)/8
x=31.125/8
x=3.89 answer.
x=(-5-36.125)/8
x=-41.125/8
x=-5.14 answer.
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