SOLUTION: The length of a rectangle is 1 cm more than 5 times its width. If the area of the rectangle is 89 cm2, find the dimensions of the rectangle to the nearest thousandth. I wonder

Algebra ->  Surface-area -> SOLUTION: The length of a rectangle is 1 cm more than 5 times its width. If the area of the rectangle is 89 cm2, find the dimensions of the rectangle to the nearest thousandth. I wonder       Log On


   

Question 91310: The length of a rectangle is 1 cm more than 5 times its width. If the area of the rectangle is 89 cm2, find the dimensions of the rectangle to the nearest thousandth.
I wonder if this is somewhat on track? I have been baffled for the past three hours,on this equation.
(1+2x)(5+2x)=89
3+-the square root of 93/2
the square root of 93 divided by 2=9.64divided by 2=4.862182 plus three=7.82182
and 4.862182-3=1.82182
Answer by tutor_paul(519) About Me  (Show Source):
You can put this solution on YOUR website!
Write an equation for the length in terms of width:
Equation #1: L=5W%2B1
Write an expression for the Area:
Equation #2: LW=89
Now you have 2 equations with 2 unknowns. Any time you have at lease as many equations as you have unknowns, you can find the unknowns!
===================================
Substitute the value of L from Equation #1 into Equation #2:
%285W%2B1%29W=89
5W%5E2%2BW-89=0
Apply the quadratic formula:
W=%28-1%2B-sqrt%281%5E2-%284%2A5%2A%28-89%29%29%29%29%2F%282%2A5%29
W=%28-1%2B-sqrt%281781%29%29%2F10
W=(-1+-42.2)/10
W=4.12,W=-4.32
The width needs to be positive, so throw out the negative result:
highlight%28W=4.12%29
Now, substitute this value of W into Equation #1 to find L.
I will leave that part to you :-)
Goos Luck,
tutor_paul@yahoo.com