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 don't unde

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> 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 don't unde      Log On


   

Question 91456: 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 don't understand where to even start.
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
W=WIDTH
L=LENGTH L=5W+1
AREA=L*W
89=W(5W+1)
89=5W^2+W
5W^2+W-89=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*-89])/2*5
W=(-1+-SQRT[1+1780])/10
W=(-1+-SQRT1781)/10
W=(-1+-42.2)/10
W=(-1+42.2)/10
W=41.2/10
W=4.12 ANSWER FOR THE WIDTH
L=5*4.12+1
L=20.6+1
L=21.6
PROOF
4.12*21.6=89
89=89