SOLUTION: 17) The length of a rectangle is 3 cm more than 2 times its width. If the area of the rectangle is 99 cm^2, find the dimensions of the rectangle to the nearest thousandth.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: 17) The length of a rectangle is 3 cm more than 2 times its width. If the area of the rectangle is 99 cm^2, find the dimensions of the rectangle to the nearest thousandth.      Log On


   

Question 78294: 17) The length of a rectangle is 3 cm more than 2 times its width. If the area of the rectangle is 99 cm^2, find the dimensions of the rectangle to the nearest thousandth.
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
L=3+2W
(3+2W)W=99
3W+2W^2-99=0
2W^2+3W-99=0
USING THE QUADRATIC EQUATION WE GET:
W=(-3+-SQRT[3^2-4*2*-99])/2*2
W=(-3+-SQRT[9+792])/4
W=(-3+-SQRT801)/4
W=(-3+-28.3)/4
W=(-3+28.3)/4
W=25.3/4
W=6.325 ANSWER.
L=3+2*6.325
L=3+12.65
L=15.65 ANSWER.
PROOF
6.325*15.65=99
99=99