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

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


   

Question 113765This question is from textbook Beginning Algebra
: The length of a rectangle is 3cm more than 2 times the width. If the area of the rectangle is 99 cm^2, find the dimensions of the rectangle to the nearest thousandth.
The equation I got from this is:
x(2x+3)=99
I have worked it to this point, and gotten stuck.
2x^2+3x=99
2x^2+3x-99=0
I am not sure what to do after this. I have tried to factor this, but am not able to. What do I do now? This question is from textbook Beginning Algebra

Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
L=3+2W
AREA=L*W
99=(3+2W)W
99=3W+2W^2
2W^2+3W-99=0
USING THE QUADRATIC EQUATION x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
YOU GET:
W=(-3+-SQRT[3^3-4*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 FOR THE WIDTH.
L=3+2*6.325
L=3+12.65
L=15.65 FOR THE LENGTH.
PROOF:
99=(3+2*6.325)*6.325
99=(3+12.65)*6.325
99=15.65*6.325
99~99