SOLUTION: Solve the following application problem (show your equation and then your solution). Round answers to nearest thousandths and properly label your answer as well. The length of a

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Solve the following application problem (show your equation and then your solution). Round answers to nearest thousandths and properly label your answer as well. The length of a       Log On


   

Question 184476: Solve the following application problem (show your equation and then your solution). Round answers to nearest thousandths and properly label your answer as well.
The length of a rectagle is 5 cm. more than its width. If the area of the rectange is 90 cm^2, find the dimensions.

Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
L=W+5
L*W=90
(W+5)W=90
W^2+5W-90=0
USING THE QUADRATIC EQUATION W = (-b +- sqrt( b^2-4*a*c ))/(2*a) }WE GET:
W=(-5+-SQRT(5^2-4*1*-90])/2*1
W=(-5+-SQRT[25+360])/2
W=(-5+-SQRT385)/2
W=(-5+-19.62)/2
W=(-5+19.62)/2
W=14.62/2
W=7.31 cm. IS THE WIDTH.
L=7.31+5=12.31 cm. IS THE LENGTH.
PROOF:
7.31*12.31=90
90~90