SOLUTION: The length of a rectangle is 2 cm more than 5 times its width. If the area of the rectangle is 65 cm2, find the dimensions of the rectangle to the nearest thousandth Hint: Cal

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: The length of a rectangle is 2 cm more than 5 times its width. If the area of the rectangle is 65 cm2, find the dimensions of the rectangle to the nearest thousandth Hint: Cal      Log On


   

Question 89562: The length of a rectangle is 2 cm more than 5 times its width. If the area of the rectangle is 65 cm2, find the dimensions of the rectangle to the nearest thousandth
Hint: Call the width x. Then the length is 5x + 2. Now write your equation and solve.
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
L=5W+2 & THE AREA=W(5W+2)
5W^2+2W=65
5W^2+2W-65=0
USING THE QUADRATIC EQUATION WE GET: x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
X=(-2+-SQRT[2^2-4*5*-65])/2*5
X=(-2+-SQRT[4+1300])/10
X=(-2+-SQRT1304)/10
X=(-2+-36.11)/10
X=(-2+36.11)/10
X=34.11)/10
X=3.41 ANSWER FOR THE WIDTH.
L=5*3.41+2
L=17.05+2
L=19.05 ANSWER FOR THE LENGTH.
PROOF
19.05*3.41=65
65=65