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

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


   

Question 89088: The length of a rectangle is 2 cm more than 5 times its width. If the area of the rectangle is 65 cm^2, 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 ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 2 cm more than 5 times its width. If the area of the rectangle is 65 cm^2, 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.
:
They pretty much tell you what to do, width * length = 65 sq/cm
:
x*(5x+2) = 65
5x^2 + 2x = 65
5x^2 + 2x - 65 = 0
:
Solve this using the quadratic formula; a=5, b=2, c=-65
:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
:
x+=+%28-2+%2B-+sqrt%28+2%5E2+-+4%2A5%2A-65+%29%29%2F%282%2A5%29+
:
I'll let you do the math here. Only the positive solution is wanted here
Substitute your value for x in the original equation to check it