SOLUTION: The length of a rectangle is 3 times its width. If the width is doubled and the length is increased by 1, the area is increased by 56. Represent the length in terms of the width

Algebra ->  Rectangles -> SOLUTION: The length of a rectangle is 3 times its width. If the width is doubled and the length is increased by 1, the area is increased by 56. Represent the length in terms of the width      Log On


   



Question 1076979: The length of a rectangle is 3 times its width. If the width is doubled and the length is increased by 1, the area is increased by 56.
Represent the length in terms of the width: Length = ______
Represent the area of the original rectangle using w and the expression from part a.
Area of original rectangle = ________________
Represent the width of the new rectangle. Width of new rectangle = _____
Represent the length of the new rectangle. Length of new rectangle = _______
Represent the area of the new rectangle. Area of the new rectangle = _________
Distribute in the equation from part e to get a quadratic equation: ___________________
Write the equation that compares the areas of the old and new rectangles:
___________________________________= ________________________________
Combine like terms and write the equation in standard form.
___________________________________________________________
a = _____ b = ______ c = _______
Plug the values from part i into the quadratic formula:


Simplify and solve. Show your work.


Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
width is x, length ix 3x, area is 3x^2
2x*(3x+1)=6x^2+2x, and subtract 56=3x^2, the original area
3x^2+2x-56=0
a=3, b=2, c=-56
x=(1/6)(-2+/- sqrt(4+672)); sqrt 676=26
x=(1/6)(24), take positive root only, and x=4
width 4
length is 12
area is 48
width 8 and length 13, area is 104, which is 56 more.