SOLUTION: he width of a rectangle is 2 inches more than the side of a square. The length of the rectangle is 1 inch less than twice the width of the square .If the rectangles area is 2 inche

Question 1026229: he width of a rectangle is 2 inches more than the side of a square. The length of the rectangle is 1 inch less than twice the width of the square .If the rectangles area is 2 inches2 more than the square, find the side of the square.

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
let s = the side of the square
:
The width of a rectangle is 2 inches more than the side of a square.
w = s+2
The length of the rectangle is 1 inch less than twice the width of the square
L = 2s-1
If the rectangles area is 2 inches2 more than the square, find the side of the square.
L*W = s^2 + 2
Replace L and W
(2s-1)(s+2) = s^2 + 2
FOIL
2s^2 + 4s - s - 2 = s^2 + 2
Form a quadratic equation
2s^2 - s^2 + 3s - 2 - 2 = 0
s^2 + 3s - 4 = 0
Factors to
(s+4)(s-1) = 0
the positive solution
s = 1 inch is the side of square
;
:
Check this; find the dimensions of the rectangle
w = 1+2
w = 3
L = 2(1) - 1
L = 1
3 sq/in is the area
the square would be 1 sq/in, 2 less than the rectangle
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