SOLUTION: The length of a rectangle is 7ft more than twice the width and the area of the rectangle is 99ft squared .find the dimensions of the rectangle

Click here to see ALL problems on Evaluation Word Problems

Question 1052679: The length of a rectangle is 7ft more than twice the width and the area of the rectangle is 99ft squared .find the dimensions of the rectangle
Found 2 solutions by Alan3354, addingup:
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 7ft more than twice the width and the area of the rectangle is 99ft squared .find the dimensions of the rectangle
---------------
L = 2W + 7
L*W = 99

Answer by addingup(3677) (Show Source):
You can put this solution on YOUR website!
L= 2W+7
L*W= 99
(2W+7)*W = 99
2W^2+7W = 99
W^2+7/2W = 99/2
Take 1/2 the coefficient and square it. The coefficient is 7/2 and (7/2)/2
7/2*1/2 = 7/4 Now we square it:(7/4)^2 = 49/16 add this to both sides:
W^2+7/2+49/16 = 99/2+49/16
W^2+7/2+49/16 = 841/16
Perfect square on the left (x+7/2)*(x+7/2), multiply First Outer Inner Last and you get x^2+7/2x+49/16
(x+7/2)*(x+7/2) can be written as (x+7/2)^2:
(x+7/2)^2 = 841/16
x+7/8 = sqrt 841/16
x = -9
or
x = 11/2 = 5.5 fraction or decimal, depends how you need it.
We know we're looking for a positive number, so let's try 5.5 in the formula for the area:
(2*5.5)+7*5.5 = 99
18*5.5 = 99 Correct
:
John