SOLUTION: If the length of a rectangle sign is to be 2 feet longer than the width and the area is to be 35 square feet, determine the length and width of the sign.

Question 262770: If the length of a rectangle sign is to be 2 feet longer than the width and the area is to be 35 square feet, determine the length and width of the sign.
Found 2 solutions by drk, unlockmath:
Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
Let width = x
and length = x + 2
area = LW
or
area = x(x+2) = 35
so, we get
x^2 + 2x - 35 = 0
factor to get
(x+7)(x-5) = 0
So,
x = -7 which is not possible
or
x = 5 and x + 2 = 7
length = 7 and width = 5
Answer by unlockmath(1688) (Show Source): You can put this solution on YOUR website!
Hello,
We know the area of a rectangle is width times height, right? So let's set up "x"
to represent the width so the length will be x+2. Now we can write an equation:
x(x+2)=35 Rewritten this will look like:
x^2+2x=35 Subtract 35 from both sides to be:
x^2+2x-35=0 Now we can factor this to get:
(x-5)(x+7)=0 Solve for x:
x=5
x=-7 We know this answer would not work so:
x=5 is our solution. So: Width is 5 feet and length is 7 feet.
Make sense?
Check out a new book I wrote at:
www.math-unlock.com

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