SOLUTION: the area of a rectangle is 56 square feet. its lenght is 10 feet longer than the width. find the lenght and width?

Question 418780: the area of a rectangle is 56 square feet. its lenght is 10 feet longer than the width. find the lenght and width?
Found 2 solutions by praseenakos@yahoo.com, duckness73:
Answer by praseenakos@yahoo.com(507) (Show Source): You can put this solution on YOUR website!
width = x feet
length = x + 10 feet

area = length x width

56 = x(x+18)

==> x + 14 = 0 or x-4 = 0

==> x = -14 or x = 4

since length and width cannot be a negative number, you can ignore the negative value u got for x
==> width = 4 feet
so length = 4 + 10 = 14 feet

Answer by duckness73(47) (Show Source): You can put this solution on YOUR website!
Let x = width of the rectangle
Let y = length of the rectangle
Since the area of a rectangle is the width * length, and we know that this rectangle has an area of 56 square feet:
x * y = 56
Since the length of the rectangle is 10 feet longer than the width:
y = x + 10
Using the area equation from above, we have:
x * (x + 10) = 56 which can be simplified as:
x**2 + 10x = 56
x**2 + 10x - 56 = 0 which is a quadratic equation.
Using factoring to solve this equation, we have:
(x + 14)(x - 4) = 0 which means that
x = -14 or x = 4. So the width of the rectangle is either -14 or 4 feet.
So, which is it? Since it doesn't make any sense to have a rectangle with a negative width, the width of a rectangle cannot be -14 feet, so the width must be 4. Since the length (which is y) is 10 feet longer than the width, we have
y = x + 10 = 4 + 10 = 14
So, the length (y) of the rectangle is 14 and the width (x) is 4.

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