SOLUTION: The length of a rectangle is 5 inches longer than the width. The area of the rectangle is 50 inches squared. Find the length and the width of the rectangle.

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Question 572708: The length of a rectangle is 5 inches longer than the width. The area of the rectangle is 50 inches squared. Find the length and the width of the rectangle.
Answer by mathsmiles(68) (Show Source):
You can put this solution on YOUR website!
First thing we want to do is write down what we know:

Area of a rectangle:
A = L x W
L = 5 + W (length is 5 inches longer than the width)
A = 50 (Area is 50 inches squared)

Putting this all together:
L x W = 50
Substituting for L with the above equation:
(5+W) x W = 50
Multiplying out the paren
5W + W^2 = 50
Subtract 50 from both sides and rearrange the terms a little:
W^2 + 5W - 50 = 0

Now solve:
(W _ _) (W _ _) = 0 (we need to figure out the operand and the term for each)
The negative whole number indicates these have different signs
(W - _) (W + _) = 0
We need to find two factors of 50 whose difference (subtract them) gives 5.
50 = 25 x 2 Nope
50 = 5 x 10 Yup!

Since the 5W term is positive, we need to put the 10 in the positive factor and 5 in the negative factor so ...
(W - 5)(W + 10) = 0
W -5 = 0
W = 5
W + 10 = 0
W = -10

Since we're talking about the area of a rectangle, we have to assume it has positive sides or we've just entered another dimension. :-) So,
W = 5 inches
L = 5 + 5 (5 inches longer than the width)
L = 10 inches

Checking:
A = L x W
A = 5 x 10
50 = 50 Correct!