SOLUTION: The length of a rectangular sign is 3 feet longer than the width. If the sign's area is 54 square feet, find the length and the width.

Question 78740This question is from textbook
: The length of a rectangular sign is 3 feet longer than the width. If the sign's area is 54 square feet, find the length and the width. This question is from textbook

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
The problem tells you that the Length (L) equals the Width (W) plus 3 feet. In equation
form this becomes:
.

.
The problem also tells you that the Area (A) is 54 square feet. From geometry you know that
the formula for the Area of a rectangle is that Area equals the product of the Length
times the Width. In equation form this is:
.

.
Substitute the given area to get:
.

.
This equation has two unknowns. You can't solve it unless you can eliminate one of the
unknowns. To do that you can note that L = W + 3. If you substitute W + 3 for L in the
equation you get:
.

.
And multiplying out the right side results in:
.

.
Get this into the standard quadratic form by first subtracting 54 from both sides and
then switching sides around to get:
.

.
This can be factored into:
.

.
Notice that this equation will be true if either of the 2 factors is a zero because
zero times anything is zero and therefore equal to the right side of this equation.
.
Set W+9 equal to zero and solve to find that W = -9. But what sense does a negative
width make in this case? None at all, so disregard this solution.
.
Next set W-6 equal to zero and solve to find that W = +6. That's better. The width is
6 feet, and since we know that the length is 3 feet longer than the width, the length
has to be 9 feet. Check by noting that the product of the length times the width is
9 ft times 6 ft and that equals the given area of 54 sq ft.
.
Hope this helps you to understand the problem a little better.
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