the length of a rectangular photograph is 8in. more than the width. if the area is 560 in^2 what are the dimensions of the photograph?
the width is____ in. and the length is ____ in.
length is L = W + 8
Area is length times width.
560 = (w+8)w
560 = w^2 + 8w
0 = w^2 + 8w - 560
factor to solve for w. (discard negative values) Then substitute to find L.
Be sure to check by multiplying to be sure you get the 560 in^2.
To find the dimensions of the photograph, we can set up a system of equations based on the given information.
Let's represent the width of the photograph as "w" inches. Since the length is 8 inches more than the width, we can represent the length as "w + 8" inches.
The area of a rectangle is given by the formula A = length × width. In this case, we know that the area is 560 in^2. Substituting the given values into the formula, we have:
560 = (w + 8) × w
To solve this equation, we can expand and rewrite it as a quadratic equation:
560 = w^2 + 8w
Rearranging terms, we get:
w^2 + 8w - 560 = 0
Now we can solve this quadratic equation by factoring or by using the quadratic formula. For simplicity, let's solve it by factoring.
Looking for factors of -560 with a sum of 8, we find that w = 20 satisfies the equation:
(w + 28)(w - 20) = 0
So we have two possible values for the width: w = -28 or w = 20. Since the width cannot be negative, we discard the value w = -28.
Therefore, the width of the photograph is 20 inches.
To find the length, we use the equation we initially set up: length = width + 8. Plugging in the value we found for the width:
length = 20 + 8 = 28 inches
So the dimensions of the photograph are:
Width = 20 inches
Length = 28 inches