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