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.
help? thanks!
w(w+8) = 560
20x28
To solve this problem, let's start by setting up equations based on the given information.
Given:
Length = Width + 8
Area = Length * Width = 560 in^2
Now we can use these equations to find the dimensions of the photograph.
1. Substitute the first equation into the second equation to find the dimensions:
(Width + 8) * Width = 560
Simplifying further:
Width^2 + 8Width = 560
Rearranging the equation to form a quadratic equation:
Width^2 + 8Width - 560 = 0
2. Now we need to solve the quadratic equation. We can do this by factoring, completing the square, or using the quadratic formula. In this case, let's use factoring.
Factoring the quadratic equation:
(Width - 20)(Width + 28) = 0
Setting each factor equal to zero and solving for Width:
Width - 20 = 0 or Width + 28 = 0
Solving for Width:
Width = 20 or Width = -28 (since negative width is not meaningful in this context)
Since width cannot be negative, we can discard Width = -28.
3. Now that we have the width, we can find the length by using the first equation:
Length = Width + 8
Length = 20 + 8
Length = 28
Therefore, the dimensions of the photograph are:
Width = 20 in.
Length = 28 in.