The length of a rectangular poster is 8 more inches than half its width. The area of the poster is 40 square inches. Solve for the dimensions (length and width) of the poster.
Answer
Attempt 1 out of 2
The dimensions are
inches by
inches.
Let's assume the width of the poster is x inches.
According to the problem, the length of the poster is 8 more inches than half its width. So the length can be represented as (1/2)x + 8 inches.
The area of a rectangle is found by multiplying its length and width. So we can set up the equation:
(length)(width) = area
((1/2)x + 8)(x) = 40
Now we can solve for x:
(1/2)x^2 + 8x = 40
x^2 + 16x - 80 = 0
Factoring this quadratic equation, we get:
(x + 20)(x - 4) = 0
Setting each factor equal to zero, we find:
x + 20 = 0 or x - 4 = 0
Solving for x in each equation, we get:
x = -20 or x = 4
Since the width cannot be negative, we ignore the solution x = -20.
Therefore, the width of the poster is 4 inches.
To find the length, we substitute this value of x back into the expression for the length:
length = (1/2)x + 8 = (1/2)(4) + 8 = 2 + 8 = 10 inches
So the dimensions of the poster are 10 inches by 4 inches.