Question 158038: A painting canvas has a length 6 inches longer than its width. If the area is 112 in^2, what are the dimensions of the canvas?
Answer by midwood_trail(310)   (Show Source):
You can put this solution on YOUR website! This is not a mixture problem.
This is a geometry word problem.
A painting canvas has a length 6 inches longer than its width. If the area is 112 in^2, what are the dimensions of the canvas?
Let width = x
Let length = x + 6
The area given is 112 inches^(2).
The area of a rectangle (this canvas is a rectangle) is given by the formula
A = L*W
We plug and chug.
112 = (x + 6)(x)
When I multiply the right side, a quadratic will be the result.
112 = x^2 + 6x
Subtract 112 from both sides and equate the entire polynomial to zero.
x^2 + 6x - 112 = 0
We now factor the left side of this quadratic equation.
14 times -8 = -112
So, the factor becomes (x + 14) (x - 8).
Set each factor to equal 0 and solve for x.
x + 14 = 0
x = -14....This answer for x is REJECTED because length CANNOT be negative.
x - 8 = 0
x = 8
We now know the value of x.
Since the width = x and I just found the value of x, the width is 8 inches.
The length = x + 6
The length = 8 + 6
The length = 14 inches
The dimensions of the canvas are:
width = 8 inches
length = 14 inches
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