SOLUTION: A 22-ft by 28 ft rectangular swimming pool is surrounded by a walkway of uniform width. If the total area of the walkway is 216 ft^2, how wide is the walkway?

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Question 902253: A 22-ft by 28 ft rectangular swimming pool is surrounded by a walkway of uniform width. If the total area of the walkway is 216 ft^2, how wide is the walkway?
Answer by Stitch(470) About Me  (Show Source):
You can put this solution on YOUR website!
The area of a rectangle is given by the equation A = L*W, where L is the length and W is the width.
We can find the total area of the larger rectangle (The area of the pool + the area of the walkway) by adding the two areas together.
The area of the pool is :22+%2A+28+=+616ft^2
Now we will added the two given areas together to find the total area of the pool and the walkway.
616ft^2 + 216ft^2 = 832ft^2
Now we need to write the equation of the area of the new large rectangle.
Let X be the width of the walkway.
The new length: L+=+X+%2B+22+%2B+X
The new width: W+=+X+%2B+28+%2B+X
The new area equation is:
832+=+%28X+%2B+22+%2B+X%29+%2A+%28X+%2B+28+%2B+X%29
Combine like terms
832+=+%282X+%2B+22%29+%2A+%282X+%2B+28%29
Use FOIL to simplify the right hand side of the equation.
ERROR Algebra::Solver::Engine::invoke_solver_noengine: solver not defined for name 'foil'.
Error occurred executing solver 'foil' .

Rewrite the equation
832+=+4X%5E2+%2B+100X+%2B+616
Subtract 832 from both sides
+0+=+4X%5E2+%2B+100X+-+216
Now use the quadratic equation to solve for X.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aX%5E2%2BbX%2Bc=0 (in our case 4X%5E2%2B56X%2B-165+=+0) has the following solutons:

X%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2856%29%5E2-4%2A4%2A-165=5776.

Discriminant d=5776 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-56%2B-sqrt%28+5776+%29%29%2F2%5Ca.

X%5B1%5D+=+%28-%2856%29%2Bsqrt%28+5776+%29%29%2F2%5C4+=+2.5
X%5B2%5D+=+%28-%2856%29-sqrt%28+5776+%29%29%2F2%5C4+=+-16.5

Quadratic expression 4X%5E2%2B56X%2B-165 can be factored:
4X%5E2%2B56X%2B-165+=+4%28X-2.5%29%2A%28X--16.5%29
Again, the answer is: 2.5, -16.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+4%2Ax%5E2%2B56%2Ax%2B-165+%29

X = 2.5 & -16.5
Now since we can not have a negative distance -16.5ft will not work for this problem.
So the width of the walkway is 2.5ft