SOLUTION: A gardener has a 30 foot by 20 foot rectangular plot of ground. she wants to build a brickway of uniform width on the border of the plot. if the gardener wants to have 400 square f
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-> SOLUTION: A gardener has a 30 foot by 20 foot rectangular plot of ground. she wants to build a brickway of uniform width on the border of the plot. if the gardener wants to have 400 square f
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Question 762397: A gardener has a 30 foot by 20 foot rectangular plot of ground. she wants to build a brickway of uniform width on the border of the plot. if the gardener wants to have 400 square feet of ground left for planting, how wide (to two decimal places) should she build the walkway?
I solved 2.19 but it was using trial and error.
my teacher asked for the complete process.
You can put this solution on YOUR website! Area = length x width.
If you are have a uniform walkway, then you have to subtract that distance from both sides of the length and width.
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.
Subtract 400 from both sides.
Now use the quadratic equation to solve for X.
Quadratic equation (in our case ) has the following solutons:
For these solutions to exist, the discriminant should not be a negative number.
First, we need to compute the discriminant : .
Discriminant d=6800 is greater than zero. That means that there are two solutions: .
Quadratic expression can be factored:
Again, the answer is: 22.8077640640442, 2.19223593595585.
Here's your graph:
X = 2.19 & 22.81
Now if you use 22.81 in the equation for the area of the garden you will come up with negative area. Since we can not have negative area 22.81 is not a real solution to this problem.
X = 2.19feet