SOLUTION: Please help me solve this problem: A rectangular pond measures 3m by 5m. A concrete walk of uniform width is constructed around the pond. If the walk and pond together cove

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Question 168386: Please help me solve this problem:
A rectangular pond measures 3m by 5m. A concrete walk of uniform width is constructed around the pond. If the walk and pond together cover an area of 39m^2, how wide is the walk?

Found 2 solutions by checkley77, nerdybill:
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
3*5=15 M^2 FOR THE POND.
(3+2X)(5+2X)=39
15+10X+6X+4X^2-39=0
4X^2+16X-24=0
4(X^2+4X-6)=0
USING THE QUADRATIC EQUATIONx+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+WE GET:
X=(-4+-SQRT[4^2-4*1*-6])/2*1
X=(-4+-SQRT16+24])/2
X=(-4+-SQRT40)/2
X=(-4+-6.325)/2
X=(-4+6.32)/2
X=2.32/2
X=1.16 M. IS THE WIDTH OF THE WALK.
PROOF:
(3+2*1.16)(5+2*1.16)=39
(3*2.32)(5+2.32)=39
5.32*7.32=39
39~39

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangular pond measures 3m by 5m. A concrete walk of uniform width is constructed around the pond. If the walk and pond together cover an area of 39m^2, how wide is the walk?
.
Let w = width of walk
.
width of pond w/walk = 3+2w
length of pond w/walk = 5+2w
.
39 = (3+2w)(5+2w)
39 = 15 + 6w + 10w + 4w^2
39 = 15 + 16w + 4w^2
0 = -24 + 16w + 4w^2
0 = 4w^2 + 16w - 24
0 = w^2 + 4w - 6
.
Using the quadratic equation to solve we get:
x = {1.162, -5.162}
.
We can toss out the negative answer leaving us with:
1.162 meters (width of walk)
.
Here's the details of the quadratic:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aw%5E2%2Bbw%2Bc=0 (in our case 1w%5E2%2B4w%2B-6+=+0) has the following solutons:

w%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=%284%29%5E2-4%2A1%2A-6=40.

Discriminant d=40 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+40+%29%29%2F2%5Ca.

w%5B1%5D+=+%28-%284%29%2Bsqrt%28+40+%29%29%2F2%5C1+=+1.16227766016838
w%5B2%5D+=+%28-%284%29-sqrt%28+40+%29%29%2F2%5C1+=+-5.16227766016838

Quadratic expression 1w%5E2%2B4w%2B-6 can be factored:
1w%5E2%2B4w%2B-6+=+1%28w-1.16227766016838%29%2A%28w--5.16227766016838%29
Again, the answer is: 1.16227766016838, -5.16227766016838. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B4%2Ax%2B-6+%29