SOLUTION: This is another word problem question. One leg of a right triangle is 96 inches. Find the hypothesis and the other leg if the length of the hypotenuse exceeds 2 1/2 times the other

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Question 139101: This is another word problem question. One leg of a right triangle is 96 inches. Find the hypothesis and the other leg if the length of the hypotenuse exceeds 2 1/2 times the other leg by 4 inches.
>>>i tried working it out and got 2 1/2 times leg + 4 eguals hypotenuse
and 2 1/2 divided leg - 4 eguals leg.
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
Let a=96, c=2.5b+4
a^2+b^2=c^2
96^2+b^2=(2.5b+4)^2
9216+b^2=6.25b^2+20b+16
5.25b^2+20b-9200=0
b=40" (see below)
c=2.5*40+4=104"
.
Check:
9216+1600=10816
104^2=10816
.
Ed
.
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 5.25x%5E2%2B20x%2B-9200+=+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=%2820%29%5E2-4%2A5.25%2A-9200=193600.

Discriminant d=193600 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-20%2B-sqrt%28+193600+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2820%29%2Bsqrt%28+193600+%29%29%2F2%5C5.25+=+40
x%5B2%5D+=+%28-%2820%29-sqrt%28+193600+%29%29%2F2%5C5.25+=+-43.8095238095238

Quadratic expression 5.25x%5E2%2B20x%2B-9200 can be factored:
5.25x%5E2%2B20x%2B-9200+=+%28x-40%29%2A%28x--43.8095238095238%29
Again, the answer is: 40, -43.8095238095238. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+5.25%2Ax%5E2%2B20%2Ax%2B-9200+%29