SOLUTION: The hypotenuse of a right triangle is 11 miles long. One leg of the triangle is 3 miles longer than the other leg. Find the sum ofo the lengths of the two legs of the right trian

Algebra ->  Triangles -> SOLUTION: The hypotenuse of a right triangle is 11 miles long. One leg of the triangle is 3 miles longer than the other leg. Find the sum ofo the lengths of the two legs of the right trian      Log On


   

Question 112968: The hypotenuse of a right triangle is 11 miles long. One leg of the triangle is 3 miles longer than the other leg. Find the sum ofo the lengths of the two legs of the right triangle.
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
Let a be b+3
a^2+b^2=c^2
(b+3)^2+b^2=11^2
b^2+6b+9+b^2=121
2b^2+6b-112=0
2(b^2+3b-56)=0
b=6.13217... mi (see below)
a=9.13217... mi
Check:
9.13217^2+6.13217^2=121
Ed
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B3x%2B-56+=+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=%283%29%5E2-4%2A1%2A-56=233.

Discriminant d=233 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-3%2B-sqrt%28+233+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%283%29%2Bsqrt%28+233+%29%29%2F2%5C1+=+6.13216876123687
x%5B2%5D+=+%28-%283%29-sqrt%28+233+%29%29%2F2%5C1+=+-9.13216876123687

Quadratic expression 1x%5E2%2B3x%2B-56 can be factored:
1x%5E2%2B3x%2B-56+=+%28x-6.13216876123687%29%2A%28x--9.13216876123687%29
Again, the answer is: 6.13216876123687, -9.13216876123687. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B3%2Ax%2B-56+%29