SOLUTION: Hi, I am very confused with this problem I cannot seem to apply the pythagoream formula to it. Can you Please help? Thanks:" Two boats leave San Diego at the same time. One boat

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Hi, I am very confused with this problem I cannot seem to apply the pythagoream formula to it. Can you Please help? Thanks:" Two boats leave San Diego at the same time. One boat      Log On


   

Question 10887: Hi, I am very confused with this problem I cannot seem to apply the pythagoream formula to it. Can you Please help? Thanks:" Two boats leave San Diego at the same time. One boat sails due north and the other due west. When they are 10 miles apart, the boat traveling north has sailed 2 miles farther than the boat sailing west. Find the distance each boat has traveled."
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Ok, let the distance traveled by the west-going boat = x miles. Then the distance traveled by the north-going boat is (x+2) miles.
You can apply the Pythagorean theorem to this problem.
In the right triangle you get when you diagram the paths of these two boats, the hypotenuse is their distance from each other and that is 10 miles. So, c = 10 miles. Now you can set up the equation per Pythagorus: {{c^2 = a^2 + b^2}}}
10%5E2+=+x%5E2+%2B+%28x%2B2%29%5E2
100+=+x%5E2+%2B+%28x%5E2+%2B+4x+%2B+4%29 Simplify. Subtact 100 from both sides and collect like-terms.
2x%5E2+%2B+4x+-+96+=+0 Factor out a 2 to simplify this.
x%5E2+%2B+2x+-+48+=+0 Factor this quadratic equation.
%28x+-+6%29%28x+%2B+8%29+=+0
x+-+6+=+0, then x+=+6 or
x+%2B+8+=+0, then x+=+-8 Discard this solution as the distance can't be negative.
So, x = 6 miles and x+2 = 8 miles.
The north-going boat traveled 8 miles while the west-going boat traveled 6 miles.
You can see that 6, 8, and 10 are "Pythagorean numbers becuase 6%5E2+%2B+8%5E2+=+10%5E2