SOLUTION: The flower garden has the shape of a right traingle. 51 ft. of a perennial border forms the hypotenuse of the traingle, and one leg is 21 ft. longer than the other leg. Find the le

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: The flower garden has the shape of a right traingle. 51 ft. of a perennial border forms the hypotenuse of the traingle, and one leg is 21 ft. longer than the other leg. Find the le      Log On


   

Question 325806: The flower garden has the shape of a right traingle. 51 ft. of a perennial border forms the hypotenuse of the traingle, and one leg is 21 ft. longer than the other leg. Find the lengths of the legs. What are the lengths of the legs in ft?
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
X^2+(X+21)^2=51^2
X^2+X^2+42X+441=2,601
2X^2+42X+441-2,601=0
2X^2+42X-2,160=0
2(X^2+21X-1,080)=0
2(X-24)(X+45)=0
X-24=0
X=24 ANS.
X+45=0
X=-45 ANS.