Quadratic Equations
posted by purpletree .
The longer leg of a right triangle is 3 ft longer than three times the shorter leg. The hypotenuse is 3 ft shorter than four times the shorter leg. Find the lengths of the three sides of the right triangle. (Please provide steps/explanation to solve this problem.)

shoter leg = x
longer leg = 3x+3
hypot= 4x3
to solve for hypotenuse you use
a^2 + b^2 = c^2, so
(x)^2 + (3x+3)^2 = (4x3)^2
(x)(x) + (3x+3)(3x+3) = (4x3)(4x3)
then multiply
(x^2) + (9x^2+18x+9) = (16x^224x+9)
then combine like terms
10x^2 + 18x + 9 = 16x^2  24x + 9
then i subtracted 10x^2 from both sides
and got:
18x + 9 = 6x^2  24x +9
next, i subtracted 18x from both sides:
9 = 6x^2  42x + 9
and then subtracted 9 from both sides
0 = 6x^2  42x
then pull out a 6x on the right side of the equation
0 = 6x(x7)
so
6x=0
and
x7=0
so x= 0 and 7
so the shorter leg is :
0 units or 7 units
the longer leg is :
3 units or 24 units
and the hypotenuse is :
3 units or 25
and since you cant have a hypotenuse with a negative value
i think the three lengths of your triangle are
7, 24, and 25 
Thanks for the help. :)