SOLUTION: "Suppose that one leg of a right triangle is 2 feet shorter than the other leg. the hypotenuse is 2 feet longer than the other leg. Find the lengths of all 3 sides of the right tri

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Question 739929: "Suppose that one leg of a right triangle is 2 feet shorter than the other leg. the hypotenuse is 2 feet longer than the other leg. Find the lengths of all 3 sides of the right triangle."
Theorem: a^2+b^2=c^2
which is:
x^2+(x-2)^2=(x+2)^2
here's the equation steps I have so far:
x^2+x^2-4x-4=x^2+4x+4
2x^2-4x-4=x^2+4x+4
(subtracted x^2 from each side)
x^2-4x-4=4x+4
(subtracted 4x from each side)
x^2-8x-4=4
(subtracted 4 from each side)
x^2-8x-8=0
this is where i'm stuck. I cannot factor this down. I can't figure out if I messed up on the first steps. I've retried this problem way to much and am finally asking for help. Any help would be great! Thank you!
Found 2 solutions by mananth, jim_thompson5910:
Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website!
Theorem: a^2+b^2=c^2
which is:
x^2+(x-2)^2=(x+2)^2
here's the equation steps I have so far:
x^2+x^2-4x+4=x^2+4x+4 ( there was an error in the formula) in this step
2x^2-4x+4=x^2+4x+4
(subtracted x^2 from each side)
x^2-4x+4=4x+4
(subtracted 4x from each side)
x^2-8x+4=4
(subtracted 4 from each side)
x^2-8x=0
x(x-8)=0
x=0 OR x=8
8^2+6^2 = 10^2
LHS = RHS

Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
x^2+(x-2)^2=(x+2)^2

x^2+x^2-4x +4=x^2 + 4x + 4 ... corrections are in red

2x^2 - 4x + 4 = x^2 + 4x + 4

2x^2 - 4x + 4 - x^2 - 4x - 4 = 0

x^2 - 8x = 0

x(x - 8) = 0

x = 0 or x - 8 = 0

x = 0 or x = 8

Ignore the trivial solution of x = 0

So the only useful solution is x = 8

This means that the two legs are 8 and 6 (since x-2 = 8-2 = 6)

The hypotenuse is 10 (x+2 = 8+2 = 10)