+647 Find all values of t such that t - 1, t + 1, and 4 could be the lengths of the sides of a right triangle.
+129852 If "t" has to be an integer...it is impossible to have integer sides with "4" as a hypotenuse
So.......I will assme that either
4^2 + ( t - 1)^2 = ( t + 1)^2 or 4^2 + (t + 1)^2 = (t - 1)^2
But.......for the second to be true, t = -4, which would give us negative side lengths
So.....we have
4^2 + ( t - 1)^2 = ( t + 1)^2
16 -2t = 2t
16 = 4t → t = 4
So....the other two sides are 3 and 5
If there are no restrictions on t, we have that
( t - 1)^2 + ( t + 1)^2 = 4^2
2t^2 + 2 = 16
t^2 = 7 → t = √7
So.....the sides are √7 - 1 , √7 + 1 , 4
