Find all values of t such that t - 1, t + 1, and 8 + t could be the lengths of the sides of a right triangle.
+2668 If 8+t is the hypotenuse, we have the equation (8+t)2=(t+1)2+(t−1)2
Solve for t, and use the positive, real values
If t+1 is the hypotenuse, we have the equation (1+t)2=(t+8)2+(t−1)2
Again, solve for t, and use positive, real values.
If t−1 is the hypotenuse, we have the equation (t−1)2=(t+8)2+(t+1)2
Solve for t, and use the real, positive values.
Can you take it from here?
