Hi Liz,
I am going to solve a similar problem.
Solve and check for extraneous solutions
(t^{2} + 3)^{1/2} = 2
First square both sides to get
t^{2} + 3 = 2^{2} = 4
t^{2} = 1 so
t = 1 or t = 1
Now check to see if either solution is extraneous.
If t = 1 then (t^{2} + 3)^{1/2} = (1^{2} + 3)^{1/2} = (4)^{1/2} = 2
If t = 1 then (t^{2} + 3)^{1/2} = ((1)^{2} + 3)^{1/2} = (4)^{1/2} = 2
Thus both solutions are valid and there is no extraneous solution.
Now lets try a slightly different problem.
Solve and check for extraneous solutions
(t^{2} + 3)^{1/2} = 2t
Again square both sides and solve
t^{2} + 3 = 4 t^{2}
3 = 3 t^{2}
1 = t^{2} so
t = 1 or t = 1
Again check to see if either solution is extraneous.
If t = 1 then (t^{2} + 3)^{1/2} = (1^{2} + 3)^{1/2} = (4)^{1/2} = 2 and 2t = 2 so t = 1 is a valid solution.
If t = 1 then (t^{2} + 3)^{1/2} = ((1)^{2} + 3)^{1/2} = (4)^{1/2} = 2 but 2t = 2 and hence t = 1 is an extraneous solution.
Now try your problem.
Penny
