Question: solve and check for extraneous solution

3(w + 1)^1/2 = 6

Hi Liz,

I am going to solve a similar problem.

Solve and check for extraneous solutions

(t² + 3)^1/2 = 2

First square both sides to get

t² + 3 = 2² = 4
t² = 1 so
t = 1 or t = -1

Now check to see if either solution is extraneous.

If t = 1 then (t² + 3)^1/2 = (1² + 3)^1/2 = (4)^1/2 = 2
If t = -1 then (t² + 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

(t² + 3)^1/2 = 2t

Again square both sides and solve

t² + 3 = 4 t²
3 = 3 t²
1 = t²  so
t = 1 or t = -1

Again check to see if either solution is extraneous.

If t = 1 then (t² + 3)^1/2 = (1² + 3)^1/2 = (4)^1/2 = 2 and 2t = 2 so t = 1 is a valid solution.
If t = -1 then (t² + 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