Solve radical equations Algebra 2 help ?/

1. √ 3 x + 1 = 7

2. √x + 6 = x

3. 2 √ n + 3 = n

4. √x + 4 = √ x - 1 + 1

2) √x + 6 = x
√x = x - 6
(√x)^2 = (x - 6)^2
x = x^2 - 12x + 36
0 = x^2 - 11 + 36
121/4 = x^2 - 11 + 121/4 + 36
121/4 = (x - 11/2)^2 + 36
-23/4 = (x - 11/2)^2
x - 11/2 = i√23/2 or x - 11/2 = -i√23/2
x = 11/2 + i√23/2 or x = 11/2 - i√23/2

If you meant √(x + 6) = x, then
(√(x + 6))^2 = x^2
x + 6 = x^2
0 = x^2 - x - 6
0 = (x - 3)(x + 2)
x = 3 or x = -2
Discard x = -2 because it is the extraneous solution.

Solve the others similarly.

Can you clarify where brackets should be for these problems, as I'm sure that at least the last one should have some somewhere?

Basically all problems like this though involve isolating the radical term and then squaring both sides to get rid of any roots.

I'll do the three different versions of the first problem, which hopefully will give you the general idea, please ask if you need more clarification though.

√3 x + 1 = 7
√3 x = 6
x = 6 / √3
x = 2√3

or

√(3x) + 1 = 7
√(3x) = 6
square
3x = 36
x = 12

or

√(3x + 1) = 7
3x + 1 = 49
3x = 48
x = 16