How do you solve and check for extraneous solutions in `sqrt(x+1)= x-1`?

Since you're dealing with a radical term, recognize that you need to have

`x + 1 >=0" "` and `" "x-1>=0`

This is the case because, for real numbers, you can only take the square root of a positive number; moreover, the result of this operation will always be a positive number.

If `x-1>=0`, then you automatically have `x+1 >=0`, so you can combine these two conditions

`x -1 >= 0 implies x >=1`

Square both sides of the equation to get rid of the radical term

`(sqrt(x+1))^2 = (x-1)^2`

`x+color(red)(cancel(color(black)(1))) = x^2 - 2x + color(red)(cancel(color(black)(1)))`

This is equivalent to

`x^2 - 3x = 0`

`x(x - 3) = 0`

This equation can be euqal to zero if either `x=0` or `x-3=0`, which would imply

`x-3 = 0 implies x = 3`

Since `x=0` does not satisfy the initial contidion `x>=1`, it will be an extraneous solution. The only valid solution to your original equation will thus be `x = color(green)(3)`.

Do a quick check to make sure that you got the calculations right

`sqrt(3 + 1) = 3- 1`

`sqrt(4) = 2`

`2 = 2" "color(green)(sqrt())`