Question

How do you solve and check for extraneous solutions in `sqrt(x+1) + 5 = x`?

Answer 1

The only valid solution is `x=8`
A second candidate solution `x=3` can be eliminated by checking for the validity of the given equation with `x=3` (and noting that it is not valid).

Explanation:

Given `sqrt(x+1) + 5 = x`

Subtract `5` from both sides
`color(white)("XXXX")` `sqrt(x+1) = x-5`

Square both sides (possibly generating an extraneous root at this point)
`color(white)("XXXX")` `x+1 = x^2-10x+25`

Subtract `(x+1)` from both sides (and flip sides)
`color(white)("XXXX")` `x^2-11x+24 = 0`

Factor
`color(white)("XXXX")` `(x-3)(x-8) = 0`

`rArr` `color(white)("XXXX")` `x=3` `color(white)("XXXX")`or`color(white)("XXXX")` `x=8`

Substituting `3` for `x` in the Left Side of original equation
`color(white)("XXXX")` `sqrt(3+1)+5 = 2+3`
`color(white)("XXXX")` `color(white)("XXXX")` `!=3`
The solution `x=3` is extraneous.

Substituting `8` for `x` in the Left Side of original equation
`color(white)("XXXX")` `sqrt(8+1)+5 = 3+5`
`color(white)("XXXX")` `color(white)("XXXX")` `=8`
`color(white)("XXXX")` `color(white)("XXXX")` `=x`
The solution `x=8` is valid.