Question

How do you solve and check for extraneous solutions in `sqrt(23x+3)=14`?

Answer 1

`x= 193/23` (no extraneous solutions)

Explanation:

Given `sqrt(23x+3) =14`

Square both sides:
`color(white)("XXXX")` `23x+3 = 196`

Subtract 3 from both sides
`color(white)("XXXX")` `23x=193`

Divide both sides by `23`
`color(white)("XXXX")` `x= 193/23` `color(white)("XXXX")`approximately 8.391304 according to my calculator

Verify equation with `x=193/23` to ensure this is not extraneous;
`color(white)("XXXX")` #`sqrt(23(193/23)+3)`

`color(white)("XXXX")` `=sqrt(196)`

`color(white)("XXXX")` `=14`

So `x=193/23` is a valid result