Question

How do I find the extraneous solution of `sqrt(x+4)=x-2`?

Answer 1

`x=0" is an extraneous solution"`

Explanation:

`"square both sides to 'undo' the radical"`

`(sqrt(x+4))^2=(x-2)^2`

`rArrx+4=x^2-4x+4`

`"rearrange into standard form"`

`rArrx^2-5x=0larrcolor(blue)"in standard form"`

`rArrx(x-5)=0`

`rArrx=0" or "x=5`

`color(blue)"As a check"`

`"Substitute these values into the original equation"`

`x=0tosqrt4=2" and "0-2=-2`

`2!=-2rArrx=0color(red)" is an extraneous solution"`

`x=5tosqrt9=3" and "5-2=3larr" True"`

`rArrx=5color(red)" is the solution"`