Question

How do I find the extraneous solution of `sqrt(x-3)-sqrt(x)=3`?

Answer 1

`sqrt{x-3}-sqrt{x}=3`

by subtracting `sqrt{x-3}`,

`Rightarrow -sqrt{x}=3-sqrt{x-3}`

by squaring both sides,

`Rightarrow x=9-6sqrt{x-3}+x-3`

by cancelling out `x`'s,

`Rightarrow0=6-6sqrt{x-3}`

by dividing by 6,

`Rightarrow 0=1-sqrt{x-3}`

by adding `sqrt{x-3}`,

`Rightarrow sqrt{x-3}=1`

by squaring both sides,

`Rightarrow x-3=1`

by adding 3,

`Rightarrow x=4`

Even though all implications above are valid, some of them are not reversible. As a result, `x=4` we found does not satisfy the original equation, which makes it extraneous.