Question

Identify any extraneous solution. See above pic

Answer 1

`D:z\geq 0\wedge z+5\geq 0\\D:z\geq 0\wedge z\geq -5\\D:z\geq0\\\\1+\sqrt z=√(z+5)\\1+2\sqrt z+z=z+5\\2\sqrt z=4\\\sqrt z=2\\z=4`

`4\in D` so there are no extraneous solutions.

Answer 2

There are no extraneous solutions

Reasoning:

An extraneous solution is a solution that isn't valid, it might be imaginary like the square root of a negative number.

first we want to isolate z:

1+sqrt(z)=sqrt(z+5)

^2 all ^2 all

(1+sqrt(z))(1+sqrt(z))=z+5

expand

1+2sqrt(z)+z=z+5

-1 -z -z -1

2sqrt(z)=4

/2 /2

sqrt(z)=2

^2 all ^2 all

z=4

Since there is one solution and it is a real number, there are no extraneous solutions.