Given:
[tex]2\sqrt{x-5}=2[/tex]
Required:
We need to solve for x and identify if it is an extraneous solution.
Explanation:
First find the value of x
[tex]\begin{gathered} 2\sqrt{x-5}=2 \\ \sqrt{x-5}=1 \\ x-5=1 \\ x=6 \end{gathered}[/tex]
This solution is not extraneous, because extraneous solution emerage from solving the problem but are not actually valid solution fro initial problem.
With rounding , this solution is valid for the initial problem
Final answer:
A
