Remember that an extraneous solution of an equation, is the solution that emerges from solving the equation but is not a valid solution.
 
Lets solve our equation to find out what is the extraneous solution:
[tex] \sqrt{x-3} =x-5[/tex]
[tex](\sqrt{x-3})^2 =(x-5)^2[/tex]
[tex]x-3=x^2-10x+25[/tex]
[tex]x^2-11x+28=0[/tex]
[tex](x-4)(x-7)=0[/tex]
[tex]x-4=0[/tex] and [tex]x-7=0[/tex]
[tex]x=4[/tex] and [tex]x=7[/tex]

So, the solutions of our equation are [tex]x=4[/tex] and [tex]x=7[/tex]. Lets replace each solution in our original equation to check if they are valid solutions:
- For [tex]x=7[/tex]
[tex] \sqrt{x-3} =x-5[/tex]
[tex]\sqrt{7-3} =7-5[/tex]
[tex] \sqrt{4} =2[/tex]
[tex]2=2[/tex]
We can conclude that 7 is a valid solution of the equation.

- For [tex]x=4[/tex]
[tex] \sqrt{x-3} =x-5[/tex]
[tex]\sqrt{4-3} =4-5[/tex]
[tex] \sqrt{1} =1[/tex]
[tex]1 \neq 1[/tex]
We can conclude that 4 is not a valid solution of the equation; therefore, 4 is a extraneous solution.

We can conclude that the correct answer is: D. the extraneous solution is x = 4