ANSWER
The equation has one valid solution and one extraneous solution.
A valid solution for x is 5
[tex]\sqrt[]{x-1}-5=x-8[/tex]
Add 5 to both-side of the equation
[tex]\sqrt[]{x-1}-5+5=x-8+5[/tex][tex]\sqrt[]{x-1}=x-3[/tex]Take the square of both-side
[tex]x-1=(x-3)^2[/tex]x - 1=x²-6x + 9
Rearrange
x² - 6x + 9 - x + 1 =0
x² - 7x + 10 = 0
We can solve the above quadratic equation using factorization method
x² - 5x - 2x + 10 = 0
x(x-5) - 2(x - 5) = 0
(x-5)(x-2)=0
Either x -5 =0 OR x-2 =0
Either x =5 or x=2
To check whether the equation is valid or non-extraneous, let's plug the values into the equation and see if it gives a true statement
For x =5
[tex]\sqrt[]{5-1}-5=5-8[/tex][tex]\sqrt[]{4}-5=-3[/tex][tex]-3=-3[/tex]The above is a true statement
For x =2
[tex]\sqrt[]{2-1}-5=2-8[/tex][tex]1-5=2-8[/tex]The above is not a true statement
Therefore, the equation has one valid solution and one extraneous solution.
A valid solution for x is 5