SOLUTION:
[tex]\begin{gathered} \sqrt[]{x-1}\text{ - 5 = x - 8} \\ \sqrt[]{x-1}=\text{ x -8 + 5} \\ \sqrt[]{x-1}\text{ = x -3} \end{gathered}[/tex]
Squaring both sides;
[tex]\begin{gathered} (\sqrt[]{x-1})^2=(x-3)^2 \\ x-1=x^2\text{ - 6x + 9} \\ \end{gathered}[/tex]
Re-arranging the equation;
[tex]\begin{gathered} x^2-6x-x+9\text{ + 1 = 0} \\ x^2\text{ -7x + 10 = 0} \\ x^2-5x\text{ - 2x + 10 = 0} \\ x(x-5)-2(x-5)=0 \\ (x-2)(x-5)=0 \\ x=2\text{ or x = 5} \end{gathered}[/tex]
We now verify using the two values of x to know the valid solution for x;
When x = 2
[tex]\begin{gathered} \sqrt[]{x-1}\text{ - 5 = x - 8} \\ \sqrt[]{2-1}\text{ - 5 = 2 - 8} \\ \sqrt[]{1}-5\text{ = -6} \\ 1-5\text{ = -6} \\ -4\text{ = -6 ( NOT valid or FALSE)} \end{gathered}[/tex]
When x = 5
[tex]\begin{gathered} \sqrt[]{x-1}\text{ - 5 = x - 8} \\ \sqrt[]{5-1}\text{ - 5 = 5 - 8} \\ \sqrt[]{4}\text{ - 5 = -3} \\ 2\text{ - 5 = -3} \\ -3\text{ = -3 (Valid or TRUE)} \end{gathered}[/tex]
CONCLUSION:
The valid solution is x = 5
