#### Answer

$x=5$

#### Work Step by Step

Using the properties of equality to isolate the radical expression in the given equation, $
x=\sqrt{x-1}+3
,$ results to
\begin{array}{l}\require{cancel}
x=\sqrt{x-1}+3
\\\\
x-3=\sqrt{x-1}
.\end{array}
Squaring both sides and using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x-3=\sqrt{x-1}
\\\\
(x-3)^2=\left( \sqrt{x-1} \right)^2
\\\\
(x)^2-2(x)(3)+(3)^2=x-1
\\\\
x^2-6x+9=x-1
\\\\
x^2-6x-x+9+1=0
\\\\
x^2-7x+10=0
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
x^2-7x+10=0
\\\\
(x-5)(x-2)=0
.\end{array}
Equating each factor to zero (Zero Product Property, the solutions are
\begin{array}{l}\require{cancel}
x-5=0
\\\\\text{OR}\\\\
x-2=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x-5=0
\\\\
x=5
\\\\\text{OR}\\\\
x-2=0
\\\\
x=2
.\end{array}
If $x=2,$ then
\begin{array}{l}\require{cancel}
x=\sqrt{x-1}+3
\\\\
2=\sqrt{2-1}+3
\\\\
2=\sqrt{1}+3
\\\\
2=1+3
\\\\
2=4
\text{ (FALSE)}
.\end{array}
Hence, only $
x=5
$ satisfies the original equation.