Answer
$\{8\}$
Work Step by Step
For the root to be real, x must be such that $ x\geq-1/6.$
The LHS is nonnegative, so the RHS = $ x-1 \;$ is also nonnegative.
All solutions must satisfy
$ x\geq 1\qquad (*)$
$\sqrt{6x+1}=x-1 \quad $... Square both sides
$6x+1=(x-1)^{2}$
$6x+1=x^{2}-2x+1$
$0=x^{2}-8x $
$0=x(x-8)$
$ x=0 \;$ does not satisfy (*). We discard it.
$ x=8 \;$ is a valid solution (satisfies (*) )
Check:
$\sqrt{6(8)+1}=8-1$
$\sqrt{49}=7$
The solution set is $\{8\}$.
