Answer
$\{5\}$
Work Step by Step
Rewrite the equation so the root is isolated on one side (with a + sign)
Add $\sqrt{x+11}-1$ to both sides.
$ x-1=\sqrt{x+11}$
For the root to be real, x must be such that $ x\geq-11.$
The RHS is nonnegative, so the LHS = $ x-1 \;$ is also nonnegative.
All solutions must satisfy
$ x\geq 1\qquad (*)$
Square both sides.
$(x-1)^{2}=x+11$
$ x^{2}-2x+1=x+11$
$ x^{2}-3x-10=0$
We factor this trinomial by finding two factors of $-10$ that add to $-3$; they are $-5$ and $+2$.
$(x-5)(x+2)=0$
$ x=-2 \;$ does not satisfy (*). We discard it.
$ x=5 \;$ is a valid solution (satisfies (*) )
Check:
$5-\sqrt{5+11}=1$
$5-\sqrt{16}=1$
$5-4=1$
The solution set is $\{5\}$.
