#### Answer

$k=0$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given radical equation, $
\sqrt{k^2+2k+9}=k+3
,$ square both sides of the equation. Use special products and concepts of solving quadratic equations. Finally, do checking of the solution with the original equation.
$\bf{\text{Solution Details:}}$
Squaring both sides of the equation results to
\begin{array}{l}\require{cancel}
\left( \sqrt{k^2+2k+9} \right)^2=(k+3)^2
\\\\
k^2+2k+9=(k+3)^2
.\end{array}
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 expression above is equivalent to
\begin{array}{l}\require{cancel}
k^2+2k+9=(k)^2+2(k)(3)+(3)^2
\\\\
k^2+2k+9=k^2+6k+9
.\end{array}
Using the propertis of equality to isolate the variable results to
\begin{array}{l}\require{cancel}
(k^2-k^2)+(2k-6k)=9-9
\\\\
-4k=0
\\\\
k=\dfrac{0}{-4}
\\\\
k=0
.\end{array}
Upon checking, $
k=0
$ satisfies the original equation.