Answer
$x=1$
Work Step by Step
Squaring both sides of the given equation, $
\sqrt[]{\sqrt{x+3}+\sqrt{x}}=\sqrt[]{3}
,$ results to
\begin{array}{l}\require{cancel}
\sqrt{x+3}+\sqrt{x}=3
.\end{array}
Squaring both sides again results to
\begin{array}{l}\require{cancel}
(\sqrt{x+3})^2+2(\sqrt{x+3})(\sqrt{x})+(\sqrt{x})^2=9
\\\\
x+3+2\sqrt{x(x+3)}+x=9
\\\\
(x+x)+(3-9)=-2\sqrt{x(x+3)}
\\\\
2x-6=-2\sqrt{x(x+3)}
\\\\
x-3=-\sqrt{x(x+3)}
.\end{array}
Squaring both sides again results to
\begin{array}{l}\require{cancel}
(x)^2+2(x)(-3)+(-3)^2=x(x+3)
\\\\
x^2-6x+9=x^2+3x
\\\\
(x^2-x^2)+(-6x-3x)=-9
\\\\
-9x=-9
\\\\
x=\dfrac{-9}{-9}
\\\\
x=1
.\end{array}
Upon checking, $
x=1
$ satisfies the original equation.