Answer
The solutions are $x=0$ and $x=4$
Work Step by Step
$\sqrt{1+4\sqrt{x}}=1+\sqrt{x}$
Square both sides of the equation:
$(\sqrt{1+4\sqrt{x}})^{2}=(1+\sqrt{x})^{2}$
$1+4\sqrt{x}=1+2\sqrt{x}+x$
Remove $1$ from both sides of the equation:
$4\sqrt{x}=x+2\sqrt{x}$
Take $2\sqrt{x}$ to the left side of the equation:
$4\sqrt{x}-2\sqrt{x}=x$
$2\sqrt{x}=x$
Once again, square both sides of the equation:
$(2\sqrt{x})^{2}=x^{2}$
$4x=x^{2}$
Take $4x$ to the right side:
$0=x^{2}-4x$
Rearrange:
$x^{2}-4x=0$
Take out common factor $x$ :
$x(x-4)=0$
Set both factors equal to $0$ and solve each individual equation:
$x=0$
$x-4=0$
$x=4$
The solutions are $x=0$ and $x=4$
