College Algebra (6th Edition)

The solutions are $x=0$ and $x=4$
$\sqrt{1+4\sqrt{x}}=1+\sqrt{x}$ 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 $x=4$