## Algebra and Trigonometry 10th Edition

$\sqrt {16+9\sqrt x} = 4+\sqrt x$ $(\sqrt {16+9\sqrt x})^2 = (4+\sqrt x)^2$ $16+9\sqrt x = 16+8\sqrt x+x$ $\sqrt x = x$ $x=x^2$ x(x-1)=0 x= 0 or 1 (Both satisfy the original equation)