Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 118: 56

$x=0$

We solve by factoring: $x+8\sqrt{x}=0$ $8\sqrt{x}=-x$ $(8\sqrt{x})^{2}=(-x)^{2}$ $64x=x^{2}$ $x^{2}-64x=0$ $x(x-64)=0$ $x=0$ or $x=64$ However, $x=64$ does not work the original equation, so we throw this solution out. $(64)+8\sqrt{64}\ne 0$ Thus, the only solution is: $x=0$