Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 118: 57

$x=16$

We solve by factoring: $x+\sqrt{x}=20$ $x+\sqrt{x}-20=0$ $[\sqrt{x}+5][\sqrt{x}-4]=0$ $\sqrt{x}+5=0$ or $\sqrt{x}-4=0$ $\sqrt{x}=-5$ or $\sqrt{x}=4$ $x=(-5)^2$ or $x=4^2$ $x=25$ or $x=16$ However, $x=25$ does not work in the original equation, so we throw this solution out. $25+\sqrt{25}\ne 20$ Thus, the only solution is: $x=16$