## College Algebra (10th Edition)

$x=4$
We solve by factoring: $x+\sqrt{x}=6$ $x+\sqrt{x}-6=0$ $[\sqrt{x}+3][\sqrt{x}-2]=0$ $\sqrt{x}+3=0$ or $\sqrt{x}-2=0$ $\sqrt{x}=-3$ or $\sqrt{x}=2$ $x=(-3)^2$ or $x=2^2$ $x=9$ or $x=4$ However, $x=9$ does not work in the original equation, so we throw this solution out. $9+\sqrt{9}\ne 6$ Thus, the only solution is: $x=4$