## College Algebra (10th Edition)

$x=-4$ or $x=1$
We solve by factoring: $x^{2}+3x+\sqrt{x^{2}+3x}=6$ $x^{2}+3x+\sqrt{x^{2}+3x}-6=0$ $[\sqrt{x^2+3x}+3][\sqrt{x^2+3x}-2]=0$ $\sqrt{x^2+3x}+3=0$ or $\sqrt{x^2+3x}-2=0$ $\sqrt{x^2+3x}=-3$ is not possible because a square root is never negative. So: $\sqrt{x^2+3x}=2$ $x^2+3x=(2)^2$ $x^2+3x=4$ $x^2+3x-4=0$ $(x+4)(x-1)=0$ $x=-4$ or $x=1$