## College Algebra (10th Edition)

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