## College Algebra 7th Edition

$x=9$, $x=4$
$x-5\sqrt{x}+6=0$ We make the substitution $y=\sqrt{x}$. The equation becomes: $y^{2}-5y+6=0$ $(y-3)(y-2)=0$ $y-3=0$ or $y-2=0$ $y=3$ or $y=2$ We find the $x$ solutions with the $y$ solutions: $y=\sqrt{x}$ $x=y^2$ For $y=3$: $x=3^2=9$ For y=2: $x=2^2=4$