## College Algebra 7th Edition

$x=\pm 1$ $x=\pm \sqrt{3}$
$x^{4}-4x^{2}+3=0$ First, we substitute $y=x^{2}$ into the equation: $y^{2}-4y+3=0$ Next, we factor and solve: $(y-1)(y-3)=0$ $y-1=0$ or $y-3=0$ $y=1$ or $y=3$ Finally, we convert the $y$ solutions to $x$ solutions: $y=x^2$ $x=\pm\sqrt{y}$ If $y=1$: $x=\pm \sqrt{1}=\pm 1$ If $y=3$: $x=\pm \sqrt{3}$