## College Algebra (10th Edition)

$x=\sqrt{3}$ and $x=\sqrt{2}$
We solve by factoring: $\sqrt[4]{5x^{2}-6}=x$ $(\sqrt[4]{5x^{2}-6})^{4}=x^{4}$ $5x^{2}-6=x^{4}$ $x^{4}-5x^{2}+6=0$ $(x^2-3)(x^2-2)=0$ $(x^2-3)=0$ or $(x^2-2)=0$ $x^2=3$ or $x^2=2$ $x=\pm\sqrt{3}$ or $x=\pm\sqrt{2}$ However, the negative solutions $x=-\sqrt{3}$ and $x=-\sqrt{2}$ do not work in the original equation (because the 4th root on the left side of the equation is never negative), so we throw these solutions out. Thus, the only solution are: $x=\sqrt{3}$ and $x=\sqrt{2}$