## College Algebra 7th Edition

$x=3$
$x^{1/2}-x^{-1/2}-6x^{-3/2}=0$ $x^{-3/2}(x^{2}-x-6)=0$ $x^{-3/2}(x+2)(x-3)=0$ $x^{-3/2}=0$ or $x+2=0$ or $x-3=0$ $\frac{1}{x^{3/2}}=0$ or $x=-2$ or $x=3$ The left equation has no solution. The middle solution will not work in the original equation (we can not take the square root of a negative number). Thus the only solution is $x=3$