Find all values of $a$ such that $\frac{a-3}{\sqrt{a}} = -a \sqrt{a}$.
1 answer
Manipulating the left side, $$\frac{a-3}{\sqrt{a}} = \frac{a - 3}{\sqrt{a}} \cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{a \sqrt{a} - 3 \sqrt{a}}{\sqrt{a} \cdot \sqrt{a}} = \frac{a \sqrt{a} - 3 \sqrt{a}}{a} = \sqrt{a} - \frac{3}{\sqrt{a}}.$$Multiplying through by $a$ (which is possible since if $a = 0$, the equation does not hold for any value of $a$) gives $a - 3 = a \sqrt{a} - 3 \sqrt{a}^2 = a \sqrt{a} - 3a$, so $3 \sqrt{a}^2 = 2a$. Then \[a = \frac{9}{4} \sqrt{a}^2 = \frac{9}{4} \cdot \frac{2a}{3} = \frac{3a}{2}.\]Hence $a = \boxed{0}$, which we must check against the hypotheses.