24. \( \frac{a}{b \sqrt{a}}+\frac{\sqrt{a}}{b}=3 \) \( \Rightarrow \sqrt{a}= \) ?

To solve the equation \( \frac{a}{b \sqrt{a}}+\frac{\sqrt{a}}{b}=3 \), we can start by simplifying the expressions. The first term can be rewritten as \( \frac{\sqrt{a}}{b} \) since \( a = (\sqrt{a})^2 \). This gives us: \(\frac{\sqrt{a}}{b} \cdot \sqrt{a} + \frac{\sqrt{a}}{b} = 3\). Factoring out \( \frac{\sqrt{a}}{b} \), we have: \(\frac{\sqrt{a}}{b} (\sqrt{a} + 1) = 3\). Multiplying both sides by \( b \) results in: \(\sqrt{a} (\sqrt{a} + 1) = 3b\). Now, let’s set \( \sqrt{a} = x \). The equation becomes: \( x(x + 1) = 3b \). This simplifies to a quadratic: \( x^2 + x - 3b = 0\). Applying the quadratic formula: \[ x = \frac{-1 \pm \sqrt{1 + 12b}}{2} \] Thus, \( \sqrt{a} = \frac{-1 + \sqrt{1 + 12b}}{2} \) or \( \sqrt{a} = \frac{-1 - \sqrt{1 + 12b}}{2} \). Since \( \sqrt{a} \) must be non-negative, we take: \(\sqrt{a} = \frac{-1 + \sqrt{1 + 12b}}{2}\).