f. \( 3 \sqrt{x}+\frac{5}{\sqrt{x}}=16 \)

To tackle the equation \( 3 \sqrt{x} + \frac{5}{\sqrt{x}} = 16 \), you can start by making a substitution. Let \( y = \sqrt{x} \). This turns your equation into \( 3y + \frac{5}{y} = 16 \). Multiplying through by \( y \) (assuming \( y \neq 0 \)) gives you \( 3y^2 + 5 = 16y \). Rearranging gives you a standard quadratic equation: \( 3y^2 - 16y + 5 = 0 \). You can now use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the values for \( y \), and then substitute back to find \( x \). To avoid common pitfalls, remember to check that your solution for \( y \) yields a valid non-negative value since \( y = \sqrt{x} \) must be non-negative. After solving the quadratic, take the valid roots, substitute back to find \( x \), and confirm if they satisfy the original equation. Enjoy the process—solving equations is like cracking a code!