Question:

Step 1: Let $x = \sqrt[4]{r}$. Since $r$ is a positive real number, $x > 0$. The given condition is $$ x - \frac{1}{x} = 4 $$ Step 2: Square both sides of the equation. $$ \left(x - \frac{1}{x}\right)^2 = 4^2 $$ $$ x^2 - 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = 16 $$ $$ x^2 - 2 + \frac{1}{x^2} = 16 $$ Step 3: Isolate $x^2 + \frac{1}{x^2}$. $$ x^2 + \frac{1}{x^2} = 16 + 2 $$ $$ x^2 + \frac{1}{x^2} = 18 $$ Step 4: Consider the expression to be found, $x + \frac{1}{x}$. Square this expression. $$ \left(x + \frac{1}{x}\right)^2 = x^2 + 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 $$ $$ \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} $$ $$ \left(x + \frac{1}{x}\right)^2 = \left(x^2 + \frac{1}{x^2}\right) + 2 $$ Step 5: Substitute the value of $x^2 + \frac{1}{x^2}$ from Step 3. $$ \left(x + \frac{1}{x}\right)^2 = 18 + 2 $$ $$ \left(x + \frac{1}{x}\right)^2 = 20 $$ Step 6: Take the square root of both sides. Since $x > 0$, $x + \frac{1}{x}$ must be positive. $$ x + \frac{1}{x} = \sqrt{20} $$ $$ x + \frac{1}{x} = \sqrt{4 \cdot 5} $$ $$ x + \frac{1}{x} = 2\sqrt{5} $$ Thus, the value of $\sqrt[4]{r} + \dfrac{1}{\sqrt[4]{r}}$ is $2\sqrt{5}$.