### Given that $w > 0$ and that $w - \dfrac{1}{w} = 7$, find the value of $\left(w + \dfrac{1}{w} \right)^2$.

$53$
1. We are given, $w > 0$ and $w - \dfrac{1}{w} = 5$ and we need to find the value of $\left(w + \dfrac{1}{w} \right)^2$.
2. \begin{align} & \left(w + \dfrac{1}{w} \right)^2 = w^2 + \dfrac{1}{ w^2 } + 2(w)(\dfrac{1}{w}) \\ & \left(w + \dfrac{1}{w} \right)^2 = w^2 + \dfrac{1}{ w^2 } + 2 \end{align}
Adding and subtracting $2$ on RHS, we get,
\begin{align} & \left(w + \dfrac{1}{w} \right)^2 = w^2 + \dfrac{1}{ w^2 } + 2 + 2 - 2 \\ & \left(w + \dfrac{1}{w} \right)^2 = w^2 + \dfrac{1}{ w^2 } - 2(w)(\dfrac{1}{w}) + 4 \\ & \left(w + \dfrac{1}{w} \right)^2 = \left(w - \dfrac{1}{w} \right)^2 + 4 \\ & \left(w + \dfrac{1}{w} \right)^2 = 7^2 + 4 \\ & \left(w + \dfrac{1}{w} \right)^2 = 53 \end{align}
3. Hence, the value of $\left(w + \dfrac{1}{w} \right)^2$ is $53$.