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


Answer:

^@53^@

Step by Step Explanation:
  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^@.

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