Pakistan
Given that ^@w > 0^@ and that ^@ w - \dfrac{1}{w} = 7^@, find the value of ^@\left(w + \dfrac{1}{w} \right)^2^@.
Answer:
^@53^@
- 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^@.
- ^@\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}^@ - Hence, the value of ^@\left(w + \dfrac{1}{w} \right)^2^@ is ^@53^@.
