Bahrain
If ^@x(2 - \sqrt{ 2 } ) = y(2 + \sqrt{ 2 } ) = 1^@, find the value of ^@x^2 - y^2^@.
Answer:
^@2 \sqrt{ 2 }^@
- Since ^@x(2 - \sqrt{ 2 }) = 1^@, we can say that ^@x = \dfrac{1}{ 2 - \sqrt{ 2 } } ^@.
- Now if we multiply both numerator and denominator of the given fraction by ^@2 + \sqrt{ 2 }^@, we get:
^@\dfrac{1}{ 2 - \sqrt{ 2 } } \times \dfrac{ 2 + \sqrt{ 2 } }{ 2 + \sqrt{ 2 } } ^@
^@\begin{align}\text{ The denominator } & = (2 - \sqrt{ 2 })(2 + \sqrt{ 2 }) \\ & = (2^2 - (\sqrt{ 2 })^2) \\ & = (4 - 2) \\ & = 2 \end{align}^@ - From above steps we get ^@x = \dfrac{ 2 + \sqrt{ 2 } } { 2 }^@.
- Similarly, for ^@y(2 + \sqrt{ 2 }) = 1^@, we can say that:
^@y = \dfrac{1}{ 2 + \sqrt{ 2 } }^@. - Now if we multiply both numerator and denominator by ^@2 - \sqrt{ 2 } ^@, we get:
^@\dfrac{1}{ 2 + \sqrt{ 2 } } \times \dfrac{ 2 - \sqrt{ 2 } }{ 2 - \sqrt{ 2 } }^@
^@\begin{align}\text{ The denominator } & = (2 + \sqrt{ 2 })(2 - \sqrt{ 2 }) \\ & = (2^2 - (\sqrt{ 2 })^2) \\ & = (4 - 2) \\ & = 2 \end{align}^@ - From the step ^@4^@ and ^@5^@, we get: ^@y = \dfrac{ 2 - \sqrt{ 2 } }{ 2 }^@.
- ^@\begin{align}\text{ Now the value } x^2 - y^2 & = \left( \dfrac{ 2 + \sqrt{ 2 } } { 2 }\right)^2 - \left(\dfrac{ 2 - \sqrt{ 2 } } { 2 }\right)^2 \\ & = \left(\dfrac{ 2^2 + 2 + 4 \sqrt{ 2 } } { 4 } - \dfrac{ 2^2 + 2 - 4 \sqrt{ 2 } } { 4 } \right) \\ & = \dfrac{ 2^2 + 2 + 4 \sqrt{ 2 } - 2^2 - 2 + 4 \sqrt{ 2 } } { 4 } \\ & = 2\sqrt{ 2 } \end{align}^@
