If ^@x(2 - \sqrt{ 2 } ) = y(2 + \sqrt{ 2 } ) = 1^@, find the val | Math Question and Answer

Answer:

$2 \sqrt{ 2 }$

Step by Step Explanation:

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: $y = \dfrac{ 2 - \sqrt{ 2 } }{ 2 }$ .
$\begin{align}\text{ Now the value } x^2 - y^2 & = \left( \dfrac{ 2 + \sqrt{ 2 } } { 2 }\right)^2 - \left(\dfrac{ 2 - \sqrt{ 2 } } { 2 }\right)^2 \\ & = \left(\dfrac{ 2^2 + 2 + 4 \sqrt{ 2 } } { 4 } - \dfrac{ 2^2 + 2 - 4 \sqrt{ 2 } } { 4 } \right) \\ & = \dfrac{ 2^2 + 2 + 4 \sqrt{ 2 } - 2^2 - 2 + 4 \sqrt{ 2 } } { 4 } \\ & = 2\sqrt{ 2 } \end{align}$

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