If \(x = {{\sqrt5 - \sqrt4} \over {\sqrt5 + \sqrt4}}\) and \(y = {{\sqrt5 + \sqrt4} \over {\sqrt5 - \sqrt4}}\) then the value of \(x^2 - xy + y^2 \over x^2 +xy + y^2\) = ?

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SSC CGL 2022 Tier-I Official Paper (Held On : 08 Dec 2022 Shift 2)
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  1. \(361 \over 363\)
  2. \(341 \over 343\)
  3. \(384 \over 387\)
  4. \(321 \over 323\)

Answer (Detailed Solution Below)

Option 4 : \(321 \over 323\)
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PYST 1: SSC CGL - English (Held On : 11 April 2022 Shift 1)
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Detailed Solution

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Given:

 \(x = {{\sqrt5 - \sqrt4} \over {\sqrt5 + \sqrt4}}\) and \(y = {{\sqrt5 + \sqrt4} \over {\sqrt5 - \sqrt4}}\)

Calculations:

According to question,

 \(x = {{\sqrt5 - \sqrt4} \over {\sqrt5 + \sqrt4}}\) 

⇒x = 9 - 4√5

⇒x² = 161 - 72√5

and \(y = {{\sqrt5 + \sqrt4} \over {\sqrt5 - \sqrt4}}\) 

⇒y = 9 + 4√5

⇒y² = 161 + 72√5

Now, 

 \(x^2 - xy + y^2 \over x^2 +xy + y^2\)

= (161 - 72√5 - 1 + 161 + 72√5)/(161 - 72√5 + 1 + 161 + 72√5)

= (322 - 1)/(322 + 1)

= 321/323

Hence, The Required value is 321/323.

Alternate Method

We know, if x = \(\frac{(p - q)}{(p + q)}\) and y = \(\frac{(p + q)}{(p - q)}\) then (x + y) = \(\frac{2\times (p^2 + q^2)}{(p^2 - q^2)}\)

 \(x = {{\sqrt5 - \sqrt4} \over {\sqrt5 + \sqrt4}}\) and \(y = {{\sqrt5 + \sqrt4} \over {\sqrt5 - \sqrt4}}\)  so, x + y = \(\frac{2(\sqrt5^2 + \sqrt4^2)}{(5 - 4)}\) = 18 and xy = 1

So, \(x^2 - xy + y^2 \over x^2 +xy + y^2\) = \(\frac{(x + y)^2 - 3xy}{(x + y)^2 -xy}\) = \(\frac{18^2 - 3}{18^2 -1}\) = 321/323

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