Question
Download Solution PDFIf \(x = {{\sqrt5 - \sqrt4} \over {\sqrt5 + \sqrt4}}\) and \(y = {{\sqrt5 + \sqrt4} \over {\sqrt5 - \sqrt4}}\) then the value of \(x^2 - xy + y^2 \over x^2 +xy + y^2\) = ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
\(x = {{\sqrt5 - \sqrt4} \over {\sqrt5 + \sqrt4}}\) 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
Last updated on Jun 11, 2025
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