Question

If $ x = 9 + 4\sqrt 5 $ and $ xy = 1, $ then find the value of $ \dfrac{1}{{322}}\left( {\dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}}} \right). $

Answer 1

Hint: First of all find the square of the term “x” and the term “y” with the help of finding the reciprocal and the conjugate of the terms and place in the equation and simplify for the resultant required value.

Complete step-by-step answer:
Given that: $ x = 9 + 4\sqrt 5 $
and $ xy = 1 $
Make required term “y” the subject-
Term multiplicative on one side if moved to the opposite side then it goes to the denominator.
 $ y = \dfrac{1}{x} $
Place the value of “x” in the above equation –
 $ y = \dfrac{1}{{9 + 4\sqrt 5 }} $
Find the conjugate for the above expression in the denominator –
\[y = \dfrac{1}{{9 + 4\sqrt 5 }} \times \dfrac{{9 - 4\sqrt 5 }}{{9 - 4\sqrt 5 }}\]
Simplify the above expression using the identity - $ (a + b)(a - b) = {a^2} - {b^2} $
\[y = \dfrac{{9 - 4\sqrt 5 }}{{{{(9)}^2} - {{(4\sqrt 5 )}^2}}}\]
Place the value for the square of the terms –, square of the square-root of the term is the number since the square and square root cancel each other.
\[y = \dfrac{{9 - 4\sqrt 5 }}{{81 - 80}}\]
Simplify the above equation finding the difference of the term in the denominator.
\[y = \dfrac{{9 - 4\sqrt 5 }}{1}\]
Any number upon one is the number itself.
\[y = 9 - 4\sqrt 5 \] ….. (A)
Square of the above term is –
\[\left( {{y^2}} \right) = {\left( {9 - 4\sqrt 5 } \right)^2}\] ….. (B)
Similarly, find the square of the term
 $ x = 9 + 4\sqrt 5 $
 $ {x^2} = {\left( {9 + 4\sqrt 5 } \right)^2} $ …… (C)
Now, the value of $ \dfrac{1}{{322}}\left( {\dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}}} \right). $
Take LCM (least common multiple) for the above equation –
 $ = \dfrac{1}{{322}}\left( {\dfrac{{{x^2} + {y^2}}}{{{x^2}{y^2}}}} \right) $
Square multiplied with another square can be written as the product of terms and its whole square.
 $ = \dfrac{1}{{322}}\left( {\dfrac{{{x^2} + {y^2}}}{{{{(xy)}^2}}}} \right) $
Place the given value $ xy = 1, $ in the above expression
 $ = \dfrac{1}{{322}}\left( {\dfrac{{{x^2} + {y^2}}}{{{{(1)}^2}}}} \right) $
Place value from equation (A) and (B)
 $ = \dfrac{1}{{322}}\left( {{{\left( {9 + 4\sqrt 5 } \right)}^2} + {{\left( {9 - 4\sqrt 5 } \right)}^2}} \right) $
Expand the above expression –
\[ = \dfrac{1}{{322}}\left( {\left( {{{(9)}^2} + 2(9)4(\sqrt 5 ) + {{(4\sqrt 5 )}^2}} \right) + \left( {{{(9)}^2} - 2(9)4(\sqrt 5 ) + {{(4\sqrt 5 )}^2}} \right)} \right)\]
Place the values for the square of the terms, square is the term which is multiplied with itself twice.
\[ = \dfrac{1}{{322}}\left( {\left( {81 + 72\sqrt 5 + 80} \right) + \left( {81 - 72\sqrt 5 + 80} \right)} \right)\]
Simplify, like terms with the same value and the opposite sign cancel each other.
\[ = \dfrac{1}{{322}}\left( {81 + 80 + 81 + 80} \right)\]
Simplify the above expression
\[ = \dfrac{1}{{322}}\left( {322} \right)\]
Common terms from the numerator and the denominator cancels each other.
 $ = 1 $
Hence, $ \dfrac{1}{{322}}\left( {\dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}}} \right) = 1 $
So, the correct answer is “1”.

Note: Always remember that the term with the same value and the opposite sign cancel each other. Also, be good in square and square roots. Square and square root cancels each other. Remember the square of the numbers at least till twenty.