Math, asked by parikshit37, 10 months ago


if \: x = 2 -   \sqrt{5}  \div 2 +  \sqrt{5}  \: and \: y = 2 +  \sqrt{5}  \div 2 -  \sqrt{5.}   \: find \: the \: value \: of \: x{\:}^{2} and  \: y {}^{2}

Answers

Answered by abhi569
3

Answer:

161 - 72√5  &   161 + 72√5

Step-by-step explanation:

x = ( 2 - √5 ) / ( 2 + √5 )

Dividing and multiplying by 2 - √5 RHS of x:

⇒ x = ( 2 - √5 )( 2 - √5 ) / ( 2 + √5 )( 2 - √5 )

⇒ x = ( 2 - √5 )^2 / { ( 2 )^2 - ( √ 5 )^2 }               { ( a + b )( a - b ) = a^2 - b^2 }

⇒ x = ( 2 - √5 )^2 / ( 4 - 5 )

⇒ x = ( 2 - √5 )^2 / ( - 1 ) = - ( 2 - √5 )^2

⇒ x = - ( 4 + 5 - 4√5 )

⇒ x = - ( 9 - 4√5 )

⇒ x^2 = ( - { 9 - 4√5 } )^2

⇒ x^2 = ( 9 - 4√5 )^2

⇒ x^2 = 81 + 80 - 72√5

⇒ x^2 = 161 - 72√5       ...( 1 )

    y = ( 2 + √5 ) / ( 2 - √5 )

Dividing and multiplying RHS of y by ( 2 + √5 ):

⇒ y = ( 2 + √5 )( 2 + √5 ) / ( 2 - √5 )( 2 + √5 )

⇒ y = ( 2 + √5 )^2 / ( 4 - 5 )

⇒ y = - ( 2 + √5 )^2

⇒ y = - ( 4 + 5 + 4√5 )

⇒ y = - ( 9 + 4√5 )

⇒ y^2 = ( 9 + 4√5 )^2

⇒ y^2 = 161 + 72√5                    ...( 2 )

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