Question

If x= 3 - root 8 find the values of x+1/x, x^2+1/x^2, x^2- 1/x^2

Answer 1

Hey dear !! :D
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Given , x = 3 - √8

(I) x + ( 1 / x ) = ?

=> (3 - √8) + (1 / 3 - √8 )

=> (3 - √8) + { ( 3 + √8 ) / (3 + √8 ) (3 - √8) }

=> (3 - √8) + { ( 3 + √8 ) / ( 9 - 8) }

=> 3 - √8 + 3 + √8

=> 6

Ans. => x + ( 1 / x ) = 6
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Given , x = 3 - √8

(II) x² + ( 1 / x²) = ?

=> (3 - √8)² + (1 / 3 - √8 )²

=> ( 9 - 8) + { 1 / (9 - 8)}

=> 1 + 1 / 1

=> 1 + 1

=> 2

Ans. => x² + ( 1 / x²) = 2
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Given , x = 3 - √8

(III) x² - ( 1 / x²) = ?

=> (3 - √8)² - (1 / 3 - √8 )²

=> ( 9 - 8) - { 1 / (9 - 8)}

=> 1 - 1 / 1

=> 0

Ans. => x² - ( 1 / x²) = 0
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Hope it helps!!! :D

Answer 2

Solution :

[tex]-\ \textgreater \ x = 3 - \sqrt{8} \\ =\ \textgreater \ \frac{1}{x} = \frac{1}{3- \sqrt{8} } = \frac{3+ \sqrt{8} }{9-8} = 3 + \sqrt{8} [/tex]

[tex]=\ \textgreater \ x + \frac{1}{x} = ( 3 + \sqrt{8} ) + ( 3 - \sqrt{8} ) = 6 \\ \\ =\ \textgreater \ ( x + \frac{1}{x} )^2 = 36 \\ \\ =\ \textgreater \ x^2 + ( \frac{1}{x} )^2 = 34 \\ \\ =\ \textgreater \ ( x + \frac{1}{x} )( x - \frac{1}{x} ) = ( 6 )( 2\sqrt{8} ) = 12 \sqrt{8} [/tex]

Hence, your answer is :
[tex](i) ( \ x + \frac{1}{x} \ ) = 6 \\ \\ (ii) \ ( \ x^2 + \frac{1}{x^2} \ ) = 34 \\ \\ (iii) \ ( \ x^2 - \frac{1}{x^2} \ ) = 12 \sqrt{8} [/tex]