Question

If
$x = 3 + \sqrt{2}$
find the value of
$x + \frac{1}{ {x}^{2} }$
​

Answer 1

Looks like you are asking for the value of x^2 + 1 / x^2.

Answer:

( 550 + 288√2 ) / 49

Step-by-step explanation:

Given,

    x = 3 + √2

⇒ x = 3 + √2

⇒ 1 / x = 1 / ( 3 + √2 )

Multiplying and diviiding RHS by ( 3 - √2 ):

⇒ 1 / x = ( 3 - √2 ) / ( 3 + √2 )( 3 - √2 )

⇒ 1 / x  = ( 3 - √2 ) / [ ( 3 )^2 - ( √2 )^2 ]           { Using ( a + b )( a - b ) = a^2 - b^2 }

⇒ 1 / x = ( 3 - √2 ) / [ 9 - 2 ]

⇒ 1 / x = ( 3 - √2 ) / 7

Therefore,

⇒ x^2 + 1 / x^2

⇒ ( 3 + √2 )^2 + [ ( 3 - √2 ) / 7 ]^2

⇒ [ 49( 3 + √2 )^2 + ( 3 - √2 )^2 ] / 49

⇒ [ 49( 9 + 2 + 6√2 ) + ( 9 + 2 - 6√2 ) ] / 49

⇒ [ 49( 9 ) + 9 + 49( 2 ) + 2 + 49 ( 6√2 ) - 6√2 ] / 49

⇒ [ 9( 49 + 1 ) + 2( 49 + 1 ) + 6√2( 49 - 1 ) ] / 49

⇒ [ 9( 50 ) + 2( 50 ) + 6√2( 48 ) ] / 49

⇒ [ 450 + 100 + 288√2 ] / 49

⇒ ( 550 + 288√2 ) / 49

Answer 2

Step-by-step explanation:

$\bold{Given - } \\ x = 3 + \sqrt{2} \\ \\ \bold{To \: find - } \\ {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ Now, \\ According \: to \: the \: question \\ \\ x = 3 + \sqrt{2} \\ And \\ \frac{1}{x} = \frac{1}{3 + \sqrt{2} } \\ \\ \bold{Rationalising \: the \: denominator} \\ \\ \frac{1}{(3 + \sqrt{2} )} \times \frac{(3 - \sqrt{2} )}{(3 - \sqrt{2}) } \\ = \frac{3 - \sqrt{2} }{9 - 2} \\ = \frac{3 - \sqrt{2} }{7} \\ = \fbox\bold{\frac{1}{x} = \frac{3 - \sqrt{2} }{7} } \\ \\ {(3 + \sqrt{2}) }^{2} + (\frac{3 - \sqrt{2} }{7} ) {}^{2} \\ = 11 + 6 \sqrt{2} \: + \frac{11 - 6 \sqrt{2} }{49} \\ = \frac{49(11 + 6 \sqrt{2}) + (11 - 6 \sqrt{2} ) }{49} \\ = \frac{539 + 294 \sqrt{2} + 11 - 6 \sqrt{2} }{49} \\ =\frac{550 + 288 \sqrt{2} }{49} \\ \\ Hence, \\ The \: value \: of \: \bold{ {x}^{2} + \frac{1}{ {x}^{2} } } \: is \: \bold{ \frac{550 + 288 \sqrt{2} }{49} }$

$\bold{Formula \: Used - } \\ \\ {(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} \\ {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} \\ (a + b)(a - b) = {a}^{2} - {b}^{2}$