Question

If x=2+/3 find value of x²+1/x²

Answer 1

Answer:

$given \\ \\ x = 2 + \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \\ \\ = \frac{2 - \sqrt{3} }{ {2}^{2} - { (\sqrt{3}) }^{2} } \\ \\ = 2 - \sqrt{3} \\ \\ then \: x + \frac{1}{x} = 4 \\ \\ now \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ = {(x + \frac{1}{x}) }^{2} - 2x( \frac{1}{x} ) \\ \\ = {4}^{2} - 2 \\ \\ = 16 - 2 \\ \\ = 14$

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Answer 2

$\huge\mathfrak\blue{Answer:}$

Given:

• We have been given that
• $x = 2 + \sqrt{3}$

To Find:

• Find the value of ( x² + 1/x² )

Solution:

We have given that

$x = 2 + \sqrt{3}$

Now we will find value of 1/x

$= > \dfrac{1}{x} = \dfrac{1}{2 + \sqrt{3} }$

On Rationalizing the denominator

$= > \dfrac{1}{x} = \dfrac{1}{2 + \sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} }$

$= > \dfrac{1}{x} = \dfrac{2 - \sqrt{3} }{(2 + \sqrt{3} )(2 - \sqrt{3} )}$

Using [ ( a + b )( a - b ) = a² - b² ]

$= > \dfrac{1}{x} = \dfrac{2 - \sqrt{3} }{ {(2)}^{2} - {( \sqrt{3}) }^{2} }$

$= > \dfrac{1}{x} = \dfrac{2 - \sqrt{3} }{4 - 3}$

$= > \dfrac{1}{x} = 2 - \sqrt{3}$

Now finding the value of ( x + 1/x ) :

$= > x + \dfrac{1}{x} = (2 + \sqrt{3} ) + (2 - \sqrt{3} )$

$= > x + \dfrac{1}{x} = 4$

We get the value of ( x + 1/x ) = 4

Squaring both sides we get

$= >{( x + \dfrac{1}{x} )}^{2} = {(4)}^{2}$

Using [ ( a + b ) ² = a² + b² + 2ab ]

$= > {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times x \times \dfrac{1}{x} = 16$

$= > {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 16$

$= > {x}^{2} + \dfrac{1}{ {x}^{2} } = 16 - 2$

$= > {x}^{2} + \dfrac{1}{ {x}^{2} } = 14$

The value of ( x + 1/x² ) is 14