Question

If
$x = \sqrt{6 + \sqrt{3} }$
, then find x²+1/x²-2​

Answer 1

Answer:

$\huge{\underline{\mathtt{\red{❥A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}$Answer:

{x}^{2} + \dfrac{1}{ {x}^{2} } = \boxed{14}x

2

+

x

2

1

=

14

\red\bigstar★ Given:

x = 2 + √3

\green\bigstar★ To find:

The value of x² + 1/x²

\blue\bigstar★ Solution:

x = 2 + √3

So,

\frac{1}{x} = \frac{1}{2 + \sqrt{3} }

x

1

=

2+

3

1

Now, by rationalising the denominator, we get,

\frac{1}{x} = \frac{2 - \sqrt{3} }{(2 + \sqrt{3} )(2 - \sqrt{3}) }

x

1

=

(2+

3

)(2−

3

)

2−

3

\implies \frac{1}{x} = \frac{2 - \sqrt{3} }{ {(2)}^{2} - { (\sqrt{3} }^{2} )}⟹

x1 = (2) 2 −( 32 )2− 3

{ By using (a + b)(a - b) = a² - b² }

\implies \: \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3}⟹

x1 = 4−32− 3

\implies \: \frac{1}{x} = \frac{2 - \sqrt{3} }{1}⟹

x1 = 12− 3

\implies \: \frac{1}{x} = 2 - \sqrt{3}⟹

x1 =2− 3

\therefore∴ 1/x = 2 - √3

Now, x² + 1/x² = ( x + 1/x )² - 2

So, x² + 1/x² = ( 2 + √3 + 2 - √3 )² - 2

( By substituting the values of x and 1/x )

\implies \: {x}^{2} + \frac{1}{ {x}^{2} } = {(4)}^{2} - 2⟹x

2 + x 21 =(4) 2 −2

( √3 and -√3 get cancelled)

\implies \: {x}^{2} + \frac{1}{ {x}^{2} } = 16 - 2⟹x

2 + x 21 =16−2

\implies \: {x}^{2} + \frac{1}{ {x}^{2} } = 14⟹x

2 + x 21 =14

\therefore \: \boxed{{x}^{2} + \frac{1}{ {x}^{2} } = 14}∴

x 2 + x 21 =14

★ Concepts used in the answer :

a² - b² = ( a + b )( a - b )

Substituting the values.

Rationalising the denominator.

\red\bigstar★ Some identities:

(a + b)(a - b) = a² - b²

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

(a + b)³ = a³ + b³ + 3ab² + 3a²b

a³ + b³ = ( a + b )(a² - ab + b²)

a³ - b³ = (a - b) ( a² + b² + ab)