Question

$if x = 2 + \sqrt{3} find \: the \: value \: of \: x ^{2} + \frac{1}{x ^{2} }$
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Answer 1

$\pink\bigstar$ Answer:

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

$\red\bigstar$ Given:

• x = 2 + √3

$\green\bigstar$To find:

• The value of x² + 1/x²

$\blue\bigstar$ Solution:

x = 2 + √3

So,

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

Now, by rationalising the denominator, we get,

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

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

{ By using (a + b)(a - b) = a² - b² }

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

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

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

$\therefore$ 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 )

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

( √3 and -√3 get cancelled)

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

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

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

$\pink\bigstar$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)