Question

if x = √5+2 find the value of x2 + 1/x2

Answer 1

Hey mate!

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Given :

x = √5 + 2

To find :

x² + 1 / x²

Solution :

x = √5 + 2

⇒ 1 / x = 1 / √5 + 2

⇒ 1 / x = 1 / √5 + 2 ×  √5 - 2 /  √5 - 2

⇒ 1 / x =  √5 - 2 / ( √5 )² - ( 2 ) ²

⇒ 1 / x =  √5 - 2 / 5 - 4

⇒ 1 / x =  √5 - 2

Now,

x + 1 / x =  √5 + 2 +  √5 - 2

x + 1 / x = 2 √5

Again,

On squaring both sides, we have ;

( x + 1 / x )² = ( 2 √5 ) ²

⇒ x² + 1 / x² + 2 = 20

⇒ x² + 1 / x² = 20 - 2

⇒ x² + 1 / x² = 18

Hence,

The value of x² + 1 / x² = 18.

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Thanks for the question!

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Answer 2

$\underline{\bold{Given:-}}$

$x = \sqrt{5} + 2$

$\underline{\bold{To\:find:-}}$

${x}^{2} + \frac{1}{ {x}^{2} }$

$\underline{\bold{Solution:-}}$

$x = \sqrt{5} + 2 \\ \\ \frac{1}{x} = \frac{1}{ \sqrt{5} + 2 } \\ \\ \bold{On \: rationalising \: them} \\ \\ \frac{1}{x} = \frac{1}{ \sqrt{5} + 2 } \times \frac{ \sqrt{5} - 2}{ \sqrt{5} - 2}$

$\bold{Using \: identity }$

$\bold {{a}^{2} - {b}^{2} = (a + b)(a - b)}\\ \\ \frac{1}{x} = \frac{ \sqrt{5} - 2 }{ { (\sqrt{5}) }^{2} - {2}^{2} } \\ \\ \frac{1}{x} = \frac{ \sqrt{5} - 2 }{5 - 4} \\ \\ \frac{1}{x} = \sqrt{5} - 2$

Now,

$x + \frac{1}{x} = \sqrt{5} + 2 + \sqrt{5} - 2 \\ \\ x + \frac{1}{x} = 2 \sqrt{5} \\ \\ \bold{On \: squaring \: both \: sides} \\ \\ {(x + \frac{1}{x}) }^{2} = {(2 \sqrt{5} )}^{2} \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times x \times \frac{1}{x} = 4 \times 5 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 20 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 20 - 2 \\ \\\boxed{\bold{ {x}^{2} + \frac{1}{ {x}^{2} } = 18}}$

⭐Hope it may help you⭐