Question

$if \: x = 2 + \sqrt{3} \: then \: find \: the \: value \: of \: \sqrt{ {x} } + \frac{1}{ \sqrt{x} } . \\ answer \: is \: \sqrt{6} . \: please \: solve \: it.$

Answer 1

$\underline{\underline{\large{\mathfrak{Solution : }}}}$



$\sf Let \: \sqrt{x} \: + \: \dfrac{1}{\sqrt{x}} \: = \: y$



$\textsf{Squaring both sides : }$



$\sf \implies {\left(\sqrt{x} \: + \: \dfrac{1}{\sqrt{x}} \right)}^2 \: = \: y^2$



$\sf \implies {(\sqrt{x})}^2 \: + \: {\left(\dfrac{1}{\sqrt{x}}\right)}^2 \: + \: 2 \: \times \: \sqrt{x} \: \times \: \dfrac{1}{\sqrt{x}} = \: y^2$



$\sf \implies x \: + \: \dfrac{1}{x} \: + \: 2 \: = \: y^2$



$\underline{\textsf{Put the value of x }}$



$\sf \implies ( 2 \: + \: \sqrt{3} )\: + \: \dfrac{1}{2 \: + \: \sqrt{3}} \: + \: 2 \: = \: y^2$



$\sf \implies \dfrac{(2 \: + \: \sqrt{3})^2 \: + \: 1}{2 \: + \: \sqrt{3}} \: + \: 2 \: = \: y^2$



$\sf \implies \dfrac{ 2^2 \: + \: (\sqrt{3})^2 \: + \: 2 \: \times \: 2 \: \times \: \sqrt{3} \: + \: 1 }{2 \: + \: \sqrt{3}} \: + \: 2 \: = \: y^2$



$\sf \implies \dfrac{4 \: + \: 3 \: + \: 4\sqrt{3} \: + \: 1}{2 \: + \: \sqrt{3}} \: + \: 2 \: = \: y^2$



$\sf \implies \dfrac{8 \: + \: + \: 4\sqrt{3}}{2 \: + \: \sqrt{3}} \: + \: 2 \: = \: y^2$



$\sf \implies \dfrac{4( 2 \: + \: \sqrt{3})}{2 \: + \: \sqrt{3}} \: + \: 2 \: = \: y^2$



$\sf \implies 4 \: + \: 2 \: = \: y^2$



$\sf \implies 6 \: = \: y^2$



$\sf \quad \therefore y \: = \: \pm \sqrt{6}$

Answer 2

Step-by-step explanation:

Hey mate ^_^

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Given :

x = 2 + √3

To find :

x² + 1 / x²

Solution ;

x = 2 + √3

⇒ 1 / x = 1 / 2 + √3 × 2 - √3 / 2 - √3

⇒ 1 / x = 2 - √3 / 2² - √3²

⇒ 1 / x = 2 - √3 / 4 - 3

⇒ 1 / x = 2 - √3

Now,

x + 1 / x = 2 + √3 + 2 - √3

⇒ x + 1 / x = 2 + 2

⇒ x + 1 / x = 4

And, on squaring both sides.

( x + 1 / x ) ² = (4)²

⇒ x² + 1 / x² + 2 = 16

⇒ x² + 1 / x² = 16 - 2

⇒ x² + 1 / x² = 14.

Hence,

x² + 1 / x² = 14.

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Thanks for the question!

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