Question

if x=2+√3 find x²+1/x²​
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Answer 1

Answer:

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Step-by-step explanation:

14

Answer 2

Given :

$x = 2 + \sqrt{3}$

To find :

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

Formula used :

${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

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

Solution :

$\implies \: x = 2 + \sqrt{3}$

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

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

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

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

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

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

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

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

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

we know :-

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

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

$\implies16 - 2= {x}^{2} + \dfrac{1}{{x}^{2}}$

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

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

Answer : 14

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$\large\bf\blue{Additional-Information}$

$\implies{(a+b)^2 = a^2 + b^2 + 2ab}$

$\implies{(a-b)^2 = a^2 + b^2 - 2ab}$

$\implies{(a+b)^3 = a^3 + b^3 + 3ab(a + b)}$

$\implies{(a-b)^3 = a^3 - b^3 - 3ab(a-b)}$

$\implies{(a^3+b^3)= (a+b)(a^2 - ab + b^2)}$

$\implies{(a^3-b^3)= (a-b)(a^2 + ab + b^2)}$

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