Question

How to solve questions of type :-
If x = 2+√3, find the value of x² + 1/x².
Please give step by step explanation. ​

Answer 1

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$\huge\sf\underline\red{answered\:by\:KIMU}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:x = 2 + \sqrt{3}$

Consider,

$\rm :\longmapsto\:\dfrac{1}{x}$

$\rm \: = \: \: \dfrac{1}{2 + \sqrt{3} }$

On rationalizing the denominator, we get

$\rm \: = \: \: \dfrac{1}{2 + \sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} }$

We know,

$\boxed{ \rm \: (x + y)(x - y) = {x}^{2} - {y}^{2}}$

using this identity, we get

$\rm \: = \: \: \dfrac{2 - \sqrt{3} }{ {(2)}^{2} - {( \sqrt{3} )}^{2} }$

$\rm \: = \: \: \dfrac{2 - \sqrt{3} }{4 - 3}$

$\rm \: = \: \: \dfrac{2 - \sqrt{3} }{1}$

$\rm \: = \: \: 2 - \sqrt{3}$

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

Hence,

$\rm :\longmapsto\:x + \dfrac{1}{x}$

$\rm \: = \: \: 2 + \sqrt{3} + 2 - \sqrt{3}$

$\rm \: = \: \: 4$

So,

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

On squaring both sides, we get

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times x \times \dfrac{1}{x} = 16$

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 16$

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } = 16 - 2$

$\bf :\longmapsto\: {x}^{2} + \dfrac{1}{ {x}^{2} } = 14$

Additional Information :-

More Identities to know:

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)