Question

$\mathfrak{Find \: the \: solution \: of}$
$\mathsf{ {(2 + \sqrt{3} )}^{ {x}^{2} - 2x + 1} + {(2 - \sqrt{3} )}^{ {x}^{2} - 2x - 1} = \dfrac{4}{2 - \sqrt{3} } }$
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Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

1 - √2 ,  1  , 1  + √2  are the solutions

Step-by-step explanation:

Let say x² - 2x  = n

(2 + √3)ⁿ⁺¹  + (2 - √3)ⁿ⁻¹  = 4/(2 - √3)

=> (2 + √3)ⁿ(2 + √3)  + (2 - √3)ⁿ/(2-  √3)  = 4/(2 - √3)

Multiplying both sides by (2 - √3)

=> (2 + √3)ⁿ(2 + √3)(2 - √3) +  (2 - √3)ⁿ = 4

=>  (2 + √3)ⁿ + (2 - √3)ⁿ = 4

2 + √3  = 1/(2 - √3)    or 2 - √3 = 1/(2+ √3)

=>   (2 + √3)ⁿ + 1/(2 + √3)ⁿ = 4

let say  (2 + √3)ⁿ = a

=> a + 1/a  = 4

=> a² + 1 = 4a

=> a² - 4a + 1 = 0

=> a = (4 ± √16 - 4)/2

=> a = 2 ± √3

a = 2 + √3

=>  (2 + √3)ⁿ  = (2 + √3)

=> n = 1

=>  x² - 2x  = 1

=> x² - 2x  - 1 = 0

=> x  = (2 ± √4 + 4)/2

=> x =  1 ± √2

a = 2 -√3

=>  (2 + √3)ⁿ  = (2 - √3)

=> (2 + √3)ⁿ  = 1/ (2 +√3)

=>  (2 + √3)ⁿ  = (2 +√3)⁻¹

=> n = - 1

=>  x² - 2x  = -1

=> x² - 2x  + 1 = 0

=> (x - 1)² = 0

=> x =  1

1 - √2 ,  1  , 1  + √2  are the solutions

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