Question

Hii


if x = 2 + √3


find x² + 1/x² and x³ + 1/x³


plzz, it is branliest question...

Answer 1

Hi there !

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Given :

x = 2 + √3

To Find :

(i) x² + 1 / x²

(ii) x³ + 1 / x³

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x = 2 + √3

1 / x = 1 / 2 + √3

1 / x = 1 / 2 + √3 × 2 - √3 / 2 - √3

1 / x = 2 - √3 / (2)² - (√3)²

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  

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Question I :

x + 1/x = 4

On squaring both sides ..

( x + 1/x )² = (4)²

x² + 1 / x² + 2 = 16

x² + 1 / x² = 16 - 2

x² + 1 / x² = 14  [Ans]

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Question II :

x + 1 / x = 4

On cubing both sides ..

( x + 1/x )³ = (4)³

x³ + 1/x³ + 3 ( x + 1/x) = 64

x³ + 1/x³ + 3 (4) = 64

x³ + 1/x³ + 12  = 64

x³ + 1/x³ = 64 - 12

x³ + 1/x³ = 52  [Ans]

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Thanks for the question !

☺️☺️

Answer 2

Here is your solution

$\bold{\purple{\boxed{\boxed{\huge\underline{Given:-}}}}}$

x = 2 + √3

$\huge\boxed{\red{\bold{Now}}}$

$\frac{1}{x} = \frac{1}{2 + \sqrt{3} } \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{2 {}^{2} - ( \sqrt{3}) {}^{2} } \\ \\ \frac{1}{x} = 2 - \sqrt{3} \\ \\ now \\ x + \frac{1}{x} = 2 + \sqrt{3} + 2 - \sqrt{3} \\ \\ x + \frac{1}{x} = 4 \\\\ now \: Q.no \: (1) \\ \\ x + \frac{1}{x} = 4 \\ both \: side \: squaring \\ (x + \frac{1}{x} ) {}^{2} = 4 {}^{2} \\ \\ x {}^{2} + \frac{1}{x {}^{2} } + 2 = 16 \\ x {}^{2} + \frac{1}{x {}^{2} } = 16 - 2 \\ x {}^{2} + \frac{1}{x {}^{2} } = 14 \\ \\ Q.no \: (2) \\ \\ x + \frac{1}{x} = 4 \\ \\ \: on \: cubic \: both \: side. \\( x + \frac{1}{x} ) {}^{3} = 4 {}^{3} \\ \\ x {}^{3} + \frac{1}{ x{}^{3} } + 3(x + \frac{1}{x} ) = 64 \\ x {}^{3} + \frac{1}{x {}^{3} } + 3(4) = 64 \\ x {}^{3} + \frac{1}{x {}^{3} } = 64 - 12 \\ \\ x{}^{3} + \frac{1}{x {}^{3} } = 64 - 12 \\ x {}^{3} + \frac{1}{x {}^{3} } = 52$

$\boxed{\red{\bold{Hope\: it\: helps\: you }}}$