Question

please solve this question ❔❔​

Answer 1

Given:

$x$ = $2$ + √$3$

To find the value of:

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

$\frac{1}{x}$ = $\frac{1}{2}$ + √$3$ × $2$ - √$3$/ $2$ - √$3$

$\frac{1}{x}$ = $2$ - √$3$/ $2^2$- √$3$

$\frac{1}{x}$ = $2$ - √$3$ / $4$ - $3$

$\frac{1}{x}$ = $2$ - √$3$

Now,

$\frac{x + 1}{x}$ = $2$ + √$3$+ $2$ - √$3$

$\frac{x + 1}{x}$ = $4$

By squaring both sides,

( $\frac{x + 1}{x}$ ) ² = ( $4$)²

$\frac{x^2 + 1}{x^2}$ + $2$ = $16$

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

$\frac{x^2 + 1}{x^2}$ = $14$

Hence,

$\frac{x^2 + 1}{x^2}$ = $14$

Answer 2

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