Question

if K is equal to 2 +
$\sqrt{3}$
then find the value of (i) X+ 1 upon X (ii) X square + 1 upon X square

Answer 1

hi!!here is your answer.......______

Answer 2

Solution:-

given by :-

$x = 2 + \sqrt{3} \\ \\ 1. \\ x + \frac{1}{x} \\ 2 + \sqrt{3} + \frac{1}{2 + \sqrt{3} } \\ \frac{4 + 2 \sqrt{3} + 2 \sqrt{3} + 3 + 1}{2 + \sqrt{3} } \\ \frac{7 + 4 \sqrt{3} + 1 }{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \frac{14 + 8 \sqrt{3} - 7 \sqrt{3} - 12 + 2 - \sqrt{3} }{2 - 3} \\ 4 ans \\ 2. \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } \\ ({2 + \sqrt{3}) }^{2} + \frac{1}{ ({2 + \sqrt{3}) }^{2} } \\ \frac{(4 + 3 + 4 \sqrt{3} )(4 + 3 + 4 \sqrt{3}) + 1 }{4 + 3 + 4 \sqrt{3} } \\ \frac{16 + 12 + 16 \sqrt{3} + 12 + 9 + 12 \sqrt{3} + 16 \sqrt{3} + 12 \sqrt{3} + 48 + 1 }{7 + 4 \sqrt{3} } \\ \frac{98 + 32 \sqrt{3} + 24 \sqrt{3} }{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4\sqrt{3} } \\ \frac{686 + 224 \sqrt{3} + 168 \sqrt{3} - 392 \sqrt{3 } - 128 \sqrt{3} - 96 \sqrt{3} }{49 - 48} \\ 686 + 224 \sqrt{3}$