Question

if X = 2-√3
find
${x}^{2} - \frac{1}{ {x}^{2} }$

Answer 1

heya..... here's Your answer.....
i) x²
(2-√3)²
(2)²-(√3)²
4-3
=1
ii) 1/x²
1/1
=0
hope you have got needed answer..... mark me as brainliest plz plz plz.....

Answer 2

Hi...☺

Here is your answer...✌
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$x = 2 - \sqrt{3} \\ \\ {x}^{2} = (2 - \sqrt{3} ) {}^{2} \\ \\ {x}^{2} = {(2)}^{2} + ( { \sqrt{3} )}^{2} - 2 \times 2 \times \sqrt{3} \\ \\ {x}^{2} = 4 + 3 - 4 \sqrt{3 } \\ \\ {x}^{2} = 7 - 4 \sqrt{3} \\ \\ and \: \\ \\ \frac{1}{ {x}^{2} } = \frac{1}{7 - 4 \sqrt{3} } \\ \\ On \: rationalising \: the \: denominator \\ We \: get, \\ \\ \frac{1}{ {x}^{2} } = \frac{1}{7 - 4 \sqrt{3} } \times \frac{7 +4 \sqrt{3} }{7 + 4 \sqrt{ 3} } \\ \\ \frac{1}{ {x}^{2} } = \frac{7 + 4 \sqrt{3} }{ {(7)}^{2} - {( 4\sqrt{3})}^{2} } \\ \\ \frac{1}{ {x}^{2} } = \frac{7 + 4 \sqrt{3} }{49 - 48} \\ \\ \frac{1}{ {x}^{2} } = 7 + 4 \sqrt{3} \\ \\ Now, \\ \\ {x}^{2} - \frac{1}{ {x}^{2} } = 7 - 4 \sqrt{3} - (7 + 4 \sqrt{3} ) \\ \\ = 7 - 4 \sqrt{3} - 7 - 4 \sqrt{3} \\ \\ = - 8 \sqrt{3}$