Question

please help me out of this question​

Answer 1

Solution:-

Given

$\sf \to \: x = 2 - \sqrt{3}$

To find the value

$\sf \to \: {x}^{2} + \dfrac{1}{ {x}^{2} } \: \: and \: \: x {}^{2} - \dfrac{1}{ {x}^{2} }$

Now we can write

$\sf \to \: \dfrac{1}{x} = \dfrac{1}{2 - \sqrt{3} } \times \dfrac{2 + \sqrt{3} }{2 + \sqrt{3} } = 2 + \sqrt{3}$

Now we have to Find

$\sf \to \: \bigg({x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times x \times \dfrac{1}{x} \bigg) - 2 \times x \times \dfrac{1}{x}$

$\sf \to \: \bigg(x + \dfrac{1}{x} \bigg) ^{2} - 2$

$\sf \to \: (2 - \sqrt{3} + 2 + \sqrt{3} ) {}^{2} - 2$

$\sf \to \: (4) {}^{2} - 2$

$\sf \to \: 16 - 2 = 14$

Now we have to find the value of

$\sf \to \: {x}^{2} - \dfrac{1}{ {x}^{2} } = \bigg(x - \dfrac{1}{x} \bigg) \bigg(x + \dfrac{1}{x} \bigg)$

Put the value

$\sf \to \:( 2 - \sqrt{3} - 2 - \sqrt{3} )(2 - \sqrt{3} + 2 + \sqrt{3} )$

$\sf \to \: ( - 2 \sqrt{3} )(4)$

$\sf \to \: - 8 \sqrt{3}$

Answer :-

$\sf \to \: {x}^{2} + \dfrac{1}{ {x}^{2} } \: \: = 14 \: \: and \: \: x {}^{2} - \dfrac{1}{ {x}^{2} } = - 8 \sqrt{3}$

Option;- C is correct