Question

If x=3-√8, then (x-1/x)^2​

Answer 1

Given :

$\rm x = 3 - \sqrt{8}$

To find :

$\rm { \bigg(x + \dfrac{1}{x} \bigg) }^{2}$

Solution :

$\rm x = 3 - \sqrt{8} \\ \\ \rm \rightarrow \frac{1}{x} = \frac{1}{3 - \sqrt{8}} \\ \\ \rm \rightarrow \frac{1}{x} = \frac{1}{3 - \sqrt{8}} \times \frac{3 + \sqrt{8}}{3 + \sqrt{8}} \\ \\ \rm \rightarrow \frac{1}{x} = \dfrac{3 + \sqrt{8}}{9 - 8} \\ \\ \rm \rightarrow \frac{1}{x} = \dfrac{3 + \sqrt{8}}{1} \\ \\ \rm \rightarrow \frac{1}{x} =3 + \sqrt{8}$

$\rm \rightarrow{ \bigg(x + \dfrac{1}{x} \bigg) }^{2} = {x}^{2} + \dfrac{1}{{x}^{2}} +( 2 \times x \times \dfrac{1}{x}) \\ \\ \rm \rightarrow{ \bigg(x + \dfrac{1}{x} \bigg) }^{2} = {x}^{2} + \dfrac{1}{{x}^{2}} +2 \\ \\ \\ \\ \rm \rightarrow{ \bigg(x + \dfrac{1}{x} \bigg) }^{2} = {(3 - \sqrt{8} )}^{2} + {(3 + \sqrt{8} )}^{2} +2 \\ \\ \rm \rightarrow{ \bigg(x + \dfrac{1}{x} \bigg) }^{2} = 9 + 8 - 6 \sqrt{8} + 9 + 8 + 6 \sqrt{8} +2 \\ \\ \rm \rightarrow{ \bigg(x + \dfrac{1}{x} \bigg) }^{2} = 9 + 8 + 9 + 8 +2 \\ \\ \rm \rightarrow{ \bigg(x + \dfrac{1}{x} \bigg) }^{2} = 36$

Answer :

36