Question

If x = (3-√13)/2, what is the value of x² + 1/x² ?​

Answer 1

Question:-

If $\sf x = \dfrac{3 - \sqrt{13}}{2}$ , what is the value of x² + 1/x² ?

Answer:-

Given:

$\sf \: x = \dfrac{3 - \sqrt{13} }{2}$

We have to find:

x² + 1/x²

We know that,

(a + b)² - 2ab = a² + b²

So,

$\implies \sf \bigg( x + \dfrac{1}{x} \bigg) ^{2} - 2 \times x \times \dfrac{1}{x} = {x}^{2} + \dfrac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \bigg( \frac{3 - \sqrt{13} }{2} + \frac{1}{ \frac{3 - \sqrt{13} }{2} } \bigg) ^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \bigg( \frac{3 - \sqrt{13} }{2} + \frac{2}{3 - \sqrt{13} } \bigg) ^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \bigg( \frac{(3 - \sqrt{13} ) ^{2} + 4}{2(3 - \sqrt{13} )} \bigg)^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} }$

(a - b)² = a² + b² - 2ab

$\implies \sf \: \bigg( \dfrac{ {3}^{2} + ( \sqrt{13}) ^{2} - 2(3)( \sqrt{13} ) + 4 }{2(3 - \sqrt{13} )} \bigg) ^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \: \bigg( \frac{9 + 13 - 6 \sqrt{13} + 4}{2(3 - \sqrt{13} )} \bigg) ^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \: \bigg( \frac{26 - 6 \sqrt{13} }{2(3 - \sqrt{13} )} \bigg) ^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \: \bigg( \frac{2(13 - 3 \sqrt{13} )}{2(3 - \sqrt{13}) } \bigg) ^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \: \bigg( \frac{ - \sqrt{13} (3 - \sqrt{13} )}{ 3 - \sqrt{13} } \bigg) ^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \sf \: {( - \sqrt{13} )}^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \sf \implies 13 - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ \\ \implies \boxed{ \sf 11 = {x}^{2} + \frac{1}{ {x}^{2} }}$