Question

${x + \frac{1}{x} }^{2} - \frac{3}{2} (1 - \frac{1}{x} ) - 4$
please solve it is from quadratic equation chapter​

Answer 1

To solve -

Solve the following quadratic equation for x -

$\sf{ \bigg{ x + \dfrac{ 1 }{ x } } ^ 2 - \dfrac{ 3 }{ 2 } \times \bigg { x - \dfrac{ 1 }{ x } } - 4 }$

Solution -

$\sf{ \bigg{ x + \dfrac{ 1 }{ x } } ^ 2 - \dfrac{ 3 }{ 2 } \times \bigg { x - \dfrac{ 1 }[ x } } - 4 } \\ \\ \sf{ \implies { x^2 + dfrac{ 1 }{ x^2 } - 2 - \dfrac{ 3 }{ 2 } \times \bigg { x - \dfrac{ 1 }{ x } } } }} \\ \\ \sf{ \implies { \bigg{ x - \dfrac{ 1 }{x } } ^ 2 - \dfrac{ 3 }{ 2 } \times \bigg { x - \dfrac{ 1 }{ x } } } } \\ \\ \sf{ \bold { Let \: \bigg{ x - \dfrac{ 1 }[ x } } be \: k }} \\ \\ \sf{ \implies { k^2 - \dfrac{ 3 }{ 2 } k = 0 }} \\ \\ \sf{ \implies { k^2 = \dfrac{3}{ 2 } k }} \\ \\ \sf{ \implies { k = 0 or \dfrac{3}{ 2 } }} \\ \\ \sf{ \bold { Case \: 1 - }} \\ \\ \sf{ \implies { x^2 = 1 } \\ \\ \sf{ \implies { x = \pm 1 }} \\ \\ \sf{ \bold { Case \: 2 }} \\ \\ \sf{ \implies { x - \dfrac{ 1 }[ x } = \dfrac{ 3 }{ 2 } }} \\ \\ \sf{ \implies { \dfrac{ x^2 - 1 }{ x } = \dfrac{ 3 }[ 2 } }} \\ \\ \sf{ \implies { 2 ( x^2 - 1 ) = 3x }} \\ \\ \sf{ \implies { 2x^2 - 3x - 2 = 0 }} \\ \\ \sf{ \bold { Here , \: one \: root \: is \: discarded \: . }} \\ \\ \sf{ \bold { Hence \: x \approx 3.3 \: and \pm 1 }}$