Question

i x-1/x=5, then find x+1/x

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Answer 1

$x + \frac{1}{x} = 5 \\ \\ \frac{1}{x} = 5 - x ...(1)\\ \\ now \: x - \frac{1}{x} \\ \\ x - 5 + x \: (from \: equation \: 1) \\ \\ = \: - 5$

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Answer 2

Given :

$\sf \: x - \dfrac{1}{ x} = 5$

To find :

$\sf \: x + \dfrac{1}{ x}$

Identity used :

• (a-b)² = a² + b² - 2ab
• (a+b)² = a² + b² + 2ab

Solution :

We are given that

$\sf \: x - \dfrac{1}{ x} = 5$

Now squaring both side

$\sf \implies \: \bigg(\: x - \dfrac{1}{ x} \bigg)^{2} = {5}^{2}$

$\sf \implies \: \bigg(\: {(x)}^{2} + \bigg(\dfrac{1}{ x} { \bigg)}^{2} \: - 2 \times (x) \times \bigg( \dfrac{1}{x} \bigg) \: \: \bigg) \: = \: {5}^{2}$

$\sf \implies \: \: {(x)}^{2} + \bigg(\dfrac{1}{ x} { \bigg)}^{2} \: - 2 = 25$

$\sf \implies \: \: {(x)}^{2} + \bigg(\dfrac{1}{ x} { \bigg)}^{2} \: = 25 \: + 2$

$\sf \implies \: \: {(x)}^{2} + \bigg(\dfrac{1}{ x} { \bigg)}^{2} \: = 27 \: \: \: \: \: equation \: 1$

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$\sf \: x + \dfrac{1}{ x}$

Squaring both side

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

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

$\sf \bullet \: put \: value \: of \: \bigg(\: {(x)}^{2} + \bigg(\dfrac{1}{ x} { \bigg)}^{2} \: \bigg) \: from \: equation \: 1 \\ \\ \sf \implies \: \bigg(\: x + \dfrac{1}{ x} \bigg)^{2} = 27 + 2$

$\sf \implies \: \bigg(\: x + \dfrac{1}{ x} \bigg)^{2} = 29$

$\sf \implies \: \bigg(\: x + \dfrac{1}{ x} \bigg) \: = \: \sqrt{29}$

$\sf \implies \: \bigg(\: x + \dfrac{1}{ x} \bigg) \: = \: \pm \: 29$

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ANSWER : ± 29