Question

if x-1/x=1/54 the find the value of x+1/x

Answer 1

Given that,

$x - \dfrac{1}{x} = \dfrac{15}{4}$

Squaring both sides,

$\longrightarrow { \bigg(x - \dfrac{1}{x} \bigg)}^{2} = { \bigg( \dfrac{15}{4} \bigg) }^{2}$

Expanding L.H.S. using (a - b)² = a² + b² - 2ab with a = x and b = 1/x,

$\longrightarrow {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 \bigg(x \times \dfrac{1}{x} \bigg) = \dfrac{225}{16}$

$\longrightarrow {x}^{2} + \dfrac{1}{ {x}^{2} } - 2 = \dfrac{225}{16}$

$\longrightarrow {x}^{2} + \dfrac{1}{ {x}^{2} } = \dfrac{225}{16} + 2$

$\longrightarrow {x}^{2} + \dfrac{1}{ {x}^{2} } = \dfrac{225 + 32}{16}$

$\longrightarrow {x}^{2} + \dfrac{1}{ {x}^{2} } = \dfrac{257}{16} \quad \quad \dots(1)$

Now, we know that,

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

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

From (1),

$\longrightarrow { \bigg(x + \dfrac{1}{x} \bigg)}^{2} = \dfrac{257}{16} + 2$

$\longrightarrow { \bigg(x + \dfrac{1}{x} \bigg)}^{2} = \dfrac{257 + 32}{16}$

$\longrightarrow { \bigg(x + \dfrac{1}{x} \bigg)}^{2} = \dfrac{289}{16}$

Taking square root on both sides,

$\longrightarrow \sqrt{{ \bigg(x + \dfrac{1}{x} \bigg)}^{2}} = \sqrt{ \dfrac{289}{16} }$

$\longrightarrow \sqrt{{ \bigg(x + \dfrac{1}{x} \bigg)}^{2}} = \sqrt{ { \bigg(\pm \dfrac{17}{4} \bigg) }^{2} }$

$\longrightarrow x + \dfrac{1}{x} = \pm \dfrac{17}{4}$

Hence,

$\longrightarrow \underline{ \underline{ x + \dfrac{1}{x} = \pm \dfrac{17}{4} }}$

Answer 2

Step-by-step explanation:

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