Verified answer

GIVEN:

[tex]x + \frac{1}{x} = 9 \\ [/tex]

TO FIND:

[tex]x - \frac{1}{x} \\ [/tex]

SOLUTION:

[tex]x + \frac{1}{x} = 9

squaring \: both \: sides \: we \: get, \\ ⟹ {(x + \frac{1}{x})}^{2} = {9}^{2} \\ ⟹ {x}^{2} + {( \frac{1}{x} )}^{2} + (2 \times x \times \frac{1}{x} ) = 81 \\ ⟹ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 81 \\ ⟹ {x}^{2} + \frac{1}{ {x}^{2} } = 81 - 2 \\ ⟹{x}^{2} + \frac{1}{ {x}^{2} } = 79(i)[/tex]

we have the formula,

[( a- b )² = a²+b² - 2ab]

[tex]( {x - \frac{1}{x}) }^{2} = {x}^{2} + ({\frac{1}{x} )}^{2} - 2 \times x \times \frac{1}{x} \\ [/tex]

using equation (i) putting value ,we get:

[tex]( {x - \frac{1}{x}) }^{2} = 79 - 2 \\ (as \: x \: and \frac{1}{x} cancelled \: out) \\ ( {x - \frac{1}{x}) }^{2} = 77 \\ ( x - \frac{1}{x}) = \sqrt{77} [/tex]

HENCE,(x-1/x) =±√77.