Question

If (X+1/X)=√7 then find X²+1/X²and X⁴+1/X⁴​

Answer 1

Given,

$: \implies x + \frac{1}{x} = \sqrt{7}$

To Find,

$: \implies {x}^{2} + \frac{1}{ {x}^{2} } = \: ?$

$: \implies {x}^{4} + \frac{1}{ {x}^{4} } = \: ?$

Solution,

$: \implies x + \frac{1}{x} = \sqrt{7} \: \: \: ...(Given) \\ \\ : \implies x + \frac{1}{x} = \sqrt{7} \: \: \: ...(squaring \: \: both \: \: side)\\ \\ : \implies {(x + \frac{1}{x} )}^{2} = {( \sqrt{7}) }^{2} \\ \\ ( {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy) \\ \\ : \implies {(x)}^{2} + {( \frac{1}{x}) }^{2} + 2(x)( \frac{1}{x} ) = 7 \\ \\ : \implies {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times x \times \frac{1}{x} = 7 \\ \\ : \implies {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 7 \\ \\ : \implies {x}^{2} + \frac{1}{ {x}^{2} } = 7 - 2 \\ \\ : \implies {x}^{2} + \frac{1}{ {x}^{2} } = 5$

$: \implies {x}^{2} + \frac{1}{ {x}^{2} } = 5 \\ \\ : \implies {x}^{2} + \frac{1}{ {x}^{2} } = 5 \: \: \: ...(squaring \: \: both \: \: sides) \\ \\ : \implies{ ( {x}^{2} + \frac{1}{ {x}^{2} } )}^{2} = {(5)}^{2} \\ \\( {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy) \\ \\ : \implies {( {x}^{2}) }^{2} + {( \frac{1}{ {x}^{2} }) }^{2} + 2( {x}^{2} )( \frac{1}{ {x}^{2} } ) = 25 \\ \\ : \implies {x}^{4} + \frac{1}{ {x}^{4} } + 2 \times {x}^{2} \times \frac{1}{ {x}^{2} } = 25 \\ \\ : \implies {x}^{4} + \frac{1}{ {x}^{4} } + 2 = 25 \\ \\ : \implies {x}^{4} + \frac{1}{ {x}^{4} } = 25 - 2 \\ \\ : \implies {x}^{4} + \frac{1}{ {x}^{4} } = 23$

Required Answer,

x² + 1/x² = 5

x² + 1/x² = 5x⁴ + 1/x⁴ = 23