Question

It's x = 3-root13/2.. Please solve this neatly

Answer 1

It is given that the value of x is $\dfrac{3-\sqrt{13}}{2}$



$\implies x = \dfrac{3 - \sqrt{13} }{2} \\ \\ \\ \implies \dfrac{2}{3 - \sqrt{13} } = \dfrac{1}{x}$


By Rationalizing { divide and multiply by 3 + √13 on left hand side }

$\implies \dfrac{2}{3 - \sqrt{13} } \times \dfrac{3 + \sqrt{13} }{3 + \sqrt{13} } = \dfrac{1}{x} \\ \\ \\ \implies \dfrac{1}{x} = \dfrac{2(3 + \sqrt{13} ) }{(3 - \sqrt{13} )(3 + \sqrt{13} )}$



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From the properties of expansion, we know :
( a - b )( a + b ) = a² - b²
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$\implies \dfrac{1}{x} = \dfrac{2(3 + \sqrt{13} )}{3 {}^{2} - ( \sqrt{13} ) {}^{2} } \\ \\ \\ \implies \dfrac{1}{x} = \frac{2(3 + \sqrt{13}) }{9 - 13} \\ \\ \\ \implies \dfrac{1}{x} = \dfrac{ 2(3 + \sqrt{13} )}{ - 4} \\ \\ \\ \implies \dfrac{1}{x} = \dfrac{ - ( 3 + \sqrt{13}) }{2}$



Now,
• x = ( 3 - √13 ) / 2

= > x² = { ( 3 - √13 ) / 2 }²

= > x² = ( 9 + 13 - 6√13 ) / 4

= > x² = ( 22 - 6√13 ) / 4 ----: ( 1 )


• 1 / x = - ( 3 + √13 ) / 2

= > ( 1 / x )² = { - ( 3 + √13 ) / 2 }²

= > 1 / x² = ( 9 + 13 + 6√13 ) / 4

= > 1 / x² = ( 22 + 6√13 ) / 4 ----: ( 2 )




Then, Adding ( 1 ) and ( 2 ),

$\implies x {}^{2} + \dfrac{1}{x {}^{2} } = \dfrac{22 - 6 \sqrt{13} }{4} + \dfrac{22 + 6\sqrt{13} }{4} \\ \\ \\ \implies x {}^{2} + \dfrac{1}{x {}^{2} } = \frac{22 - \sqrt{13} + 22 + 6 \sqrt{13} }{4} \\ \\ \\ \implies x {}^{2} + \dfrac{1}{ {x}^{2} } = \dfrac{44}{4} \\ \\ \\ \implies {x}^{2} + \dfrac{1}{ {x}^{2} } = 11$



Hence, the numeric value of x² + 1 / x² is 11.