Question

If x = 4-√15 then find
$\sf x + \dfrac{1}{x} =?$​

Answer 1

To find :

• Value of $\sf x + \dfrac{1}{x} =?$

Solution :-

Given ,

x = 4 - √15

$\therefore\sf \dfrac{1}{x} = \dfrac{1}{4 - \sqrt{15}}$

$\dag\;{\underline{\frak{Rationalising\;the\;denominator}}}$

$: \implies\sf \dfrac{1}{x} = \dfrac{1(4 + \sqrt{15})}{(4 - \sqrt{15})(4 + \sqrt{15})}$

$: \implies\sf \dfrac{1}{x} = \dfrac{4 + \sqrt{15}}{4^2 - (\sqrt{15})^2}$

$: \implies\sf \dfrac{1}{x} = \dfrac{4 + \sqrt{15}}{16 - 15}$

$: \implies\sf \dfrac{1}{x} = \dfrac{4 + \sqrt{15}}{1}$

$: \implies\underline{\boxed{\sf{\dfrac{1}{x} = 4 + \sqrt{15} }}}$

_____

Now,

$: \implies\sf x + \dfrac{1}{x} = 4 - \sqrt{15} + 4 + \sqrt{15}$

$: \implies\underline{\boxed{\bold{x + \dfrac{1}{x} = 8 }}}$

$\boxed {\sf {\purple {Required \ value \ is\ 8.}}}$

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