Question

ifx=
$\sqrt{7 + 4 \sqrt{3} } then x + 1 \div x$
​

Answer 1

Answer :-

$\implies\sf x = \sqrt{7 + 4\sqrt3}$

$\implies\sf x = \sqrt{4 + 3 + 2\times 2\times \sqrt3}$

$\implies\sf x = \sqrt{(2)^2 + (\sqrt3)^2 + 2\times 2\times \sqrt3}$

$\implies\sf x = \sqrt{(2 + \sqrt{3})^2}$

$\implies\sf x = 2 + \sqrt3$

Calculating value of 1 / x :-

$\implies\sf \dfrac{1}{x} = \dfrac{1}{2 + \sqrt3}$

$\implies\sf \dfrac{1}{x} = \dfrac{1}{2 + \sqrt3} \times \dfrac{2 - \sqrt3}{2 - \sqrt3}$

$\implies\sf \dfrac{1}{x} = \dfrac{2 -\sqrt3}{(2 +\sqrt3)(2-\sqrt3)}$

$\implies\sf \dfrac{1}{x} = \dfrac{2-\sqrt3}{4-3}$

$\implies\sf \dfrac{1}{x} = \dfrac{2-\sqrt3}{1}$

$\implies\sf \dfrac{1}{x} = 2 - \sqrt3$

$\implies\sf x + \dfrac{1}{x} = 2 + \sqrt3 + 2 - \sqrt3$

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

$\boxed{\bf Value\: of\: x + \dfrac{1}{x} = 4}$