Question

solve this question from cha 1 , 9th class​

Answer 1

$\mathrm{Given : x = \dfrac{3 + \sqrt{5}}{2}}$

$\mathrm{\implies \dfrac{1}{x} = \dfrac{1}{\dfrac{3 + \sqrt{5}}{2}}}$

$\mathrm{\implies \dfrac{1}{x} = \dfrac{2}{3 + \sqrt{5}}}$

$\mathrm{Multiply\;and\;Divide\;with\;3 - \sqrt{5}}$

$\mathrm{\implies \dfrac{1}{x} = \dfrac{2(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5)}}}$

$\mathrm{\implies \dfrac{1}{x} = \dfrac{2(3 - \sqrt{5})}{(3)^2 - (\sqrt{5})^2}}}$

$\mathrm{\implies \dfrac{1}{x} = \dfrac{2(3 - \sqrt{5})}{9 - 5}}}$

$\mathrm{\implies \dfrac{1}{x} = \dfrac{2(3 - \sqrt{5})}{4}}}$

$\mathrm{\implies \dfrac{1}{x} = \dfrac{3 - \sqrt{5}}{2}}}$

$\mathrm{Consider : \left(x + \dfrac{1}{x}\right)^2}$

$\mathrm{\implies \left(x + \dfrac{1}{x}\right)^2 = x^2 + \dfrac{1}{x^2} + 2(x)\left(\dfrac{1}{x}\right)}$

$\mathrm{\implies \left(x + \dfrac{1}{x}\right)^2 = x^2 + \dfrac{1}{x^2} + 2}$

$\mathrm{\implies x^2 + \dfrac{1}{x^2} = \left(x + \dfrac{1}{x}\right)^2 - 2}$

$\mathrm{\implies x^2 + \dfrac{1}{x^2} = \left(\dfrac{3 + \sqrt{5}}{2} + \dfrac{(3 - \sqrt{5})}{2}}\right)^2 - 2}$

$\mathrm{\implies x^2 + \dfrac{1}{x^2} = \left(\dfrac{3 + \sqrt{5} + 3 - \sqrt{5}}{2}\right)^2 - 2}$

$\mathrm{\implies x^2 + \dfrac{1}{x^2} = \left(\dfrac{6}{2}\right)^2 - 2}$

$\mathrm{\implies x^2 + \dfrac{1}{x^2} = (3)^2 - 2}$

$\mathrm{\implies x^2 + \dfrac{1}{x^2} = 9 - 2}$

$\mathrm{\implies x^2 + \dfrac{1}{x^2} = 7}$