Question

if x=2+√3 then find the value of x+1\x and x-1\x and x^2+1\x^2​

Answer 1

Solution :

Given the value of x is 2 + √3.

Hence the value of 1/x should be 1/(2 + √3)

By solving the value of 1/x , we get :

$:\implies \bf{\dfrac{1}{x} = \dfrac{1}{2 + \sqrt{3}}} \\ \\ \\ :\implies \bf{\dfrac{1}{x} = \dfrac{1}{2 + \sqrt{3}} \times \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}}} \\ \\ \\ :\implies \bf{\dfrac{1}{x} = \dfrac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})}} \\ \\ \\ :\implies \bf{\dfrac{1}{x} = \dfrac{2 - \sqrt{3}}{2(2 - \sqrt{3}) + \sqrt{3}(2 - \sqrt{3})}} \\ \\ \\ :\implies \bf{\dfrac{1}{x} = \dfrac{2 - \sqrt{3}}{(4 - 2\sqrt{3} + 2\sqrt{3} - 3)}}$

$:\implies \bf{\dfrac{1}{x} = \dfrac{2 - \sqrt{3}}{4 - \not{(2\sqrt{3}} + \not{2\sqrt{3}} - 3)}} \\ \\ \\ :\implies \bf{\dfrac{1}{x} = \dfrac{2 - \sqrt{3}}{(4 - 3)}} \\ \\ \\ :\implies \bf{\dfrac{1}{x} = \dfrac{2 - \sqrt{3}}{1}} \\ \\ \\:\implies \bf{\dfrac{1}{x} = 2 - \sqrt{3}} \\ \\ \\ \boxed{\therefore \bf{\dfrac{1}{x} = 2 - \sqrt{3}}}$

Hence the value of 1/x is 2 - √3.

Value of x + 1/x :

By substituting the value of x and 1/x in the equation , we get :

$:\implies \bf{x + \dfrac{1}{x}} \\ \\ \\ :\implies \bf{x + \dfrac{1}{x} = 2 + \sqrt{3} + 2 - \sqrt{3}} \\ \\ \\ :\implies \bf{x + \dfrac{1}{x} = 2 + \not{\sqrt{3}} + 2 - \not{\sqrt{3}}} \\ \\ \\ :\implies \bf{x + \dfrac{1}{x} = 2 + 2} \\ \\ \\ :\implies \bf{x + \dfrac{1}{x} = 4} \\ \\ \\ \boxed{\therefore x + \dfrac{1}{x} = 4}$

Hence the value of x + 1/x is 4.

Value of x - 1/x :

By substituting the value of x and 1/x in the equation , we get :

$:\implies \bf{x - \dfrac{1}{x}} \\ \\ \\ :\implies \bf{x - \dfrac{1}{x} = (2 + \sqrt{3}) - (2 - \sqrt{3})} \\ \\ \\ :\implies \bf{x - \dfrac{1}{x} = 2 + \sqrt{3} - 2 + \sqrt{3}} \\ \\ \\ :\implies \bf{x - \dfrac{1}{x} = \not{2} + \sqrt{3} - \not{2} + \sqrt{3}} \\ \\ \\ :\implies \bf{x - \dfrac{1}{x} = \sqrt{3} + \sqrt{3}} \\ \\ \\ :\implies \bf{x - \dfrac{1}{x} = 2\sqrt{3}} \\ \\ \\ \boxed{\therefore x - \dfrac{1}{x} = 2\sqrt{3}}$

Hence the value of x - 1/x is 2√3.

Value of x² + 1/x² :

By squaring on both the sides of the equation , x + 1/x , we get :

$:\implies \bf{x + \dfrac{1}{x} = 4} \\ \\ \\ :\implies \bf{\bigg(x + \dfrac{1}{x}\bigg)^{2} = 4^{2}} \\ \\ \\ :\implies \bf{x^{2} + \dfrac{1}{x^{2}} + 2 \times x \times \dfrac{1}{x} = 16} \\ \\ \\ :\implies \bf{x^{2} + \dfrac{1}{x^{2}} + 2 \times \not{x} \times \dfrac{1}{\not{x}} = 16} \\ \\ \\ :\implies \bf{x^{2} + \dfrac{1}{x^{2}} + 2 = 16} \\ \\ \\ :\implies \bf{x^{2} + \dfrac{1}{x^{2}} = 16 - 2} \\ \\ \\ :\implies \bf{x^{2} + \dfrac{1}{x^{2}} = 14} \\ \\ \\ \boxed{\therefore \bf{x^{2} + \dfrac{1}{x^{2}} = 14}}$

Hence the value of x² + 1/x² = 14.