Question

If x = 2+✓3 find the value of x2 + 1/x2​

Answer 1

Step-by-step explanation:

x=2+√3

1/x=1/2+√3

1/x=2-√3/4-3

1/x=2-√3

x²+1/x²

=(x+1/x)²-2

=(4)²-2

=14

Answer 2

Given :-

x= 2+√3

To find :-

$\sf\bullet x^2+\dfrac{1}{x^2}$

Formula used !

$\bigstar{\boxed{\sf{(a+b)^2=a^2+b^2+2ab}}}$

$\underline{\textit{According \ to \ Question :- }}$

$\longmapsto\sf x= 2+\sqrt{3}$

$\longmapsto\sf \dfrac{1}{x}= \dfrac{1}{2+\sqrt{3}}\\ \\ \dag \ \ \ \sf Rationalize \ this !\\ \\ \longmapsto\sf \dfrac{1}{x}=\dfrac{1}{2+\sqrt{3}}\times \dfrac{2-\sqrt{3}}{2-\sqrt{3}}\\ \\\longmapsto\sf\dfrac{1}{x}= \dfrac{2-\sqrt{3}}{(2)^2-(\sqrt{3})^2}\\ \\ \longmapsto\sf\dfrac{1}{x}=\dfrac{2-\sqrt{3}}{4-3}\\ \\ \longmapsto\sf\dfrac{1}{x}= 2-\sqrt{3}$

$\longmapsto\sf \bigg(x+\dfrac{1}{x}\bigg)^2= x^2+\dfrac{1}{x^2}+2\times \cancel{x}\times \dfrac{1}{\cancel{x}}\\ \\ \longmapsto\sf \bigg(x+\dfrac{1}{x}\bigg)^2= x^2+\dfrac{1}{x^2}+2\\ \\ \\ \longmapsto\sf x+\dfrac{1}{x}= 2+\cancel{\sqrt{3}}+2-\cancel{\sqrt{3}}\\ \\ \longmapsto\sf x+\dfrac{1}{x}=4$

Now

$\longmapsto\sf \bigg(x+\dfrac{1}{x}\bigg)^2= x^2+\dfrac{1}{x^2}+2\\ \\ \longmapsto\sf (4)^2=x^2+\dfrac{1}{x^2}+2\\ \\ \longmapsto\sf 16-2=x^2+\dfrac{1}{x^2}\\ \\ \longmapsto\sf 14= x^2+\dfrac{1}{x^2}$

$\bigstar{\boxed{\sf{x^2+\dfrac{1}{x^2}=14}}}$