Question

If x= 2+√3, find the value of x+1/x

Answer 1

Given : x=2+\sqrt{3}x=2+3

To Find : find the value of x+\frac{1}{x}x+x1

Solution:

x+\frac{1}{x}x+x1

Since x=2+\sqrt{3}x=2+3

To find \frac{1}{x}x1

x=2+\sqrt{3}x=2+3

\frac{1}{x} =\frac{1}{2 +\sqrt{3} } \times \frac{2 - \sqrt{3} }{2 -  \sqrt{3}}x1=2+31×2− 32−3

\frac{1}{x} =\frac{2 -\sqrt{3} }{(2) {}^{2} - ( \sqrt{3}) {}^{2} }x1=(2)2−(3)22−3

\frac{1}{x}= \frac{2 -\sqrt{3}}{4 - 3}x1=4−32−3

\frac{1}{x}= 2- \sqrt{ 3}x1=2−3

Substitute the values

x+\frac{1}{x}x+x1

2+\sqrt{3}+2- \sqrt{ 3}2+3+2−3

44

Hence  the value of x+\frac{1}{x}x+x1 is 4