If $$x + \frac{1}{x}$$ = 4 , then find the value of $$x ^2+ \frac{1}{x^2}$$​​

Correct option is B

Given: 

​$$x+\frac{1}{x}=4$$​​

Solution:

The given equation: $$x+\frac{1}{x}=4$$ 

​Square both sides of the equation: 

​$$(x+\frac{1}{x})^2=4^2$$ 

​$$x^2+2\times x \times\frac{1}{x}+\frac{1}{x^2}=16$$ 

​$$x^2+2+\frac{1}{x^2}=16$$ 

​Subtract 2 from both sides :  $$x^2+\frac{1}{x^2}$$ 

​$$x^2+\frac{1}{x^2}=16 -2$$​​

​$$x^2+\frac{1}{x^2}$$ =14

Thus, correct answer is (b) 14