A new approach to find value of $x^2+\frac{1}{x^2}$ When I was teaching in a college class, I write this question on board. 
If we now $\bf{x+\dfrac{1}{x}=4}$ show the value of  $\bf{x^2+\dfrac{1}{x^2}=14}$
Some student asks me for a multi idea to show or prove that.
I tried these : $$ $$1 :$$\left(x+\frac{1}{x}\right)^2=4^2\\x^2+\frac{1}{x^2}+2x\times \frac{1}{x}=16\\x^2+\frac{1}{x^2}=16-2 $$
 2:solving quadratic equation ,and putting one of roots$$x+\frac{1}{x}=4\\\frac{x^2+1}{x}=4\\x^2+1=4x\\x^2-4x+1=0\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-4)\pm \sqrt{(-4)^2-4(1)(+1)}}{2}=\\x=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}\\x=2+\sqrt3 \to x^2=4+4\sqrt3+3=7+4\sqrt3\\x^2+\frac{1}{x^2}=7+4\sqrt3+\frac{1}{7+4\sqrt3}=\\7+4\sqrt3+\frac{1}{7+4\sqrt3}\cdot\frac{7-4\sqrt3}{7-4\sqrt3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{49-16\cdot 3}=\\7+4\sqrt3+\frac{7-4\sqrt3}{1}=14$$
3: visual approach .assume side length of a square is $x+\frac{1}{x}=4$
now I am looking for new idea to proof.Any hint will be appreciated.(more visual proof - geometrical - trigonometrical - using complex numbers ...)

Not sure how to show what the continued fraction equals but here's my attempt. Again similar in spirit to the algebraic method.
Solving for $x$ in the first equation we have: $x = 4 - \frac{1}{x}$.
Substituting this into the second equation:
$
(4 - \frac{1}{x})^{2} + \frac{1}{(4 - \frac{1}{x})^{2}} = 16 - \frac{8}{4-\frac{1}{x}} + \frac{1}{ 16 - \frac{8}{4-\frac{1}{x}}} =  16 - \cfrac{8}{4-\cfrac{1}{4-\cfrac{1}{4-\cdots}}} + \frac{1}{16 - \cfrac{8}{4-\cfrac{1}{4-\cfrac{1}{4-\cdots}}}}
$
And that big mess equals 14.
