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 ...)
 A: Here is a mildly disguised version of the same idea.
Let $x=e^u$
Then we have $$\cosh u=2$$
And we want $$x^2+\frac {1}{x^2}=2\cosh 2u=2(2\cosh^2 u-1)=2(8-1)=14$$
A: 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.
A: Some obfuscation using linear algebra:
Write $x + \frac{1}{x} = a$ and let 
$$ p(\lambda) = \left( \lambda - x \right) \left( \lambda - \frac{1}{x} \right) = \lambda^2 - a\lambda + 1$$ be a polynomial whose roots are $x$ and $\frac{1}{x}$ and consider the companion matrix
$$ A = \left( \begin{matrix} 0 & -1 \\ 1 & a \end{matrix} \right). $$
The eigenvalues of $A$ are $x$ and $\frac{1}{x}$ so $\mathrm{tr}(A) = x + \frac{1}{x} = a$. The eigenvalues of $A^2$ are $x^2$ and $\frac{1}{x^2}$ so
$$ x^2 + \frac{1}{x^2} = \mathrm{tr}(A^2) = \mathrm{tr} \left( \begin{matrix} -1 & -a \\ a & a^2 - 1 \end{matrix} \right) = a^2 - 2. $$
A: You can simplify this by noting $$x^2+\frac1{x^2}=x^2+x^{-2}=e^{2\ln(x)}+e^{-2\ln(x)}=\cos(2i\ln(x))$$
Now you can solve for $x$ very easily.
$$x^2+\frac1{x^2}=y$$$$\cos(2i\ln(x))=y$$$$x=e^{\frac{\arccos(y)}{2i}}$$
A: Here's a trigonometric approach:
Let $x=\tan\theta=\sin\theta/\cos\theta$.  Then
$$4=x+{1\over x}={\sin\theta\over\cos\theta}+{\cos\theta\over\sin\theta}={1\over\sin\theta\cos\theta}={2\over\sin2\theta}$$
so $\sin2\theta=1/2$.  Now
$$x^2+{1\over x^2}={\sin^2\theta\over\cos^2\theta}+{\cos^2\theta\over\sin^2\theta}={1\over\cos^2\theta}-1+{1\over\sin^2\theta}-1={1\over\sin^2\theta\cos^2\theta}-2\\
={4\over\sin^22\theta}-2=16-2=14$$
A: Take a look for $$x^2-4x+1=0 \\x_1.x_2=1 \\x^2+1=4x \\\div x
\\x+\frac{1}{x}=4 \\$$so we can write an equation with $x_1^2,x_2^2 ,roots$  
$$s'=x_1^2+x_2^2=s^2-2p=16-4\\x_1^2.x_2^2=1^2 \\\to z^2-14z+1=0 \\z=x^2 \\\to x^4+\frac{1}{x^4}=14$$
