Substituting the value $x=2+\sqrt{3}$ into $x^2 + 1/x^2$ My teacher gave me a question which I am not able to solve:

If $x=2+\sqrt{3}$ then find the value of $x^2 + 1/x^2$

I tried to substitute the value of x in the expression, but that comes out to be very big. 

Here is a slightly weird way of doing it. $x$ looks like the quadratic formula, so if we can cook up a quadratic equation that it satisfies, we won't actually have to square it. The solutions to
$$ y^2+2by+c = 0 $$
are
$$ y = -b \pm \sqrt{b^2-c} $$
(because $a=1$ and the $b$ has a $2$ multiplying it that cancels the $2$ normally present). In this case, the number not in the square root is $2$, so $b=-2$. Then $b^2=4$, so to agree with the number inside the square root, we have $ 4-c=3 $, so $c=1$. Therefore $x$ satisfies the quadratic equation
$$ x^2 -4x+1 = 0 \tag{1} $$
From here, we can read off that
$$ x^2 = 4x-1, $$
and if we divide (1) by $x$, we also find
$$ \frac{1}{x} = 4-x, \tag{2} $$
and dividing (1) by $x^2$ gives
$$ \frac{1}{x^2} = \frac{4}{x}-1 = 4(4-x)-1 = 15-4x, $$
applying (2) to the $1/x$ term. Therefore,
$$ x^2 + \frac{1}{x^2} = 4x-1 + 15-4x = 14. $$
