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. 

Since everyone else has answered with various shortcut methods, let me just show you how you could simply substitute in directly and get an answer cleanly without things ever getting 'too big'.
We know that $x=2+\sqrt{3}$, so we can square this using the usual $a^2+2ab+b^2$ binomial formula: $x^2=2^2+2(2)(\sqrt3)+(\sqrt3)^2$ $= 4+4\sqrt3+3$ $=7+4\sqrt{3}$.
Now, $\dfrac1{x^2}$ is $\dfrac1{7+4\sqrt3}$; we can use the usual method for rationalizing the denominator to handle this, by multiplying by $\dfrac{7-4\sqrt3}{7-4\sqrt3}$.  This gives $\dfrac1{x^2}$ $=\dfrac{7-4\sqrt3}{(7+4\sqrt3)(7-4\sqrt3)}$ $=\dfrac{7-4\sqrt3}{7^2-(4\sqrt3)^2}$ $=\dfrac{7-4\sqrt3}{7^2-4^2\cdot 3}$ $=\dfrac{7-4\sqrt3}{49-48}$ $=7-4\sqrt3$.
The most important lesson here is that whenever you have an expression of the form $x=a+b\sqrt{c}$, then the powers of $x$ — both positive and negative — will never get too messy; they'll all always be of the form $p+q\sqrt{c}$ for some rational $p$ and $q$.  In this case, $p$ and $q$ were integers even for the negative powers of $x$; that won't always be the case, but they'll always be rational.
