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. 
 A: HINT : 
$$\frac{1}{x}=\frac{1}{2+\sqrt 3}=\frac{2-\sqrt 3}{(2+\sqrt 3)(2-\sqrt 3)}=2-\sqrt 3$$
A: 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,
$$ x^2 + \frac{1}{x^2} = 4x-1 + 15-4x = 14. $$
A: Hint: Try to find a simple expression for $1/x$.
A: Hint:
$$(2-\sqrt{3})(2+\sqrt{3})=1$$
$$(2-\sqrt{3})(x)=1$$
so
$$\frac{1}{x}=2-\sqrt{3}$$
$$(x+\frac{1}{x})^2=x^2+2+\frac{1}{x^2}$$
$$x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2$$
$$x^2+\frac{1}{x^2}=(2+\sqrt{3}+2-\sqrt{3})^2-2=14$$
A: Siddhant, when you said that you tried to substitute the value of $x$, I could understand why your calculations went too far, and what you meant by 'very big'.
But here's a very short and easy way of solving your question. 
Write $x^2 + \frac {1}{x^2} = (x+\frac1x)^2 - 2$ and then solve. 
$$\implies 2+\sqrt3+ \frac {1}{2+\sqrt3} = \frac {(2+\sqrt3)^2 + 1}{2+\sqrt3} $$
$=4$
Now, what remains of your question is $4^2 - 2$ which is $14$.
A: 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.
A: By brute-force:
$$(2+\sqrt3)^2+\frac1{(2+\sqrt3)^2}=\frac{(2+\sqrt3)^4+1}{(2+\sqrt3)^2}.$$
Then
$$(2+\sqrt3)^2=7+4\sqrt3,\\
(2+\sqrt3)^4=(7+4\sqrt3)^2=97+56\sqrt3,$$
$$\frac{(2+\sqrt3)^4+1}{(2+\sqrt3)^2}=\frac{98+56\sqrt3}{7+4\sqrt3}=\frac{14\cdot7+14\cdot4\sqrt3}{7+4\sqrt3}=14.$$
A: By the Vieta formulas, $2\pm\sqrt3$ are the roots of the quadratic equation $x^2-4x+1=0$, or $$x+\frac1x=4.$$ Squaring
$$x^2+2+\frac1{x^2}=16.$$
