How to find the square root of a polynomial? Here, by the word polynomial, I am referring to those polynomials, who have a square root in simple algebraic expressions. 
But how to get the intuition, on how to expand the polynomial, so that we can factor it, and then find its square root? Or is there any other method also? 
Like $\sqrt{\frac{x^2}4+\frac1{x^2}-\frac1x+\frac x2-\frac34}$, 
At first look, we get instinct to first take LCM of all and add them. But after that, the numerator becomes $x^4+2x^3-3x^2-4x+4$. Now, (I don't want the answer) how to get that instinct on factoring this polynomial? Or is there a method? 
Remark: Is there any faster, non-rigorous way also, of finding the square root?
 A: Hint: Let $t=\dfrac x2-\dfrac1x$ .
A: Here is a way:
Let the square root be $(x^2 + ax + 2b)$.  You want to know the coefficient of the $x$-term, and you know the constant will be either be $-2$ or $2$.
Square this trinomial:
$(x^2 + ax + 2b)^2 = $
$x^4 + 2ax^3 + (4b + a^2)x^2 + 4abx + 4b^2 = $
$x^4 + 2x^3 - 3x^2 - 4x + 4$
Equate coefficients, solve for $a$ and $b$, and you will have your answer.
A: Given that your polynomial
$$
x^4+2x^3-3x^2-4x+4
$$
has integer coefficients, and the highest coefficient is one and the lowest four, if there are any rational solutions, they must be one of
$$
\pm \frac{1}{1},\quad \pm \frac{2}{1},\quad \pm\frac{4}{1}.
$$
Here, you should run over different divisors of $4$ and $1$.
You will surely succeed with that in this case, since you will find two double roots...
A: evaluate your polynomial at $y-2$ and see what happens. this is to get rid of power 3 here. and hopefully simplify things.

edit
let $f(x)=x^4+\color{red}2x^3-3x^2-4x+4$. now evaluate at $y-\color{red}2$, i.e. plug $y-\color{red}2$ for $x$:
$$f(y-\color{red}2)=(y-\color{red}2)^4+2(y-\color{red}2)^3...=?$$
A: The example you gave is not a polynomial as there are negative powers involved (e.g. 1/x^2). I guess that you mean to find a polynomial which, if you take the square root, yields another polynomial.
This is only possible if all the linear factors are in even multiplicity, i.e. the polynomial has 2n values for every root. If that is the case, you can take the square root as you described.
response to comment
Ok as you do not want the answer, this is what I would do:
- Find the roots
As a hint, x = 1 is a root. Use factor division on the polynomial to reduce its degree to a cubic equation and continue to find the roots.
- Check that all roots come in even frequencies, if not, we cannot take a "nice" square root.
