Extract the Square Root of $ a^2-\left(\frac{\left(3a \sqrt{a}\right)}{2}\right)-\left(\frac{\left(3 \sqrt{a}\right)}{2}\right)+\frac{41a}{16}+1 $ 
Extract the Square Root of $$ a^2-\left(\frac{\left(3a \sqrt{a}\right)}{2}\right)-\left(\frac{\left(3 \sqrt{a}\right)}{2}\right)+\frac{41a}{16}+1 $$

I found a method to do this in a book but couldn't understand how to do it. Can anyone explain it?

 A: This appears to be checking that $16a^2 - 24a^{\frac{3}{2}} + 41a - 24a^{\frac{1}{2}} + 16$ is a perfect square of a linear expression of the form $c_1 a + c_2 a^{\frac{1}{2}} + c_3$ for certain constants $c_1, c_2, c_3$ using some technique.  Note that squaring that linear expression gives
$$\left( c_1 a + c_2 a^{\frac{1}{2}} + c_3 \right)^2 = c_1^2 a^2 + 2c_1 c_2 a^{\frac{3}{2}} + \left(c_2^2 + 2c_1 c_3\right) a \; + 2c_2 c_3 a^{\frac{1}{2}} + c_3^2$$
Equating the coefficients of the terms with the original equation gives the following set of equations:
$$\begin{align}
c_1^2 & = 16 \tag{1}\label{eq1} \\
2c_1 c_2 & = -24 \tag{2}\label{eq2} \\
c_2^2 + 2c_1 c_3 & = 41 \tag{3}\label{eq3} \\
2c_2 c_3 & = -24 \tag{4}\label{eq4} \\
c_3^2 & = 16 \tag{5}\label{eq5} \\
\end{align}$$
From \eqref{eq1} and \eqref{eq5}, we can see that $c_1$ and $c_3$ are $\pm 4$. From \eqref{eq2} and \eqref{eq4}, if $c_2$ is non-zero, then $c_1 = c_3$.  Checking using the positive case for $c_1$ gives from these $2$ equations that $c_2$ is $-3$.  This can be confirmed using \eqref{eq3}.  Thus, the linear expression is what the book shows as the denominator, i.e., $4a - 3a^{\frac{1}{2}} + 4$.  The negative case of $c_1 = c_3 = -4$ gives the other square root of the equation.
As for what the book is doing, it seems to be some sort of short hand for what I just did above.  If I took the time & effort, I could possibly figure it out, but it's getting too late for me to do that tonight.
A: Let see an easier example: take square root of $a^2-4a+4$.
By the method given in the source:
$\hspace{1cm}$
Note: In your example, you practically do the same operations. Also, you double $(-3)\cdot 2=-6$ and multiply $8\cdot 4=32$. Also, note in your example, the minus sign is already considered, so you don't subtract, but add.
A: Let $x=\sqrt{a}$, then
$$ a^2-\left(\frac{\left(3a \sqrt{a}\right)}{2}\right)-\left(\frac{\left(3 \sqrt{a}\right)}{2}\right)+\frac{41a}{16}+1 $$
$$ = {16x^4 -24x^3+41x^2 -24x+16\over 16}$$
$$ = ({4x^2+bx+4 \over 4})^2$$
where you have to find $b$. This is easy to do, just put some walue for $x$ , say $x=1$ in both expresions:
$$ {25\over 16} = {(8+b)^2\over 16}\implies b+8= \pm 5$$
Trying both values we see that $b=-3$ works and you are done. 
A: $$a^2-\left(\frac{\left(3a \sqrt{a}\right)}{2}\right)+\frac{41a}{16}-\left(\frac{\left(3 \sqrt{a}\right)}{2}\right)+1$$
For convenience, let $x=\sqrt a$. Then we get
$$x^4-\dfrac 32x^3 +\frac{41}{16}x^2-\dfrac 32x+1$$
Use of the square root algorithm would produce this.
\begin{array}{r}
   && x^2 & -\dfrac 34x & +1\\
   &&---  &--- &--- &--- &---\\
   &\mid&x^4 &-\dfrac 32x^3 &+ \dfrac{41}{16}x^2 &-\dfrac 32x &+1 \\
&x^2 \mid & x^4 \\
&\mid&--- &--- &---\\
&\mid &&-\dfrac 32x^3 &+ \dfrac{41}{16}x^2 \\
&2x^2-\dfrac 34x \mid &&-\dfrac 32x^3 &+\dfrac{9}{16}x^2 \\
&\mid &&--- &--- &--- &--- \\
&\mid && &2x^2 &-\dfrac 32x &+1 \\
&2x^2-\dfrac 32x+1 \mid && &2x^2 &-\dfrac 32x &+1 \\
&&& &--- &--- &--- \\
&&&&&&0
\end{array}
Hence
$$a^2-\frac 32a\sqrt a+\frac{41}{16}a-\frac 32\sqrt a+1
  = \left(a - \dfrac 34\sqrt a + 1\right)^2$$
