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?


$$a^2-\left(\frac{\left(3a \sqrt{a}\right)}{2}\right)+\frac{41a}{16}-\left(\frac{\left(3 \sqrt{a}\right)}{2}\right)+1$$
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
$$a^2-\frac 32a\sqrt a+\frac{41}{16}a-\frac 32\sqrt a+1
  = \left(a - \dfrac 34\sqrt a + 1\right)^2$$
