$\sqrt{{(a+\frac{1}{2})}^2+\frac{3}{4}}$ is irrational if $a$ is

$\sqrt{{(a+\frac{1}{2})}^2+\frac{3}{4}}$

$=\sqrt{{(a+\frac{1}{2})}^2+\frac{3}{4}}$

$=\sqrt{a^2+\frac{1}{4}+a+\frac{3}{4}}$

$=\sqrt{a^2+\frac{1}{4}+a+\frac{3}{4}}$

Case-1:

To make it perfect square another ‘$a$’has to be add

$a^2+2a+1={(a+1)}^2$

 So, $a$ cannot be -1

Case-2:

If is zero then it is a perfect square

$\therefore$ $\sqrt{{(a+\frac{1}{2})}^2+\frac{3}{4}}$ is irrational only when, $a\in Z-\{-1,0\}$