Is root of $x^2$ = $x$ ?

$\sqrt{\left(-\frac 12\right)^2} = \sqrt{(-1)^2\left(\frac 12\right)^2} = \sqrt{1\times \left(\frac 12\right)^2} = \sqrt{\left(\frac 12\right)^2} = \boxed{\dfrac 12}$

Because Square Roots Are Always Tricky

If, $f(y) = \sqrt{x^2}$ , then $f(y) =$ $\begin{cases} x \quad \text{if x > 0} \\ -x \quad \text{ if x < 0} \\ \end{cases}$ As $-\dfrac12 < 0$ the answer will be : $-\left(-\dfrac12 \right) = \dfrac12$

The function $\sqrt{x^2}$ is the same as $|x|$ , so $\sqrt{ \left( - \dfrac{1}{2} \right)^2 }=\left|-\dfrac12\right|=\dfrac12$

Relevant wiki: Definition of Absolute Value

We know that $\sqrt{x^2}=|x|$ which $|x|=x$ if $x$ positive and $|x|=-x$ if $x$ negative. So $\sqrt{(-\frac{1}{2})^2}=|(-\frac{1}{2})|=-(-\frac{1}{2})=\boxed{\large{\frac{1}{2}}}$