How does $\frac12\sqrt{n^2 + \frac{4m^3}{27}}$ become $\sqrt{\frac{n^2}4 + \frac{m^3}{27}}$?

Question

$$\begin{align} &\frac n2 \pm \frac12 \,\sqrt{n^2 + \frac{4m^3}{27}}\\[6pt] =\; &\frac n2 \pm \phantom{\frac12}\sqrt{\frac{n^2}4 + \frac{m^3}{27}} \end{align}$$

Can someone please explain me how this radical multiplied by $1/2$ becomes this? You can see $1/2$ disappeared, and division by $4$ appeared under square root.

Thanks.

Answer 1

By one of the laws of exponents $$a^mb^m=(ab)^m,$$

we have $$\frac12\sqrt{k}=\sqrt{\bigg(\frac12\bigg)^2}\sqrt{k}=\sqrt{\frac14}\sqrt{k}=\sqrt{\frac14k}.$$

Answer 2

Since it is $$\frac{1}{2}\sqrt{n^2+\frac{4m^3}{27}}=\sqrt{\frac{1}{4}\left(n^2+\frac{4m^3}{27}\right)}=\sqrt{\frac{n^2}{4}+\frac{4m^3}{4\times27}}$$

Answer 3

Remember these: $$\sqrt{a}\sqrt{b} = \sqrt{ab}$$ $$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$$

And note that: $$\frac{1}{2} = \frac{\sqrt{1}}{\sqrt{4}}$$

So: $$\frac{\sqrt{1}}{\sqrt{4}}\sqrt{n^2+\frac{4m^3}{27}} = \frac{\sqrt{1}\sqrt{n^2+\frac{4m^3}{27}}}{\sqrt{4}} = \frac{\sqrt{1(n^2+\frac{4m^3}{27})}}{\sqrt{4}} = \sqrt{\frac{1(n^2+\frac{4m^3}{27})}{4}} = \sqrt{\frac{n^2}{4}+\frac{m^3}{27}} $$