The value of $\sqrt{15+8\mathrm{i}}+\sqrt{15-8\mathrm{i}}$ is equal to

Suppose that $\sqrt{15+8i}=x+iy$

Squaring on both sides 

$x^2-y^2+2ixy=15+8i$

it implies that $x^2-y^2=15$ and $xy=4$

The possible values of $x,y$ are $x=4,y=1$

Hence, $\sqrt{15+8i}=4+i$ and $\sqrt{15-8i}=4-i$

Therefore, $\sqrt{15+8\mathrm{i}}+\sqrt{15-8\mathrm{i}}=4+i+4-i=\boxed{8}$

$$\begin{gathered} \boxed{\text{ SQUARE ROOT OF COMPLEX NUMBER } \\ \sqrt{a+ib}=\pm \left\{\frac{\sqrt{\left|z\right|+a}}{2}+i\frac{\sqrt{\left|z\right|-a}}{2}\right\}\text{ for }b>0 \\ =\pm \left\{\frac{\sqrt{\left|z\right|+a}}{2}-i\frac{\sqrt{\left|z\right|-a}}{2}\right\}\text{ for }b<0\text{ where }\left|z\right|=\sqrt{a^2+b^2}} \end{gathered}$$