The equation   ${\cos }^{8}x+b{\cos }^{4}x+1=0$  will have a solution if  $b\in (-\infty ,-k]$, then $\frac{k+96}{2}=$

We have ${\cos }^{8}x+b{\cos }^{4}x+1=0$

$$\begin{gathered} \Rightarrow b=-\left({\cos }^{4}x+\frac{1}{{\cos }^{4}x}\right)\le -2 \\ \Rightarrow b\le -2 \\ \Rightarrow b\in (-\infty ,-2] \\ \Rightarrow k=2 \\ \therefore \frac{k+96}{2}=49 \end{gathered}$$