Remember that two vectors uu and vv are orthogonal if and only if

\(\displaystyle{u}⋅{v}={0}\)

So let's substitute in the expressions from the question and see what we end up with.

\(\displaystyle⟨{3},{1}+{b}⟩⋅⟨{5},{1}−{b}⟩={0}{\left({3}\right)}{\left({5}\right)}+{\left({1}+{b}\right)}{\left({1}−{b}\right)}={0}\)

\((3)(5)+(1+b)(1−b)=0\)

\(\displaystyle{15}+{1}−{b}^{{2}}={0}\)

\(\displaystyle{b}=±{4}\)

So the two possible values of bb are 4 and −4.