Solution

Let $$a$$ and $$r$$ be the first term and the common ratio of the G.P. respectively.
$$\displaystyle \therefore

a=1\\ { a }_{ 3 }={ ar }^{ 2 }={ r }^{ 2 }\\ { a }_{ 5 }={ ar }^{ 4 }={

r }^{ 4 }$$
Given $$a_3+a_5 =90$$
$$ \therefore { r }^{ 2 }+{ r }^{ 4 }=90\\ \Rightarrow { r }^{

4 }+{ r }^{ 2 }=90\\ \Rightarrow { r }^{ 4 }+{ r }^{ 2 }-90=0\\\displaystyle

\Rightarrow { r }^{ 2 }=\frac { -1+\sqrt { 1+360 } }{ 2 } =\frac {

-1\pm \sqrt { 361 } }{ 2 } =\frac { -1\pm 19 }{ 2 } =-10\quad or\quad

9\\ \therefore r=\pm 3\quad \quad \quad (\text{Taking real roots})$$