<s>[INST] Algebra confusion: $A^2 + B^4$? So i am currently trying to solve a practice question and i have hit a stump.

The first term of a GP is $1$ and the sum of the third and fifth term is $90$. Find the common ratio.

Using $T_n = a \cdot r^{n-1}$ I got that $a = 1$
Hence
$ar^2 + ar^4 = 90 \implies r^2 + r^4 = 90$
But this is where i got stumped. The answer is given as $ \pm \;3$
Any help is highly appreciated. 
 [/INST] All of your calculations are right. We find ourselves needing to solve $r^2+r^4=90$.
Rearranging gives $r^4+r^2-90=0$. This can be thought of as a quadratic in $r^2$. Let $y=r^2$ then $y^2=r^4$ and so $r^4+r^2-90=0$ becomes $y^2+y-90=0$.
It turns out that $y^2+y-90$ factorises to gives $(y-9)(y+10)$.
Since $y$ was shorthand for $r^2$ we see that $r^4+r^2-90 \equiv (r^2-9)(r^2+10)$.
The solutions to $(r^2-9)(r^2+10) =0$ are $r^2=9$ and $r^2 = -10$. The equation $r^2=9$ has solutions $r=\pm 3$, while the equation $r^2=-10$ has no real solutions.
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