If $a+\frac1a=\sqrt3$ then $a^4+\frac1{a^4}=\ ?$ 
If $a+\frac1a=\sqrt3$ then $a^4+\frac1{a^4}=\ ?$

Can someone please explain to me how to solve this?
because I tried everything I know and it didn't work.
P.S: I'm in 8th grade so no quadratic formula.

We have these identities. (Do you see why these are true? Hint: FOIL. Or, if you learned $``{(a+b)^2=a^2+2ab+b^2}
``$, use that.)
$$\left(a+\frac1a\right)^2=\left(a^2+\frac1{a^2}\right)+2$$
$$\left(a^2+\frac1{a^2}\right)^2=\left(a^4+\frac1{a^4}\right)+2$$
Now, plug in $(a+1/a)=\sqrt3$.
$$\left(\sqrt3\right)^2=\left(a^2+\frac1{a^2}\right)+2$$
So $3=(a^2+1/a^2)+2$, and $a^2+1/a^2=1$.
$$(1)^2=\left(a^4+\frac1{a^4}\right)+2$$
So $1=(a^4+1/a^4)+2$, or $a^4+1/a^4=-1$ and we have solved the problem.
