Finding the recurrence relation(with square roots)

Question

I came across a very peculiar recurrence relation :

$\sqrt {T(n)} = \sqrt {T(n-1)} + 2 \sqrt {T(n-2)} $

And Initial Condition $T(0) = T(1)= 1$

Any helps on how to find it

Answer 1

Denote $A(n)=\sqrt{T(n)}$ and solve for A formula.

For example try to find in $A(n) = \lambda^{n}$ form.