Despite having solved quadratic quations for years I can't seem to be able to get the same result than maple on this one (not as simplified as Maple's), so I wonder if someone could not explain:
I'm trying to solve this by hand:
I would think one could solve it by solving first for $ w^{2} $ (with the standard quadratic solution equation), and then doing the square root of that solution i.e. $w = \sqrt{w^{2}}$, yet the result I get doing this is much more complicated than the roots that Maple spits out.
Anyone has an idea?
Thanks
Hint:
Write the polynomial in canonical form: $$(w^2-w_0^2)^2-w_0^4\Bigl(\Bigl(1+\frac1{2Q^2}\Bigr)^2-1\Bigr)=(w^2-w_0^2)^2-\frac{w_0^4}{2Q^2}\frac{4Q^2+1}{2Q^2},$$ whence $$w^2=w_0^2\pm\frac{w_0^2}{2Q^2}\sqrt{4Q^2+1}.$$