Given $\tan A + \tan B = 3x$ and $\tan A \tan B = 2x^2$, find $\tan A - \tan B$ Given $$\tan A + \tan B = 3x$$ and $$\tan A \tan B = 2x^{2},$$ how can one find $\tan A - \tan B$? I have tried substitution, but failed to find the answer.
Edit: Can this problem be solved using the formulas for sums and differences of tangents?

We know that $$(a-b)^2=a^2+b^2-2ab=(a+b)^2-4ab$$ $$\implies a-b=\pm \sqrt{(a+b)^2-4ab}$$ Now, we have 
$$\tan A-\tan B=\sqrt{(\tan A+\tan B)^2-4\tan A\tan B}$$ $$=\sqrt{(3x)^2-4(2x^2)}$$ $$=\sqrt{x^2}$$$$=\pm x=|x|$$
