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?

$tanA+tanB=3x$
Squaring both sides,we get
$tan^2A+tan^2B+2tanA tanB=9x^2$
Subtract $4tanA tanB$ from both sides,
$tan^2A+tan^2B-2tanA tanB=9x^2-4tanA tanB$
$(tan A-tan B)^2=9x^2-4tanA tanB$
Put $tanA tanB$ in above equation,
$(tan A-tan B)^2=9x^2-8x^2=x^2$
$tan A-tan B=\pm x$
