If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4,$ then $(a-b)^2=\;$? If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4$, what is the value of $(a-b)^2$?  I think $a^2+b^2=36$, please confirm and is it possible to to figure out one of the variables?

You know that $(a-b)^2=30$. The same strategy tells you that $(a+b)^2=42$.
Thus 
$$a+b=\pm \sqrt{42}\quad\text{and}\quad a-b=\pm\sqrt{30}.$$
Now by adding and subtracting, we can find $2a$ and $2b$. and hence $a$ and $b$. Note that there are $4$ combinations, though if we have found one solution $a=p$, $b=q$, the other three are $a=-p$, $b=-q$, and $a=q$, $b=p$, and $a=-q$, $b=-p$. 
One of the solutions is $a=\frac{\sqrt{42}+\sqrt{30}}{2}$, $b=\frac{\sqrt{42}-\sqrt{30}}{2}$.
