Using the quad formula:
\(\frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4\cdot \:5\cdot \:4}}{2\cdot \:5}\)
\(\frac{\left(11\right)\pm \sqrt{41}}{10}\)
Substituting this into \(\frac{1}{a}^2\:+\:\frac{1}{b}^2.\)
\(\frac{1}{\frac{\left(11\right)\pm \:\sqrt{41}}{10}}^2\:+\:\frac{1}{\frac{\left(11\right)\pm \:\sqrt{41}}{10}}^2\)
= \(\frac{81\pm11\sqrt{41}}{16}\)
or
approx. 9.46464646464 and 0.66035
-Vinculum
lol