+2668 We can write \({1 \over a^2} + {1 \over b^2}\) as \({a^2+b^2} \over (ab)^2\), where both a and b are the roots of the equation.
Using Vieta's, we find the denominator is \({4 \over 5} ^2 = {16 \over 25}\)
Using the quadratic formula, we find that the 2 roots are: \({ 11 \pm \sqrt{41} \over 10}\).
Plugging it in, we get this mess: \(\huge{{1 \over { 11 + \sqrt{41} \over 10 }^2} + {1 \over { 11 - \sqrt{41} \over 10 }^2}} \over \huge{ 16 \over 25}\)
Simplifying the numerator, we get \(\large{{33 \over 10} \over {16 \over 25} }\). This simplifies to \(\color{brown}\boxed{ 81 \over 16}\)
