plz help

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

Note

1/a^2   +   1/b^2  =     [ a^2 + b^2 ] /  [ ab]^2

5x^2   -11x + 4   =  0

By Vieta

Sum of the roots =  11/5

a + b = 11/5       square both sides

a^2 + 2ab  + b^2   =  121/ 25 

a^2 + b^2  =  121/25 - 2ab      (1)

And product of the  roots  =  4/5

So

ab = 4/5    ⇒  [ab]^2  = 16/25

2ab = 8/5

So (1)  becomes a^2 + b^2 =  121/25 - 8/5  =   121/25 - 40 /25  =   81/25

So

1/a^2  + 1/b^2   =     [ a^2 + b^2]  / [  ab]^2   =   [ 81/25] / [ 16/25]  =   81/16