+579 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
+128406 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
