Find the sums of the squares of the solutions of the equation 3z^2 - 5z + 8 = 0.

The roots of 3z^2 - 5z + 8 = 0 are \(\frac{5 \pm i \sqrt{71}}{6}\)

 

Then \((\frac{5 + i \sqrt{71}}{6})^2 + (\frac{5 - i \sqrt{71}}{6})^2 = -\frac{53}{36}\)

Let the roots be x and y. Note that \(x^2 + y^2 = ( x+ y)^2 - 2xy\)

 

By Vieta's, \(x + y = -{b \over a} = {5 \over 3}\) and \(xy = {c \over a} = {8 \over 3}\)

 

So, \(x^2 + y^2 = {25 \over 9} - {16 \over 3} = {25 \over 9} - {48 \over 9} = \color{brown}\boxed{-{ 23 \over 9}}\)