Find the sum of the squares of the solutions to 2x^2 + 4x - 1 = x^2 - 3x.
if you move things around you will find this is a quadratic, and you will get x^2 + 7x - 1 = 0 (i believe) and then you can simply plug it in to the quadratic formula
is my answer below correct?
Simplifying gives \(x^2 + 7x - 1 = 0\), so the roots are \(x = {-7 \pm \sqrt{7^2-4*1*-1} \over 2}\)\(= {-7 \pm \sqrt{53} \over 2}\). \(({{-7 + \sqrt{53}} \over 2}) ^2 + ({{-7 - \sqrt{53}} \over 2})^2 = 51\)
