Find the sum of the squares of the solutions to \(2x^2+4x-1=0\).
\(\begin{array}{|rcll|} \hline 2x^2+4x-1 &=& 0 \quad | \quad :2 \\ x^2+2x-\dfrac{1}{2} &=& 0 \\ \hline \end{array}\)
By Vieta:
\( 2 = -(x_1+x_2) \\ -\dfrac{1}{2} = x_1*x_2 \)
\(\begin{array}{|rcll|} \hline x_1^2+x_2^2 &=& (x_1+x_2)^2 - 2x_1*x_2 \\ x_1^2+x_2^2 &=& (-2)^2 - 2*\left(-\dfrac{1}{2}\right) \\ x_1^2+x_2^2 &=& 4 + 1 \\ \mathbf{x_1^2+x_2^2} &=& \mathbf{5} \\ \hline \end{array}\)