Given the quadratic equation:
[tex]2x^2-3x+4=0[/tex]We solve this equation to find the roots. The general solution for a quadratic equation is given by the expression:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4\cdot a\cdot c}}{2a}[/tex]Looking at the equation, we identify:
[tex]\begin{gathered} a=2 \\ b=-3 \\ c=4 \end{gathered}[/tex]Then, using the general solution formula:
[tex]\begin{gathered} x=\frac{-(-3)\pm\sqrt[]{3^2-4\cdot2\cdot4}}{2\cdot2}=\frac{3\pm\sqrt[]{9-32}}{4} \\ \Rightarrow x_1=\frac{3+\sqrt[]{23}i}{4} \\ \Rightarrow x_2=\frac{3-\sqrt[]{23}i}{4} \end{gathered}[/tex]Now, we need to know the sum of the squares of x₁ and x₂. Then:
[tex]\begin{gathered} x^2_1+x^2_2=\frac{1}{16}(9+2\cdot\sqrt[]{23}\cdot i-23+9-2\cdot\sqrt[]{23}\cdot i-23) \\ \Rightarrow=\frac{1}{16}(18-46)=-\frac{28}{16}=-\frac{7}{4} \end{gathered}[/tex]