Rozwiązanie:
[tex]f(x)=-x^{2}+2x+3[/tex]
Na początek obliczmy:
[tex]$x_{1}+x_{2}=-\frac{b}{a} =\frac{-2}{-1} =2[/tex]
[tex]$x_{1} \cdot x_{2}=\frac{c}{a} =-\frac{3}{1} =-3[/tex]
[tex]a)[/tex]
[tex](x_{1}-x_{2})^{2}=(x_{1}+x_{2})^{2}-4x_{1}x_{2}=2^{2}-4 \cdot (-3)=4+12=16[/tex]
[tex]b)[/tex]
[tex]$\frac{x_{2}}{x_{1}} +\frac{x_{1}}{x_{2}} =\frac{x_{2}^{2}+x_{1}^{2}}{x_{1} \cdot x_{2}} =\frac{(x_{1}+x_{2})^{2}-2x_{1}x_{2}}{x_{1} \cdot x_{2}} =\frac{2^{2}-2 \cdot (-3)}{-3}=\frac{4+6}{-3} =-\frac{10}{3}[/tex]
[tex]c)[/tex]
[tex](x_{1}-2)^{2}+(x_{2}-2)^{2}=x_{1}^{2}-4x_{1}+4+x_{2}^{2}-4x_{2}+4=x_{1}^{2}+x_{2}^{2}-4(x_{1}+x_{2})+8=[/tex]
[tex]=(x_{1}+x_{2})^{2}-2x_{1}x_{2}-4(x_{1}+x_{2})+8=2^{2}-2 \cdot (-3)-4 \cdot 2+8=[/tex]
[tex]=4+6-8+8=10[/tex]