Given:
[tex]4x^2-7x-1=0[/tex]Solve:
Quadratic formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Where,
[tex]ax^2+bx+c=0[/tex]Compaire the equation then:
[tex]\begin{gathered} ax^2+bx+c=0 \\ 4x^2-7x-1=0 \\ a=4,b=-7,c=-1 \end{gathered}[/tex]So roots of equation is:
[tex]\begin{gathered} x=\frac{-(-7)\pm\sqrt[]{(-7)^2-4(4)(-1)}}{2(4)} \\ x=\frac{7\pm\sqrt[]{49+16}}{8} \\ x=\frac{7\pm\sqrt[]{65}}{8} \end{gathered}[/tex]So value of G and H is:
[tex]\begin{gathered} G=\frac{7+\sqrt[]{65}}{8};H=\frac{7-\sqrt[]{65}}{8} \\ G=\frac{7}{8}+\frac{\sqrt[]{65}}{8};H=\frac{7}{8}-\frac{\sqrt[]{65}}{8} \end{gathered}[/tex]So:
[tex]\begin{gathered} =G^2+H^2 \\ =(\frac{7}{8}+\frac{\sqrt[]{65}}{8})^2+(\frac{7}{8}-\frac{\sqrt[]{65}}{8})^2 \\ =(\frac{7}{8})^2+(\frac{\sqrt[]{65}}{8})^2+2(\frac{7}{8})(\frac{\sqrt[]{65}}{8})+(\frac{7}{8})^2+(\frac{\sqrt[]{65}}{8})^2-2(\frac{7}{8})(\frac{\sqrt[]{65}}{8}) \\ =2(\frac{49}{64}+\frac{65}{64}) \\ =2(\frac{114}{64}) \\ =\frac{114}{32} \\ =3.5625 \end{gathered}[/tex](B)
If roots is a and b the equation is:
[tex]x^2-(a+b)x+ab=0[/tex]Then equation is:
[tex]G^2+H^2=3.5625[/tex][tex]\begin{gathered} G^2H^2=(\frac{7}{8}+\frac{\sqrt[]{65}}{8})^2(\frac{7}{8}-\frac{\sqrt[]{65}}{8})^2 \\ =(0.875+1.00778)^2(0.875-1.00778)^2 \\ =(3.54486)(0.01763) \\ =0.0624 \end{gathered}[/tex]So equation is:
[tex]x^2-3.5625x+0.0624[/tex]