Respuesta :
To solve these equations let us consider x^2 = y Then x^4 = y^2.
Now,
(c). 6y^2 - 5y + 1 = 0
[tex]\begin{gathered} 6y^2-5y+1=0 \\ By\text{ using middle term splitting we have} \\ 6y^2-3y-2y+1=0 \\ 3y(2y-1)-1(2y-1)=0 \\ (3y-1)(2y-1)=0 \\ y=\frac{1}{3}\text{ and y = }\frac{1}{2} \end{gathered}[/tex]Now, the solution will be
[tex]\begin{gathered} x^2=y \\ x^2=\frac{1}{3}\text{ and x}^2=\frac{1}{2} \\ x=\pm\frac{1}{\sqrt{3}}\text{ and x = }\pm\frac{1}{\sqrt{2}} \end{gathered}[/tex]Hence this is the required solution.
(d). y^2 - 4y - 5 = 0
[tex]\begin{gathered} y^2-4y-5=0 \\ by\text{ using middle term splitting} \\ y^2-5y+y-5=0 \\ y(y-5)+1(y-5)=0 \\ (y-5)(y+1)=0 \\ y=5\text{ and y = -1} \end{gathered}[/tex]Now the solution will be
[tex]\begin{gathered} x^2=y \\ x^2=5\text{ and x}^2=-1 \\ x=\pm\sqrt{5}\text{ and x=}\pm\sqrt{-1}=\pm1i \end{gathered}[/tex]Hence this is the required solution.
