Question

1. Solve the equations: (c) 6x^4 - 5x^2 + 1 = 0; d) x^4 - 4x^2 - 5 = 0;

Answer 1

To solve these equations let us consider x^2 = y Then x^4 = y^2.

Now,

(c). 6y^2 - 5y + 1 = 0

`\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=(1)/(3)\text{ and y = }(1)/(2) \end{gathered}`

Now, the solution will be

`\begin{gathered} x^2=y \\ x^2=(1)/(3)\text{ and x}^2=(1)/(2) \\ x=\pm(1)/(√(3))\text{ and x = }\pm(1)/(√(2)) \end{gathered}`

Hence this is the required solution.

(d). y^2 - 4y - 5 = 0

`\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}`

Now the solution will be

`\begin{gathered} x^2=y \\ x^2=5\text{ and x}^2=-1 \\ x=\pm√(5)\text{ and x=}\pm√(-1)=\pm1i \end{gathered}`

Hence this is the required solution.