227k views
3 votes
1. Solve the equations: (c) 6x^4 - 5x^2 + 1 = 0; d) x^4 - 4x^2 - 5 = 0;

1 Answer

4 votes

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.

User Widjajayd
by
8.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.