Question

Solve the equations:
$2x^4+x^3-6x^2+x+2=0$.

Answer 1

We have $2x^4+x^3-6x^2+x+2=0$

Dividing by $x^2$, we get

$2x^2+x-6+\frac{1}{x}+\frac{2}{x^2}=0⟹2(x^2+\frac{1}{x^2})+(x+\frac{1}{x})-6=0$

$⟹2{(x+\frac{1}{x})}^2+(x+\frac{1}{x})-10=0$

Substituting $(x+\frac{1}{x})=y$ in the above equation, we get

$2y^2+y-10=0⟹(2y+5)(y-2)=0⟹y=2,-\frac{5}{2}$

For $y=2,x+\frac{1}{x}=2⟹x^2-2x+1=0⟹(x-1{)}^2=0⟹x=1,1$

For $y=-\frac{5}{2},x+\frac{1}{x}=-\frac{5}{2}⟹2x^2+5x+2=0⟹(x+2)(2x+1)=0⟹x=-2,-\frac{1}{2}$

$\therefore$ the roots of the equation are $1,1,-2,-\frac{1}{2}$