Question

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

Answer 1

Step 1: Reduce the given equation into a quadratic equation

The given equation is , $2x^4+x^3-6x^2+x+2=0$

Dividing by $x^2$,we get $2x^2+x-6+\frac{1}{x}+\frac{2}{x^2}=0$

Rearrange, $2x^2+\frac{2}{x^2}+x+\frac{1}{x}-6=0$

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

Adding and subtracting $4$ we get,$2\left(x^2+\frac{1}{x^2}+2\right)+\left(x+\frac{1}{x}\right)-6-4=0$

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

Let $x+\frac{1}{x}=y$,

we get $2y^2+y-10$

Step 2: Solve the above quadratic equation

Using the factorization method,

$2y^2+5y-4y-10$

$y(2y+5)-2(2y+5)$

$\left(y-2\right)(2y+5)$

$$\begin{gathered} y-2=0or2y+5=0 \\ y=2ory=\frac{-5}{2} \end{gathered}$$

Step 3: Put the value of $y=2ory=\frac{-5}{2}$ in (1) and solve for $x$

For $y=2$ we have

$$\begin{gathered} x+\frac{1}{x}=2 \\ \Rightarrow \frac{x^2+1}{x}=2 \\ \Rightarrow x^2+1=2x \\ \Rightarrow x^2-2x+1=0 \\ \Rightarrow {\left(x-1\right)}^2=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow {\left(x-1\right)}^2=0 \\ \Rightarrow x=1,1 \end{gathered}$$

For $y=\frac{-5}{2}$ we have

$$\begin{gathered} x+\frac{1}{x}=\frac{-5}{2} \\ \Rightarrow \frac{x^2+1}{x}=\frac{-5}{2} \\ \Rightarrow 2\left(x^2+1\right)=-5x \\ \Rightarrow 2x^2+2+5x=0 \end{gathered}$$

$2x^2+5x+2=0$ using the factorization method we get,

$$\begin{gathered} \Rightarrow 2x^2+5x+2=0 \\ \Rightarrow \left(x+2\right)\left(2x+1\right)=0 \\ \Rightarrow x=-2,\frac{-1}{2} \end{gathered}$$

So the roots of the equation are $x=1,1,-2,\frac{-1}{2}$