Question

The roots of the equation $2x^4+x^3-11x^2+x+2=0$ is/are

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{-3-\sqrt{5}}{2}$
• D. $\frac{-5+\sqrt{3}}{2}$

Answer 1

Answer: (A), (C)

$\frac{-3-\sqrt{5}}{2}$

Given equation
$2x^4+x^3-11x^2+x+2=0$
Dividing the given equation by $x^2$, as $x=0$ is not the root of the equation.
$2x^2+x-11+\frac{1}{x}+\frac{2}{x^2}=0\Rightarrow 2(x^2+\frac{1}{x^2})+(x+\frac{1}{x})-11=0\Rightarrow 2({(x+\frac{1}{x})}^2-2)+(x+\frac{1}{x})-11=0\Rightarrow 2{(x+\frac{1}{x})}^2+(x+\frac{1}{x})-15=0$

Assuming $x+\frac{1}{x}=t$
$2t^2+t-15=0$
$\Rightarrow (t+3)(2t-5)=0\Rightarrow t=-3\text{or}\frac{5}{2}$
Now,
$x+\frac{1}{x}=-3$
$\Rightarrow x^2+3x+1=0\Rightarrow x=\frac{-3\pm \sqrt{9-4}}{2}=\frac{-3\pm \sqrt{5}}{2}$

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