Question

Which of the following is/are the roots of the equation $2x^4-3x^3-x^2-3x+2=0$ ?

• A. $\frac{1}{2}$
• B. $2$
• C. $-2$
• D. $-\frac{1}{2}$

Answer 1

Answer: (A), (B)

$2$

Given equation is $2x^4-3x^3-x^2-3x+2=0$
This is a biquadratic equation of the form $a{x}^4+b{x}^3+c{x}^2+bx+a=0,a\ne 0$
On dividing throughout by $x^2$ we get,

$\Rightarrow 2x^2-3x-1-\frac{3}{x}+\frac{2}{x^2}=0$

$\Rightarrow 2(x^2+\frac{1}{x^2})-3(x+\frac{1}{x})-1=0$

$\Rightarrow 2((x+\frac{1}{x}{)}^2-2)-3(x+\frac{1}{x})-1=0$

According to the condition of Type 1 of biquadratic equations, let's assume $x+\frac{1}{x}=t\Rightarrow t\in (-\infty ,-2]\cup [2,\infty )$
Now, our equation becomes:
$2t^2-5t+2t-5=0$
$\Rightarrow (2t-5)(t+1)=0$
$t=-1,\frac{5}{2}$
But $t=-1$ is rejected as $t\in (-\infty ,-2]\cup [2,\infty )$

$\therefore t=x+\frac{1}{x}=\frac{5}{2}$

$\Rightarrow x^2+1=\frac{5}{2}x$

$\Rightarrow 2x^2-5x+2=0$
$\Rightarrow 2x^2-4x-x+2=0$
$\Rightarrow 2x(x-2)-1(x-2)=0$
$\Rightarrow (x-2)(2x-1)=0$
$\Rightarrow x=\frac{1}{2},2$ which are the two roots of the given biquadratic equation.